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Composite-modulus large sieve for property P
certified_mass_selection: A positive-mass extraction into jointly certified short blocks would prove E12(iii). A maximal matching pairs all but a bounded reciprocal mass locally, while an exact four-element example prevents those local pairs from being certified together in any eligible representation.
evidence/: Exact finite normalization, occupancy and obstruction checks, retained admissible-window witnesses, and certified two-block harmonic upper bounds, with exploratory solver maxima kept at their finite scope.
The remaining question in Problem 12 asks whether every infinite with property P has . Here property P means that no divides for distinct both greater than .
This note supplies a precise composite-modulus sieve inequality and tests the proposed use of its multiplicity factor. It then adds a residue-occupancy bound with exponential gain, a convergence rate in terms of a greedy sequence of coprime elements, convergence for every pairwise coprime set, and a bound in terms of the primes that elements share. A gluing construction shows that the infinite question is equivalent to a uniform bound for finite property-P sets, and a Fourier argument shows that early elements can coexist with a heavy late part only through small smooth guard moduli. With a smooth-part normalization this gives a bound for the reciprocal sum of every property-P subset of . An exact signed identity gives another sufficient criterion, whose energy premise is unproved and implies the uniform bound directly. The note assembles these links into one argument whose single open step is a joint guard budget. It bounds certified full congruence blocks by plus an explicit counting error under the block-shape and size hypotheses stated below. The error must also be bounded; certification alone does not control arbitrary sets. It identifies the missing multiscale charge and refutes proposed normalization potentials with explicit families. These are author-recorded arguments by Plasma AI, without independent review. The large-sieve, Brun--Titchmarsh, and prime-number-theorem inputs are used at statement level; no result here resolves the reciprocal sum question.
The exact inequality
Let be integers. Write
and, for , put
The empty case has . Define also
An infinite property-P set contains neither 1 nor 2. The first would forbid any two larger elements. For 2, an infinite tail has two distinct elements of the same parity. Thus every is at least 3, and . The inequality is
Moreover,
Consequently, as , (1) gives
The error is uniform in and . In particular, the multiplicity in (3) is an explicit divisor count, not an unspecified loss.
Proof and the distinct-elements convention
Set and . For , let count the elements of congruent to modulo . If , at most one of is nonzero. Otherwise elements in the two classes would be distinct forbidden larger summands. A self-inverse class can contain at most one element, but it need not be empty. There are such classes, so at most classes are occupied.
Cauchy--Schwarz and finite Fourier orthogonality give
For the desired result is immediate. For , sum (4) over . A reduced fraction with and appears among the fractions , , , exactly times: it appears once for each multiple of , with . Hence
Apply the analytic large sieve in the form of Schoen, On a Problem of Erdős and Sárközy, Lemma 1, equation (4), printed p. 192: the sum over reduced fractions of denominator at most is at most for . Its hypotheses impose no coprimality on the elements of . Bounding by in (5) and dividing by proves (1). The cited large sieve is stated without proof in Schoen, with a reference to Montgomery; its proof is not audited here.
If then , proving the first identity in (2). Also . Listing the distinct positive integers of in increasing order bounds their reciprocal sum by the th harmonic number. Thus for , which proves the second statement and (3).
This is the same analytic input as Schoen's pairwise-coprime theorem, but with collisions of fractions counted explicitly. For pairwise coprime , . The proof above also allows the singleton self-inverse residues permitted by the distinct-elements formulation. It does not silently impose the source's stronger convention allowing equal larger summands. All fractions in (4)--(5) are nonzero modulo 1; repeated zero frequencies would not satisfy the multiplicity bound.
The finite property-P set tests the exceptional classes. For the modulus and , three residue classes are occupied and the left side of (4) is 3. This equals , whereas replacing by 1 would demand 9. This local check is an exact calculation, not an infinite counterexample.
Why removing a common divisor is insufficient
Property P is preserved by dilation and by taking a subset of multiples of and dividing by . Thus an arbitrary dilation already shows that one cannot expect for every property-P set. Even imposing does not fix the problem.
Let be the primes other than 3, and set
There are infinitely many such primes by the usual elementary argument: if only finitely many primes were modulo 4, the number four times their product minus one would have a new prime divisor modulo 4. The set in (6) has gcd one, since it contains 3 and 98.
For the element 3, any sum of two elements of the tail is 1 modulo 3, since . For and distinct larger tail elements , divisibility by would give . Neither nor equals , so this would make a square modulo , which is impossible. Therefore (6) has property P.
All but one of its elements are even. For ,
Indeed, the prime 2 divides the tail elements; every odd prime divides at most one member of . This example has a convergent reciprocal sum, by comparison of with . It refutes an automatic multiplicity gain even in the gcd-one setting; it does not refute the target.
Nor does merely weighting the original moduli remove this obstruction. For arbitrary nonnegative weights , the same proof gives (1) with and . In (6), write for the total weight of the even elements. Then and , so whenever . There is no growing sieve gain from this weighted first step alone.
What the optimal weighted gain detects
The qualitative information in this weighting step has a simple exact description. For a finite , define
The denominator is positive for nonzero weights, since every is at least 3. The maximum is over primes dividing some member of . For , the weighted proof above gives .
For an infinite set of integers at least 3, the values stay bounded as if and only if some finite set of primes meets the prime divisors of every element of . This assertion does not require property P; its use here is a test of what the one-scale inequality can detect.
Indeed, if a set of primes meets every element, then
Thus . Conversely, if no finite prime set meets all elements, choose successively , each avoiding the prime divisors of its predecessors. They are pairwise coprime. Give the first elements weight one and all other elements weight zero. Once contains them the denominator in (W) is one, while the numerator is at least . Hence the gain is unbounded.
This dichotomy is not a summability theorem. For a prime , the normalized set again has property P, and
In the finite-prime case, if diverges, at least one of the finitely many prime-divisible subseries diverges, so normalization by that prime preserves divergence and property P. Repeating this observation need not terminate: the selected subset and prime may change at every passage. In the other case, the weighted gain itself gives no rate for and no summable estimate for the dyadic masses; Proposition R below supplies a rate through residue occupancy instead. A proof must relate the gain or the successive normalizations to those actual masses; an unbounded gain or a factor in isolation is insufficient.
Residue occupancy beyond a product of moduli
The large sieve in (1) gains only linearly in the weighted gain. Counting residue classes directly gains exponentially, at the price of working only beyond the least common multiple of the moduli used.
Let have property P, let be finite and nonempty, and put and . Call a class modulo self-inverse when ; there are of them. Call a set of classes modulo admissible for when its image modulo every contains no self-inverse class and no pair , and let be the largest density of an admissible set . Then, for every ,
Indeed, every element of above is larger than every . For each , a self-inverse class modulo holds at most one of these elements, since two of them would be distinct larger summands with sum . Remove these at most elements. Two remaining elements in classes and modulo would be distinct forbidden summands, so the remaining elements occupy an admissible set of classes modulo . Each class holds at most integers of .
When the elements of are pairwise coprime, the Chinese remainder theorem places an admissible set inside the product of its images, and an admissible image modulo has at most classes. Thus
with equality in the first inequality, and at most one element of is even. For pairwise coprime elements, (1) gives a factor of about beyond ; (10) gives beyond their product.
A shared prime weakens (10) in the way the example (6) predicts. Let be an odd prime and pairwise coprime integers, each coprime to . Then
Write . For a nonzero class modulo , let and be the images modulo of the admissible classes in the fibres over and . A sum of one class from each fibre is divisible by , so admissibility modulo makes and disjoint, and . The two fibres together hold at most of their classes; since is odd, these fibres contain no self-inverse class. In the fibre over 0, a class is admissible exactly when is admissible for , the self-inverse classes included. This proves the upper bound in (11). Taking every class over a set of residues containing one of each pair , together with an optimal set in the fibre over 0, attains it. For , (11) gives . That is close to one half, whereas the product in (10), applied formally to these two moduli, would give : the common prime does the work of both.
A rate when coprime elements recur
For a nonempty set of positive integers, define its greedy coprime sequence by and , where , for as long as such an element exists. It is increasing and pairwise coprime; the strict inequality also excludes selecting 1 twice when the definition is applied to a normalized finite set.
Proposition R. Let have property P and an infinite greedy coprime sequence. Then, for every ,
Proof. Apply (9) and (10) to . Since , the count satisfies for . Partial summation over gives
and gives the first bound. The only element of in is . Moreover and , so summing the first bound over gives the second.
So the unbounded-gain branch of the dichotomy above has an explicit rate. The reciprocal sum converges when the coprime elements recur often enough that , for instance when . A divergent property-P set therefore satisfies one of two alternatives. Either its greedy coprime sequence stops at some , and every other element shares a prime factor with : this is the finite-prime-cover branch. Or for infinitely many , and for each such every element of in has a prime factor dividing . The bound (12) leaves room for divergence only on the ranges for such .
Pairwise coprime sets
Proposition C. Every pairwise coprime set with property P has .
Proof. A composite is at least the square of its least prime factor, and pairwise coprime elements have distinct least prime factors. So the composite elements contribute less than . The last constant follows by summing the primes up to 1000 and bounding the rest by .
Let be the primes in and . They form a property-P set of odd primes. Apply (10) to . The only self-inverse class modulo an odd prime is 0, and no prime other than lies in it, so (9) needs no exceptional elements. Every prime of above lies in a set of reduced classes modulo with .
For , the primes of in exceed , hence lie in with and . In both cases . The Brun--Titchmarsh inequality in the form of Montgomery and Vaughan gives for and . For this is at most . Partial summation over each class of then bounds the contribution of this range by
The sum over is less than . The primes of in contribute at most , by the same inequality with modulus 1. For a finite prime list, stop after the range containing its last prime; a one-prime list uses just the initial range, and an empty list costs zero. These ranges cover every prime of , so the primes contribute less than and .
The composite part uses no property of ; property P enters only through (10) for the primes. The pairwise coprime class is the one in which Schoen and Baier bound the counting function along a sequence of . Proposition C answers the reciprocal-sum question inside that class. It is author-recorded, its Brun--Titchmarsh input is used at its published statement, and it was not compared with literature beyond the sources this lead cites.
The same argument works inside admissible classes, with a discount.
Proposition C′. Let be pairwise coprime with product , where . Let be a pairwise coprime property-P set whose elements exceed and lie in classes admissible for . Then .
Proof. A composite satisfies , where is its least prime factor, and these primes are distinct. So the composite elements contribute at most . A prime above is a unit modulo each , and no unit is self-inverse modulo , so an admissible set of units modulo has at most classes.
For the primes of in , let be the product of the smallest moduli, so . The primes lie in at most classes modulo , and for the Brun--Titchmarsh inequality gives . Partial summation bounds their contribution by .
For the primes above , let be the primes of , put and , and split at the points . The primes in are at least . By (9) and (10) inside , together with the admissible classes modulo , they lie in at most reduced classes modulo . The argument of Proposition C, with in place of , bounds this range by , and the sum over by . Adding the three parts gives the bound.
Shared primes
Proposition C extends from pairwise coprime sets to all property-P sets, at the cost of a factor depending only on the primes that two elements share.
Proposition S. Let have property P, and let be the set of primes dividing at least two elements of . Write each as , where is the largest divisor of composed of primes in , and put . Then
Proof. A prime dividing and for would divide two elements, hence lie in , which no prime factor of does. So the parts are pairwise coprime. For let and . Dividing every element of a subset of by a common divisor preserves property P, so is a pairwise coprime property-P set, and . If , then has at most one other element, since 1 divides every sum. If and , then has at most one odd element above 2 and, being pairwise coprime, no other even one. In both cases . Otherwise and Proposition C gives . Summing over gives the first inequality, interpreted in the extended nonnegative reals; it is strict for nonempty when the right side is finite. The elements of are distinct integers composed of primes in , which gives the second.
Three consequences follow. If no prime above divides two elements, in particular if every two elements have greatest common divisor at most , then . The example (6) has , so (13) bounds its reciprocal sum by . Applied to the truncations , (13) shows that a divergent property-P set has : infinitely many primes are shared, with divergent reciprocal sum. Finally, the prime elements of any property-P set form a pairwise coprime property-P set, so they contribute less than by the proof of Proposition C.
The bound (13) is far from sharp. It charges every shared part with a full pairwise coprime set, whereas property P also limits how the sets for different coexist; that interaction is the same unresolved question as below.
The remaining multiscale obligation
Write and . The dyadic criterion says that the question is exactly whether converges. The supported links compose as follows. The weighted sieve detects a finite prime cover; the occupancy bound supplies an exponential saving after the product scale; Proposition R integrates that saving when its explicit recurrence condition holds; and Propositions C and S handle coprime sets and controlled shared primes. What remains is a bound on the mass accumulated through arbitrary normalizations and through the scales preceding the occupancy estimate. Corollary H below bounds the mass of by ; the target is to remove that doubly logarithmic growth. The subsection "The whole argument and its open step" below is the canonical sketch of the current route, with its one missing estimate and the reasons it is open.
One proposed induction statement is still open:
(H) There are constants and such that, for every , all pairwise coprime with product , and every finite property-P set above occupying admissible classes for these moduli, .
Its case is the uniform finite-set assertion
This would imply E12(iii) by truncation and monotone convergence. Proposition E below proves the converse: unbounded finite examples can be glued into one infinite property-P set with divergent reciprocal sum. So (U) is equivalent to E12(iii). Neither (H) nor (U) follows from the results above. Proposition C′ supplies (H) only when is itself pairwise coprime.
Gluing finite sets into one infinite set
Lemma E1. Let and let be a finite property-P set of positive integers. There is a finite property-P set whose elements are all congruent to 1 modulo , with .
Proof. For let be its largest divisor composed of primes dividing . For each occurring , divide the corresponding part of by and call the quotient . It has property P, its elements are coprime to , and
Thus some has . The empty set is immediate, and for nonempty this follows by bounding each by their finite maximum in the displayed identity.
The elements of lie in the reduced classes modulo , so some reduced class holds a subset with . Choose with and put . Subsets and dilations of property-P sets have property P, every element of is congruent to 1 modulo , and .
This normalization keeps the full reciprocal Euler factor: the smooth parts have weights , and their sum is exactly . The quotient can contain 1 or 2; neither is discarded. For the constant can be sharpened to by maximizing jointly over and the reduced class. Indeed, the inverses permute the reduced residues in , whose sum is , and , where is the -class of . The coefficients in that sum are bounded in total by .
Proposition E. Suppose that finite property-P sets have unbounded reciprocal sums. Then there is an infinite property-P set with .
Proof. Build finite blocks in turn. Before building , let be the least common multiple and the largest element of , with and . Choose an integer with and . By hypothesis there is a finite property-P set with . Lemma E1 turns it into a property-P set , with every element congruent to 1 modulo and . Put . Then has property P, its elements are congruent to 1 modulo , they exceed and are at least 3, and .
Let . Every element of exceeds every element of an earlier block, so is infinite and . Let be elements of with , and let . Then and lie in blocks with . There are three cases.
- If , then because has property P.
- If and with , then , so . A relation would give . But divides both and , so it would divide 1, which is false. The same holds with and exchanged.
- If and with , then divides and , so . Since , .
Thus has property P.
Consequently the following are equivalent: every infinite property-P set has a convergent reciprocal sum; the assertion (U); and, for any one fixed , a uniform bound on over the finite property-P sets contained in the class . The first implies the second by Proposition E. The second implies the first by truncation, and the third trivially. The third implies the second by Lemma E1. Elements 1 and 2 change none of these bounds by more than , so it does not matter whether they are admitted. In particular Erdős and Sárközy's two statements, convergence for each set and a bound by one absolute constant, are equivalent.
The gluing uses congruence blocks: block needs mass at least before restriction, which grows without bound. It therefore turns the infinite question into the finite one without supplying the finite examples.
The class form changes the local interaction. If every element is congruent to 1 modulo and with , then with and , so . Elements whose sums stay below never interact with . Choosing divisible by every prime up to also removes every prime up to from the elements. The finite question therefore needs to be solved only for sets without small prime factors and with interaction ratio at least . This reduction does not bound the masses of such sets.
The convention allowing equal larger summands can also be reached by an explicit reduction. Call a set primitive if no two distinct elements divide one another. Write for the largest reciprocal sum of any property-P subset of , and let be the corresponding maximum over odd primitive property-P sets, with a singleton 1 allowed. Then
Indeed, let be any property-P set and let be its elements minimal under divisibility. Every has a proper divisor in ; assign it to one such divisor . No receives two elements, since it would divide their sum and both would exceed . The assigned element is at least , so . The set is primitive. For each , divide the elements of of exact 2-adic valuation by , obtaining an odd primitive property-P set . Thus
For odd primitive sets, with would imply , which is impossible. They therefore satisfy the stronger version of P that also forbids equal larger summands. A uniform bound on this subclass would prove (U) and hence the full infinite conclusion. In the signed identity below its early/late pair sums have no diagonal exceptions: . This removes the diagonal error only after the reduction (O); it does not remove denominator multiplicities or give a uniform bound on the remaining signed sum.
Large elements do not reduce the problem to one interval
Put a finite property-P set with . Since , (U) holds if and only if for one , and then for every . Taking intervals with gives . A tempting sharper target, that as for the cardinality version of the problem, is false.
Let be a finite property-P set, , and
Then has property P. In a relation forces , so , whereas . For and , while . For and , . Triples inside are excluded by dilation. As , . Taking , then letting and taking the supremum over , gives . Since is nonincreasing, for every ; the conclusion is immediate when , and follows by rearrangement otherwise.
The same argument gives a recursion for classes. If every element of is , is an odd prime with , and is a property-P set above inside the class , then $qB'\cup C$ is a property-P set inside . The window supplies mass about ; for the upper windows in that recursion, replace by . For one fixed , iterating the lower branch of multiplies each inherited mass by . If each new upper window has limiting mass , the limiting total contribution after iterations is ; the initial mass has coefficient . This geometric calculation controls that particular recursion. It does not bound arbitrary mixtures of guard primes or the independent windows at different class moduli. These constructions show structures that a proof of (U) must accommodate; they do not decide (U).
Early elements need biased smooth guards
The following proposition shows that, apart from a small mass, an early element can coexist with a heavy late part only through a divisor with small prime factors modulo which the late part is biased. It turns the informal guard primes of the constructions above into a proved dichotomy.
Fix and let be the largest reciprocal sum of a property-P subset of . Let have property P, let and , and take an early part and the late part . Weight by when , so that . Put , let be the number of for which meets , and for put
the sum over ordered pairs, included. Thus , while a late part spread evenly over the classes modulo has close to . For an integer let , and for put . Call guarded if , and free otherwise. Elements without prime factors up to have , so they are free.
Proposition F. If is nonempty and , the free elements satisfy
Proof. Put , where , and for let . Pairing with shows that is real, and . For every , grouping the fractions by their reduced denominators gives
Let . Every element of exceeds , so property P leaves only the pairs with in . Such lie in the at most two self-inverse classes modulo , and each class holds at most one element of , since two would have sum divisible by . Each such has , so . The divisors of that divide are exactly the divisors of . Hence
For a free this gives , using and . A divisor of that does not divide has a prime power factor larger than , so . Dividing the elements of divisible by by leaves a property-P subset of , so . Therefore
Write , , and , where is the exponential sum of . Then . Split into ranges with , and apply the large sieve in the form of Schoen's Lemma 1, equation (4), cited above, with modulus bound and $L_j\subseteq [1,2^{j+1}]$. This gives
The form is a weighted sum of squares, so the triangle inequality and the Cauchy--Schwarz inequality give , the last sum over the indices with nonempty. Those indices have , so . Hence , and the bound follows.
The proposition uses property P through the early elements' relation to and through the normalized sets behind , and the large sieve only through the cited statement. Choosing a large multiple of , which is at most , and large makes the free mass small whenever is a fixed fraction of .
Without small prime factors there are no guards, and the bound closes.
Corollary R. Let and . Every property-P subset of whose elements have no prime factor at most has reciprocal sum at most 30.
Proof. Let be the largest reciprocal sum of such a set; dividing the multiples of in one of them by gives another, possibly containing 1. So the proof of Proposition F holds with in place of for subsets of such a set. Suppose and take a set attaining it. It does not contain 1, since a property-P set containing 1 has at most two elements, so its elements exceed . Let be the least element of with , put , and . Then , the gap carries at most , and so , with and hence . Every element of has and is free. The number of dyadic ranges is at most , so Proposition F gives , contradicting .
The cutoff depends on . A general set may use primes up to that cutoff as guards, so this corollary does not give (U).
If Corollary R held with a constant independent of , Lemma E1 with would give (U), since every element of a set in the class has no prime factor up to . The dependence on enters only through the term , which comes from the triangle inequality across the dyadic ranges of . That loss is real for the large sieve alone: a late part lying in one nonzero class modulo a prime with has for every , so , while with the smallest range. One way to control this coherence would be to show that property-P sets confined to one class modulo across many ranges carry mass tending to 0 as grows, a quantitative class form of (U). Such an estimate must retain the lower size cutoff: here every late element exceeds . A bound such as for all unit classes with no size cutoff is false, since the class contains the property-P singleton . Even excluding 1 and 2 leaves the same problem for a fixed singleton in a unit class. With the required size restriction the quantitative class bound is unproved. It and the rough-set bound would have to be proved together.
Consequently a nearly extremal set with large has the following shape. Cut it where the early mass first reaches about , leave a gap of ratio , whose mass is at most , and keep the rest as . Then almost all early mass is guarded: it sits on elements whose part composed of prime powers at most is a modulus modulo which the late part has less than half of its expected number of pair sums divisible by . In the example the guard is and .
To finish a proof of (U) along this line it would suffice to prove a guard accounting: for some and constant , whenever is at least a fixed fraction of , the guarded early mass is at most . Then the cut above would give , a bound on independent of . The obvious charge, for the multiples of each minimal guard , is the sibling charge that the families below defeat: their sum over many guards is unbounded. A correct charge must also use that each guard costs the late part, which is biased modulo every guard at once, and that early multiples of different guards constrain one another: in (16) many guards share one window whose total mass is at most . This guard accounting is not proved here.
An idealized model shows why such an accounting is plausible and where it must work. Split the set into windows of ratio separated by gaps of ratio ; the odd-numbered and even-numbered windows each form such a family, so this costs a factor 2. Suppose every element of a window is a multiple of one of its guard primes, every later element avoids the class 0 and lies in a fixed half of the nonzero classes modulo each earlier guard prime, and integers in a window are counted by class density. Let be the density left after the guards of the first windows and over the guards of window . That window has mass at most , while . Hence , and the window masses of each family telescope to at most . Real sets leave this model in three places. Guards need only be biased, not perfect. Guards may be composite smooth numbers, so later multiples of a guard survive in its zero class, as in . And the windows lie far below the least common multiple of the guards, where class densities do not count integers. A guard accounting must handle all three.
After Proposition E the whole question is the finite assertion (U). A proof must bound uniformly over finite property-P sets, and its charge must pay for the guarded recursions above, in which a dilated copy below a guard prime coexists with a long window in one class above it. A negative answer needs finite property-P sets of unbounded reciprocal sum, in any one class ; Proposition E then supplies the infinite set. Every explicit construction recorded here pays a factor for each guarded lower copy and passes the rest to a finer class, so none has unbounded mass.
Explicit normalization potentials fail below. Their failure does not refute (H), (U), or convergence, and does not rule out potentials retaining more information about scale and residues.
A doubly logarithmic bound
Corollary R already bounds every finite property-P set by a doubly logarithmic quantity, through the smooth-part normalization of Lemma E1. For and let be the largest reciprocal sum of a property-P subset of whose elements have no prime factor at most ; thus is the quantity of the signed criterion below. Put
with an empty product equal to 1.
Corollary H. For and every , . In particular every property-P subset of has reciprocal sum at most
Consequently every infinite property-P set satisfies , and
Proof. Write and let be a property-P subset of with no prime factor at most . For let be its -smooth part, the largest divisor of composed of primes at most . It is composed of primes in . For each occurring , the elements with , divided by , form a property-P subset of with no prime factor at most ; it may contain 1. Corollary R applies to , since , and gives . Hence
With the product is . For the explicit bound, let , and . Counting prime factors in gives , where . For the latter elementary bound, put . The primes in divide , and those in divide . Hence
Both binomial coefficients are less than for ; for the odd case the two equal central coefficients together are less than . Induction from gives . Since , it follows that . Partial summation gives
Since and , the product is at most . With this gives .
For an infinite , apply the bound to ; is . With for large and , partial summation writes the tail of the displayed series as at the endpoints plus . The integrand is at most , which is integrable, and . The elements of are at least 3, so .
Mertens' product theorem, , recorded in Granville's survey, equation (1), printed p. 13, improves the constant to ; it is not needed for the order of growth. It also shows that is bounded by a constant depending only on when for a fixed , since . This extends Corollary R from to every fixed power of .
The density-zero theorem alone gives for an infinite property-P set. Corollary H does not prove (U).
The same decomposition locates the loss, once the element 1 is separated. For let be the largest reciprocal sum of a property-P subset of with no prime factor at most , and that of a property-P subset of composed of primes at most . In the proof above, exactly when , and then has at most one other element, which exceeds ; a property-P set containing 1 has at most one larger element. Hence, for every property-P subset of and every ,
where is the set of -smooth elements of . For fixed , the -rough integers in form a property-P set of reciprocal sum tending to as , by periodic counting. Thus a bound uniform in cannot be smaller than that order for fixed . A proposed bound at growing cutoffs is
(RS) Put . There is an absolute constant such that for every .
This would bound rough property-P mass by the density scale of a bounded number of automatic windows. Rough sets have no guards composed of primes at most ; larger guard primes remain possible. Even so, (RS) would not by itself give (U) through the displayed split. For , Rankin's inequality with gives
For the last bound, the logarithm of the ratio of this Euler product to is at most an absolute constant times , which is bounded by the estimate for in the proof of Corollary H. Taking with a sufficiently large absolute bounds the displayed tail by 1. Thus the smooth part reduces to with , which is smaller than for all sufficiently large at the prescribed cutoff. The resulting recursion still pays an additive constant at each step; it does not give a uniform bound.
A proof of (U) along this line must charge the smooth guards and the rough fibres above them from one budget. For a smooth guard and a fixed fibre lying above , the inherited modulus for is . Mixed fibres retain the condition for distinct larger summands. One cannot replace these conditions by independent restrictions on each prime or by one unweighted union of the rough quotients. (RS), and the joint residue estimate that such an accounting needs, are unproved.
There are two separate obstacles to proving (RS) by the bootstrap of Corollary R. First, the normalizations behind can contain 1: in particular for each . Thus cannot be replaced throughout by when the latter is smaller than 1. For a -rough set, reduction (O) gives a primitive early set with the exact split
The first equality holds because no other early element is a multiple of . In the second case the normalized fibre contains no 1 and remains -rough. A proof at the scale $R^*N(\Delta)\asymp \delta\Delta$ must control the first group separately; omitting it is not a valid bootstrap.
Second, in Proposition F comes from moduli near and late elements just above . If one uses the existing bound and seeks early and late masses of order , that inequality requires to be of order . A direct use therefore needs of order , with a sufficiently large constant, rather than a fixed gap. Even an improved treatment of the self-quotients leaving a term of order would need of order . Neither calculation controls the rough property-P mass in the intervening window. Even an upper sieve bound of order per fixed-ratio interval would sum to order over that gap, whereas a successful cut needs order . The factor can be made smaller than by taking for large .
Thus a useful near-range test is to bound rough property-P sets inside windows whose ratio is a fixed power of , while retaining the self-quotient terms. If that ratio is and , a relation inside the window has ; then is exactly the -smooth part of and is its rough part. This identifies a more restricted arithmetic problem, not a proof of (RS). Even a positive answer still has to be combined with the joint smooth-guard budget above.
Any one of the following would close the gap: a guard charge for Proposition F bounded independently of the cutoff; the free-mass bound of Proposition F with a cutoff independent of ; or the fixed- rough-set bound. The last is equivalent to (U) by the decomposition above with a fixed cutoff, and the second implies it by the proof of Corollary R. Corollary H is an author-recorded argument without independent review; it rests on Corollary R and therefore on the large-sieve statement cited in Proposition F.
A signed harmonic criterion
The squares in the exact Fourier identity can be retained longer than in Proposition F. This gives a complete conditional implication to (U) from one unproved energy estimate, which already contains the fixed-class form of (U), as shown below. The identity also separates a residue bias from the much larger unsigned Fourier energy that the usual large sieve bounds.
Let have property P, take and with , and use the exact weights on . Put and
Pairing and makes real; . Let denote the maximal mass of any property-P subset of . Then
To prove it, write and . As in Proposition F, only the at most two singleton self-inverse residue classes contribute, so for . Grouping the frequencies by reduced denominator gives
Consequently, with the exact multiplicities ,
The last inequality uses , obtained by dividing the multiples of by . Thus shared-prime multiplicity is retained before the bound, and no normalized element is excluded. This proves (SE1).
Here is the missing assertion, stated separately from the proved identity:
(SE) There is an absolute constant such that for every integer and every finite property-P set .
Conditional conclusion. If (SE) holds, then for every , and E12(iii) follows.
Indeed, suppose and choose a maximizing . Such a set contains neither 1 nor 2, since a property-P set containing either has mass at most . Choose at the first crossing of half the mass, put and . Then
In (SE1) the factor on the left is therefore at least . Assertion (SE) would give , hence . This is the entire implication; the hypothesis (SE) itself has no proof here.
The fixed-class case of (SE) already implies (U) without (SE1). Let be any finite property-P set in the class with , and take . For one has , so and . Assertion (SE) would give . Every finite property-P set in the class would then have reciprocal sum at most , and (U) follows from the class form of the equivalence after Proposition E. Thus (SE) already requires a uniform bound on this class; no reverse implication from (U) to (SE) is established here. An energy bound of this form, linear in , bounds the mass of a coherent tail outright. One possible alternative is to keep negative energy of order on a coherent class paired with the early multiplicity at that modulus, as in , where the guard 3 has and . This calculation does not disprove (SE).
The distinction from an unsigned estimate matters already for one prime. If lies in a nonzero class modulo an odd prime , then , whereas . Its contribution to , when , is , smaller by a factor than the unsigned contribution in Proposition F. The obstruction there to discarding the number of scales does not by itself refute (SE).
Multiple coherent primes still have to be paid for. If is odd and square-free and lies in a unit class modulo , then for every one has , by the Ramanujan-sum identity when , where . For square-free this follows by the Chinese remainder theorem: each prime factor contributes . Thus the contribution of all divisors of to the negative-part sum is
When some divisors exceed , (SE2) is an upper bound for their included contribution. It can grow relative to , so paying a separate constant for each prime is insufficient. For the automatic property-P window , however, as with fixed. Since , the limit of (SE2) divided by is at most . These windows therefore do not defeat a linear charge for their coherent divisors. This calculation covers neither the other denominators nor a union of many windows.
There is also cancellation between composite and prime denominators before taking negative parts. For example, if lies in the class , then : for every sum , . With the exact multiplicities are , so these two terms cancel in the identity preceding (SE1). This is compatible with property P, for example for a late class window above 384 and of ratio at most . Replacing each signed term by its negative part loses that cancellation. Thus even the proposed signed-energy bound discards some of the structure in the exact identity.
The proposed alternative signed route estimates the weighted sum jointly, retaining its composite cancellation and the early mass associated with each coherent modulus. No uniform estimate of this kind is proved here. Bounding the energies by absolute Fourier squares returns the term proportional to the number of scales, and dropping their composite denominators loses genuine multiplicity and cancellation. Formula (SE1) and its conditional conclusion are author-recorded algebraic arguments. (SE) remains an unproved sufficient premise, with neither a proof nor an equivalence to (U) established here; its failure would not refute the exact weighted identity or (U).
The whole argument and its open step
The links above compose into one argument for (U) that is complete except for one estimate. This subsection is the canonical sketch of the route; the subsections it cites carry the proofs. Work in the odd primitive class of reduction (O). Write for its largest reciprocal sum in , so that . A bound on independent of proves (U), hence E12(iii) with the distinct-summand convention. Fix and put
Step 1, the cut. Let attain , and suppose , which exceeds . A property-P set containing 1 has at most two elements, so , and every element of is odd and at least 3. Let be the least element of with , and put , and . Then . The integers of have reciprocal sum at most , so
Step 2, the exact identity. For , no ordered pair , equal or distinct, has : distinct pairs are excluded by property P, and would give for odd , against primitivity. With exact weights , grouping the frequencies by reduced denominators as in the signed criterion gives
For even , . For odd , its bound holds because dividing the multiples of by keeps a set odd, primitive and property P; the quotient may be . The equality in (I) retains the exact multiplicities; it has not used their upper bound.
Step 3, free elements. Apply Proposition F to and with and an integer , with in place of ; its normalization step stays inside the odd primitive class. Its dyadic late mass is at least , so as required. Its number of dyadic ranges is at most , so the free elements have
In the odd primitive class the proof of Proposition F needs only negative parts. With its dyadic weights, , so for a free , and
Step 4, the open step. It remains to prove that the guarded elements of , those with in Proposition F, satisfy
with independent of , , and the number of scales. This is the joint guard budget. It is neither proved nor refuted. Given Steps 1--3 it is equivalent to (U): if every were at most , it would hold with .
Step 5, conclusion. Steps 1, 3 and 4 give , so
and truncation and monotone convergence give E12(iii). If Step 4 is instead stated through (I), as a bound , the same cut gives , using .
Steps 1--3 and 5 are author-recorded arguments without independent review. Proposition F rests on the large-sieve statement it cites. Every coefficient suffices, so the open step does not ask for the early part to lose half of the maximal mass, only a fixed fraction.
Decoupled energy bounds already contain (U). Bounding each in (I) by leaves a statement about alone, of the form in the notation of the signed criterion. For every and , that assertion for all and all finite property-P sets already implies (U). Indeed, let be a product of distinct odd primes and
If lies in the class and , then for every , so and . Since diverges, some has , and the assertion gives for every such above . With the elements below this bounds every finite property-P set in the class , which is (U) by the class form of the equivalence after Proposition E. For example has , from the exact products and . So the coefficient in (SE) is not what makes it strong: with any coefficient on , a bound of this form implies (U). The joint route instead keeps the actual multiplicities , including the cancellation in identity (C) of the next paragraph. A proof of a decoupled bound would also resolve (U); none is provided here.
The coherent case inside the identity. Suppose is odd and squarefree, every divisor of is at most , and lies in one unit class modulo . Then for , and interchanging the finite sums with gives
The coherent divisors pay for the union of the guarded early elements, once each. Keeping only the negative terms and bounding each by , as the decoupled assertions do, replaces the right side by at most . For this exceeds , while the union of the multiples of the primes of has density .
Where the current estimates lose. Split into its dyadic pieces of masses . For every pair of pieces, equal or not, the same grouping gives
and (I) is their sum. The large sieve controls each diagonal energy at denominators above by a multiple of , plus a near-range term that decays geometrically in . Free-mass control with a cutoff independent of would follow if the off-diagonal terms were negligible. They are not in general. A late part lying in one nonzero class modulo a prime at every scale has aligned pieces, which is the example after Corollary R. The signed identity pays such a coherent prime only about , but then acts as a guard. Admitting coherent guards at all primes up to about returns the Euler factor of Corollary H. Thus each estimate available here loses about for every dyadic scale of late mass, either as free mass in the unsigned bound or as guard Euler weight in the signed one. Over scales the two losses balance at and give the bound . This describes the present estimates; it is not a proof that every method must lose. The estimates leave sparse late parts occupying many scales uncontrolled; they provide no lower bound on the mass of an individual occupied scale. The constructions on the problem page demonstrate why small density alone cannot replace this missing bound.
The same balance appears in the smooth-part normalization. Corollary H charges each -smooth fibre an additive 30, and the Euler sum multiplies that constant. Paying an independent constant for every possible fibre leaves this loss. A charge proportional to the rough density, as (RS) proposes, would remove that factor, but the smooth recursion would still need a joint bound.
Mixed pairs retain additional constraints. The cut identity (I) uses only pairs from . For each , put and define with exact reciprocal weights. Property P also excludes with and . Thus, by the same grouping of frequencies,
Indeed, the zero frequency for each is , and . The early cutoff inside must stay in the expression. It cannot be replaced by one multiplicity paired with the Fourier transform of all of . These constraints require consistency between the residue supports of the early elements above and those of the late set, beyond the absence of opposite pairs within itself.
One way to retain all these constraints at once is to use the full tail for each . Write , and . In the odd primitive class, including the empty-tail case, the exact quantity is
For each , integrate from to . These intervals partition , and hence
For one has , by comparison with . A uniform estimate , with and absolute , would therefore give and close (U). No such estimate is proved. Conversely a uniform bound implies it with , so this is another full-strength formulation. Its purpose is to keep the mixed constraints available to the charge, not to claim a weaker problem. Taking absolute values separately at each again loses the reuse of the same original elements and supplies no strict coefficient.
Unclosed mechanisms. The following three mechanisms do not close Step 4.
- Pairwise window charges. Comparing one early window with one later window of bounded ratio uses only a bounded number of scales, so a constant suffices there. But a bound of the form for every pair of windows still allows windows of mass each. Summing over all later windows at once restores the cross terms above. This route reopens with an estimate for those cross terms.
- Diagonal identities alone. Using only the pairs weights the pieces by instead of . For mass spread evenly over scales this is a factor smaller, which is the loss again. It reopens with the same off-diagonal estimate.
- First-divisor events. For any finite set , the sets are disjoint, so . Here is the density of the integers for which no smaller element of divides . This bounds only when stays bounded below, and property P gives no visible lower bound. For the primes, which do not have property P, is about . The route reopens if property P can be shown to keep most above a constant.
Best proved replacement and the missing inverse statement. The uniform step remains open. The best bound proved here for general finite sets is Corollary H's . Identities (M) and (T) retain mixed constraints but are not upper estimates. The next subsection bounds full certified congruence blocks, and Lemma D extends its bound to arbitrary residue sets at odd moduli. These results suggest a concrete inverse statement, with the counting error included.
For a block satisfying the hypotheses below, assign one of these errors: for a block of ratio at most ; for a single-class block covered by Lemma W; or for an odd-modulus block covered by Lemma D. Each proved block estimate is at most plus its assigned error. Write for the sum of those errors. The unproved input is:
(IC) There are absolute constants and such that every finite odd primitive property-P set is contained in a finite certified family of full congruence blocks, each satisfying one of the three block hypotheses, whose errors satisfy .
If (IC) holds, Lemma T and the block estimates give , hence . Reduction (O), with the singleton 1 handled separately, then gives and proves E12(iii). This is a complete conditional replacement for the joint guard estimate in Step 4. No covering with that strict error allowance is constructed here. Certification alone is insufficient: the singleton representation below certifies every odd primitive set but has error exactly . Conversely, if (U) held with bound on this subclass, that representation would satisfy (IC) with . Thus (IC) is equivalent to (U), not a weaker consequence already supplied by the block estimates.
The precise missing step is to turn simultaneous biases, including the mixed constraints (M), into such a cover with a strict global error allowance, or into the signed bound of Step 4 directly. Lemma P supplies only a two-window density estimate, with its separation and tameness hypotheses. Bound (V) pays for the nontame early mass, but does not control the accumulation over windows or the errors of short blocks lying below their moduli. A proposed refinement must charge the shared prime covers and these errors once against the same original set.
A sufficient alternative is to extract a fixed fraction of the original reciprocal mass into jointly certified short blocks, each used residue containing at least two original elements. Their total extracted mass is at most , so the unproved selection statement (EX) on that page would give (U), without covering the omitted elements. A maximal matching in short windows pairs all but less than 14 units of reciprocal mass into pairs sharing a prime factor. This local construction still lacks joint certification: the exact property-P set prevents its two local pairs from being certified together under any representation with that occupancy requirement. The selection and strict signed global estimates both remain open; neither the matching nor the block bounds give a further bound for general property-P sets.
The term in Proposition F is an upper large-sieve allowance. It does not by itself classify sets that approach it or prove concentration in one residue class at a modulus near . Turning a substantial part of that allowance into certified guard structure requires an inverse estimate with quantitative counting error; none is held here. The column bound above a cutoff also remains unproved and already implies (U) when applied at any fixed modulus in the class , by Lemma E1. It is not an established substitute for this inverse step. The discriminating test is a strict error bound in (IC), or a strict signed global inequality, including sparse windows and mixed fibres. Pairwise density bounds and exact Fourier expansions alone do not supply it.
Certified congruence blocks carry bounded mass
This subsection bounds full blocks with certified residue protection by plus explicit counting errors. Theorem B covers blocks of ratio at most and single classes with its lower size cutoff. Lemma D covers arbitrary residue sets at odd moduli with . Convergence for an infinite family also requires the errors to be summable. These are scoped bounds for congruence constructions, not an inverse theorem for all sets.
For an integer and , write for with any integer . A block is a set
with integers and and a nonempty set . A block includes these specified parameters; its type is attached to this representation and is
For each divisor of , the set of with has Haar measure . The type contains the classes , of measure . Blocks with for are certified if for all , and some prime power divides , and but not .
The usual protection is certified. Suppose a prime power divides every residue in , and every later modulus, and divides no sum of residues of later blocks. Taking gives the certificate. Then no element of divides a sum of two elements of later blocks.
Lemma T. The types of certified blocks are pairwise disjoint.
Proof. Let and suppose and with and . Take a certificate for . It divides , hence ; it also divides , hence , which divides . This contradicts the certificate.
Lemma W. Let lie in a property-P set, put and , and suppose . Then
Proof. If is empty, then and the bound holds. Otherwise let be the least element of , so and . Put . Since , , so . Suppose . Let be the largest integer congruent to modulo with , and put . Then because , and with . So and are distinct elements of , both larger than , with . This contradicts property P, so .
Theorem B. Let be certified blocks. Suppose each either has , or is a single class with contained in a property-P set. Then
If every block has , property P is not used, and .
Proof. For one residue class modulo in , comparing each term after the first with an integral gives a reciprocal sum at most . A block with therefore has , and . For a single-class block, Lemma W gives
so . Here , and for every
with equality only for and . The values are checked directly. For , and . For , and . The first two bounds hold because only for , and only for . For , , which follows from for odd and . Also at , and the left side grows faster from there. So every block has , and Lemma T gives . With only blocks of ratio at most , the classes are already disjoint, which gives the second bound.
The bound does not depend on how many guards are used or how they are encoded. Block constructions of the kind described on the problem page are certified when their protection is the one stated there. In those, each block is a set of residues modulo in a window of ratio at most . Every residue of an earlier block is divisible by a guard prime that divides the later moduli, and no sum of two later residues is divisible by it. The descriptions there were read for this structure only, not re-verified. Any such construction, and any subset of one, has reciprocal sum at most . The error is at most , which is summable, for example, when . This covers one guard prime per block and guard systems encoded by the digits of the block index, whenever each pair of blocks is separated by a guard of this kind. Consequently a divergent property-P set cannot be covered by such blocks with a summable displayed error. Certification without control of that error gives no convergence conclusion, as the singleton representation below shows.
The single-class restriction can be removed for odd block moduli, while retaining an explicit counting error.
Lemma D. Let be odd, , and let have property P. Then
Consequently certified blocks with odd , , and property P inside each block satisfy , without a restriction on their ratios or on the number of residue classes.
Proof. If , the progression estimate in Theorem B gives . Otherwise let be the largest integer with . The sets of residues are pairwise disjoint. To see this, an intersection, after division by a power of 2 modulo odd , would give with and . Let be the least integer above in the class of , so . The integers and both lie in the class of . Since , we have , and , a forbidden relation.
Doubling preserves every gcd type modulo odd . Therefore . The maximality of gives
Combining this with the progression estimate proves the claimed bound. Summing and applying Lemma T proves the consequence.
The counting errors cannot be omitted. In fact every finite odd primitive set has a certified representation by singleton blocks: take , , and . These are full blocks of ratio at most . For , primitivity supplies an odd prime power dividing but not . It divides both moduli and not , so it is a certificate. The short-block error in Theorem B is now , and its inequality supplies no bound on . For example the odd primes give certified singleton blocks with divergent mass and divergent error; their union does not have property P. Thus certification alone is not a useful inverse conclusion for arbitrary property-P sets. A successful use must also pay for the counting error uniformly. Lemma D does not apply to this representation, since .
Certification is forced when the blocks are full residue classes over long windows.
Lemma S. Let lie in a property-P set , and let with and . Then .
Proof. Let be the least element of . If , the congruences and are compatible, since . Their common class modulo has an element with . Then are distinct, both exceed , and , contradicting property P.
Corollary S. Let a property-P set contain blocks with for , and suppose that for all
Then the blocks are certified.
Proof. Fix , and . The elements of in the class of are for at least consecutive integers . For each prime , the condition excludes at most one residue of modulo . It is automatic when . Otherwise it requires , or when , and is a unit modulo . So some admissible meets every condition, and that has . If no prime power divides , and without dividing , then . The class of in has length at least , because , so Lemma S gives a contradiction.
The richness condition asks earlier blocks to be long compared with later moduli, which constructions with growing moduli do not satisfy. Short early blocks can be protected through prime factors of individual early elements, which the block residues do not record. That is where an uncertified protection can enter.
Without saturation, a pair of compatible classes in two far-apart windows still cannot both be dense. This is the one-pair form of the charge that the open step needs.
Lemma P. Let have property P, let be even, and let be the least prime not dividing . Let and be residues modulo with . Let be the set of with and , where is the largest divisor of supported on the primes dividing . Let , where the endpoints are integers, and . Write , and where is the number of integers in congruent to modulo . Then
Proof. If or is empty the assertion is immediate. Otherwise, for put and . Then is coprime to , so its prime factors are at least , and because . For , , and divides both and , so . Hence exactly when . Write the elements of as with , where is the least integer in congruent to , and let be the set of these . Since is invertible modulo , property P says that has no two distinct elements whose sum is congruent to a fixed modulo . So at most one of two classes modulo is occupied, and a class with holds at most one element. As in (4), with and ,
Take weights and put . Grouping the fractions by reduced denominators and using for a prime , the large sieve cited for (1), applied to with denominators at most , gives
By linear-programming duality, the largest with every equals the least over with for every , the fractional covering number of the rough parts by primes. So . The same cover counts : for a prime the elements of congruent to modulo and divisible by form one class modulo , so
Multiplying the two bounds gives the lemma.
Lemma P gives a dichotomy for the tame early part and a later class in windows with its stated separation. Either a prime power dividing the early residue and does not divide twice the late residue, in which case their types are disjoint as in Lemma T, or the product of the two relative densities is at most plus its boundary term. It does not bound a sum over many windows. A bound for every pair still allows compatible windows of relative density . Fractional covers for different comparisons can reuse the same primes outside ; Lemma P gives no combined charge for that reuse.
The restriction on prime-power valuations in Lemma P must be retained. Its exceptional early mass has a controlled cost relative to , as follows. Work in the odd primitive class, let , take an integer and put . Call an early element tame when its largest divisor supported on primes dividing divides . If is the set of nontame early elements, then
For each prime , let be its least power exceeding . A nontame element is divisible by at least one such power. The quotient bound therefore gives ; the even-prime term is zero in this class. For each term is less than , so these terms sum to at most . For the power is , and . This proves (V). At square cutoffs , the same proof gives .
Thus Lemma P has a controlled prime-power exception when used with this choice of . Bound (V) is a harmonic mass estimate, not a bound on the relative density of the exceptional elements in every window. It does not remove the many-window accumulation problem or the counting-error problem in Theorem B.
Lemmas T, W, D, S and P, bound (V), Theorem B and Corollary S are author-recorded arguments without independent review. Lemma P uses the cited large-sieve statement and finite linear-programming duality. None of them bounds general property-P sets.
Exact normalization and ownership
For a finite nonempty property-P set , form its greedy coprime list . An unselected has a unique first index with . Assign it to the least prime dividing that gcd, and let consist of the assigned numbers divided by . Then : otherwise , coprime to all earlier , would have been selected before . Each nonempty child is a property-P set, and
Every original element has one owner. This is an identity, with no overlap loss and no dyadic rounding. Iterate at every child. Along an edge the maximum element decreases, so a finite input gives a finite tree. If is the product of edge primes on the path to node , put and . Then
No exclusion of 1 or 2 is applied to a normalized child. A property-P set containing 1 has at most two elements; one containing 2 has at most one larger element of each parity. These small cases also satisfy (14). Every selected coprime list has mass less than 10, by Proposition C or these small cases. Replacing every by 10, however, loses too much.
Why geometric sibling discounts fail
Let be the first odd primes and . Take large enough that , and put
Every subset of has property P: for in that interval, , so cannot divide . The greedy coprime list of (16) is just . Its -th branch consists of multiples of avoiding , divided by . Its first original element is , so ordering branches by their first elements gives exactly this prime order.
Write . For fixed , periodic counting and partial summation give
For example, inclusion-exclusion over the excluded primes gives an error at most in (17): the reciprocal sum of the multiples of any fixed divisor on differs from by at most , with . The lower bound follows from and telescoping .
Consequently no fixed and can bound every normalized -th branch by . Choose with , then choose large in (17). This refutes a scale-independent geometric sibling mass discount, including the proposed factor . It does not refute the product-scale occupancy identity (11). The many normalized branches here come from a single original size interval.
It also refutes a constant toll at every normalization node: the root and its first generation already have , unbounded as grows. Nevertheless,
Thus the excess is in the charging rule. These finite blocks have not been joined into an infinite property-P set.
Euler weights and a complete failed proof
A node also inherits a finite set of primes dividing none of its normalized elements. Initially . The exact inheritance along is
Division by preserves all these exclusions. Set and let be the primes dividing the selected coprime list. They are disjoint from . Flatten the primes of in that order, increasingly within each . The elementary identity gives
The sum is over nonempty children; omitting empty ones only decreases it. This includes prime 2. Define the positive leftover weight
The second equality telescopes over the finite tree, including its leaves. For (16), the first-generation sum in (20) is , tending to one: these correct weights still have no uniform strict contraction.
A direct bound also fails for every fixed . To see this, let be the product of the first primes at least 5, choose , and take
Here . The terms at least lie in an automatic property-P interval; relative to 3 they occupy residue 1 and the single residue-0 term . Thus has property P. Its greedy list is . The 3-child is nonempty, and every prime has the assigned witness . All possible root children are therefore nonempty, giving , whereas . This only disproves charging the current selected list to the current leftover weight; it does not disprove a charge that transfers a budget between generations.
Here is an end-to-end attempt with a specific missing inequality. Put
If some absolute satisfied
at every node of every finite property-P tree, then (15) and (21) would imply , hence . Truncation and the dyadic criterion would prove E12(iii). This implication is complete at author standing. Its premise (22) is false for every fixed , as the following family proves. Increasing a constant cannot repair this potential.
Semiprimes refute the two-generation charge
Let be a sufficiently large integer, , and
Here , so the two factors have unique roles. The set is contained in an automatic property-P interval. Its greedy coprime list is a maximal matching between the small primes and the large primes .
The prime number theorem implies, uniformly for once is large, that contains at least primes. For a fixed , this supplies more than possible partners . If were unmatched, each partner would have to occur in a selected edge, by maximality, but there are at most selected edges. Thus every small prime is matched. Write for the selected large primes. Every small-prime branch is nonempty, since there are more partners than elements of .
All partners are assigned to the small prime : it is their only factor occurring in the root list. After division they are primes, so are selected by that child's coprime extraction. The interval prime count above also gives . Removing the selected large primes costs at most in each such sum. Consequently, with ,
In the complete prime-by-prime sum behind (20), the only possible empty branches belong to . Each omitted term is at most . Hence
The last product bound is elementary: its reciprocal includes every term of in its Euler-product expansion. Also ; otherwise the same Euler products would stay bounded, contradicting the divergence of the harmonic series.
Equations (24)--(25) show
In particular (22) fails, even at roots with no inherited exclusions. This is a complete author-recorded obstruction using the prime number theorem as an external statement, not an inference from finite solver runs. It rules out this two-generation Euler-weight charge, with any fixed coefficients on the current toll and leftover weight. It does not rule out an argument grouping more information about the original scale. The sets (23) themselves have .
What a repair must retain
The normalization identity (14), the exact ownership rule, and the Euler accounting (19)--(21) survive. The geometric sibling discount and (22) do not. The missing input is a charge that controls the total mass (15) while grouping the many prime branches that represent one original size interval. Assigning a full coprime capacity to each branch loses that common origin; assigning only its Euler leftover weight also loses the prime-cofactor contribution in (24). Neither (H) nor any replacement providing this global charge is proved here.
Residue inheritance alone does not close the gap. Under division by , a constraint modulo becomes one modulo ; new moduli need not be coprime to old ones, and normalized elements need not exceed their least common multiple. Actual property-P tails also permit the singleton self-inverse exceptions in (9). These interfaces, as well as prime 2 and terminal sets, must be included in any use of (H). Identity (11), with its specific common-prime hypotheses, cannot be applied indiscriminately to siblings through different primes.
For a constraint and a different prime , two summands from below cannot have sum divisible by : such a sum would be at least . This protection concerns two summands in the same branch. Mixed summands from different branches do not receive it. A replacement must account for those mixed constraints as well as the delayed ones.
Harmonic mass also cannot be bounded by an absolute constant times the density of an occupied residue class. For fixed and , the set has property P. Indeed, a forbidden relation would give and , hence , whereas . As , by summing one arithmetic progression. Its ratio to the class density is therefore unbounded as grows. This refutes that density-only mass charge, not (H) or (U).
A finite two-block check tests the interaction left out of these potentials. Let contain every with and prime in the respective intervals and . They have 14 and 902 elements, and each entire block has property P. Their separate masses sum to more than , whereas
The lower bound is a retained witness checked against every forbidden triple. The upper bound is an exact finite computation, independent of solver optimality. For every one of the choices of early elements, remove late elements forbidden by an early pair and form disjoint pairs of remaining late elements whose sum is divisible by a chosen early element. At most one endpoint of each such pair can remain. Round each reciprocal upward to an integer multiple of ; subtracting the smaller rounded weight on each disjoint pair gives a rigorous upper bound. The largest of these 16,384 bounds is at most . The evidence entry point regenerates this covering of all cases and checks the matching edges. Thus joint mass is less than of the sum of the separate capacities in this exact domain. This is a finite interaction bound, not a uniform loss at arbitrary scales.
The same exhaustive calculation at other later scales gives the following certified upper bounds. The early block always has 14 candidates and mass approximately ; the separate masses in the table are rounded numerical summaries, not inputs to the certificates.
| Later interval start | Later candidates | Sum of separate masses | Joint upper bound |
|---|---|---|---|
| 625 | 57 | 0.164102 | 0.111012802 |
| 3125 | 217 | 0.146748 | 0.105285764 |
| 15625 | 902 | 0.137276 | 0.101959737 |
| 78125 | 3852 | 0.130341 | 0.100627648 |
A finite variant of this interaction test keeps the early block and enlarges the later small-prime cutoff from 5 to 7, 11 and 13 at these scales, retaining exact matching bounds. The proposed finite target is for every property-P subset of the two blocks. A feasible set above this threshold refutes that target; a matching bound below it certifies only the specified finite domain. A weaker matching bound alone decides neither outcome. Matchings of mixed forbidden triples are the proposed charges. The research obligation is to turn these losses into a bound independent of both the cutoff and the number of intervals, with each deleted element charged a bounded number of times. The present matchings reuse early constraints separately for every later block; no bounded multiplicity across blocks has been proved. A fixed finite prime cover does not ensure that (13) is finite: it does not control the primes shared by distinct elements. Indeed, every dilation has the prime cover , while preserving property P and reciprocal divergence. Proposition S handles a finite shared-prime set, or more generally a convergent sum of their reciprocals; the finite-prime-cover branch remains unresolved.
Unconditional linear weighting of the original triple constraints cannot alone supply that bound. On any finite candidate domain they have the form . The fractional assignment satisfies all of them and has unbounded harmonic objective as the domain grows. Thus a linear relaxation using only those constraints and cannot prove (U). The conditional matchings above use the chosen early elements as actual Boolean decisions. A repair must preserve that conditioning or supply stronger consequences; this is not an obstruction to nonlinear or multiscale arguments.
By Proposition E, a negative answer needs no infinite construction beyond the gluing: finite property-P sets of unbounded reciprocal sum suffice, and they may be sought inside one class . Neither unbounded finite sets nor the bound (U) is established here.
Ordinary multiplication of a finite set does not provide the required amplification. The odd pairwise coprime set has property P, since its only triple has . But contains , and , and . Both larger summands are distinct, so fails P even when the input is odd and primitive. Moreover, a product of reciprocal sums counts factorizations, whereas counts distinct integers. An amplification route must supply both a P-preserving operation and a lower bound on the reciprocal mass after these multiplicities are removed. This example rules out only the unmodified product operation.
Finite computations
The evidence entry point checks, in exact arithmetic, the property-P claims of the examples above, the occupancy values in (10) and (11) for small moduli, and the greedy coprime sequence of the truncated example (6). It also builds greedy property-P sets: the primes below , with reciprocal sum about and each region of the proof of Proposition C far below its bound, and the integers from 3 below , with 72 elements and reciprocal sum about . Both satisfy (13), as does the truncated example (6). Finally it runs SciPy's mixed-integer solver for the largest reciprocal sum of a property-P subset of , and for the admissible-class maxima quoted above. On the solver reports for , for , for , and for and . The returned sets have property P and these sums exactly; their optimality is the solver's floating-point report, not a certificate. These solver results alone bear on (U) and (H) only through consistency.
Nine additional retained witnesses use the same prescribed residue halves on , where is the product of the first odd primes and . Their exact domains, property P and rational masses are replayed from the retained inputs. The solver-reported masses for are approximately , and ; optimality is not certified. These windows meet (H)'s product-scale condition, but do not prove it.
Lemma E1, Proposition E, reduction (O), the class constructions, Proposition F, Corollary R, Corollary H, inequality (SE1) with its conditional implication, the whole argument with its cut at any fraction, the implication from decoupled energy bounds to (U), identities (C), (M) and (T), Lemmas T, W, D, S and P, bound (V), Theorem B and Corollary S are written arguments in the text; the entry point does not check them. The implication from (IC) to (U) and its converse are written arguments; (IC) itself remains unproved. The exact products behind and the cases of are stated with their values and are checked by hand in the text. Proposition F and Corollaries R and H also use the stated large-sieve interface. The proposed estimate (SE) remains unproved and has no computational warrant here.
The normalization checks cover exact ownership and both telescoping identities, including normalized sets containing 1 or 2; finite instances of (16) and (23), and of the direct-toll obstruction; the semiprime matching and discarded-cofactor bounds; and the explicit constant in Proposition C. The infinite obstruction (26) uses its written prime-number-theorem argument, not these samples. The separate two-block witness and integer matching computations certify only (27) and the four finite table bounds.
Source scope
Schoen's definitions, Lemma 1 and equations (7)--(9) were read in the primary transcription. Printed p. 194 was also inspected: the lower index in its later comparison of frequency sums is printed that way. The argument here explicitly uses , as required by (9). This is not a transcription repair or an independent audit of the paper. The bibliography and canonical result are on the source card.
The Brun--Titchmarsh input is read from H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory I: Classical Theory, Cambridge University Press, 2006, Chapter 3, Theorem 3.9, printed p. 90, and the Section 3.5 notes on Section 3.3, p. 102. For , , and , the number of primes in congruent to modulo is at most : the notes explicitly remove the error term in Theorem 3.9 using the authors' 1973 paper The large sieve, Mathematika 20, 119--134. Both cited book pages were read on page images. The uses here have for , or and , so lie in the read domain. The original 1973 statement and proof were not inspected, and this is not a proof audit of the book's theorem.
The semiprime obstruction uses the prime number theorem , read at statement depth on the page image of the same book, Chapter 6, printed p. 168. Its proof is not reconstructed here. The interval lower bound used in (24) follows by subtracting the asymptotics at and ; its uniformity for follows directly from the eventual bound in the definition of that asymptotic. No error term or prime progression theorem is required. The Euler-product estimates in (25) are proved in the text.
Corollary H's explicit bound includes an elementary proof of . Mertens' product theorem is used at statement level only for the sharper constant and for the remark on cutoffs . Its statement is recorded in A. Granville, Harald Cramér and the distribution of prime numbers, Scand. Actuar. J. 1995, no. 1, 12--28, equation (1), printed p. 13, as extracted on the linked source card. That card was read at statement scope; no proof or new page image was checked for this use.
The erdosproblems.com discussion, accessed 16 October 2026, supplies congruence constructions and Tao's inverse-theorem proposal; it supplies no inverse theorem. The known AlphaProof results settle parts (i) and (ii). The reciprocal-sum claim on the proof-claims tab is the already rejected claim whose growth axiom is false. The same day's inspected erdosproblems.ai, Palomar, arXiv and formal-conjectures records yielded no additional full claim for part (iii). That is a bounded source finding, not certification of openness.