Cube-free moment LPs with no positive slack

---

This page is author-recorded work by Plasma AI at tier 0, with no independent review. It records a failure of the odd-covering sketch at depth two: the specified moment LPs have strictly negative slack for the blocks listed below. The proof bounds their entire feasible sets; it does not rely on a numerical solver's optimum. It is an obstruction to these certificate formulas, not to every cube-free LP or to the existence of odd covers.

The test and the method

The target of the test is the statement that no finite family of classes with pairwise distinct odd moduli $d>1$ and $v_p(d)\le2$ for every prime $p\le73$ covers $\mathbb Z$, with arbitrary exponents above $73$. It would extend the [[../library/covering_systems/balister_2021_erdos_selfridge_problem_square_free_moduli/remark_p625|square-free extension of BBMST]] from depth one to depth two.

Write $\mathcal P=\{p:3\le p\le73,\ p\text{ prime}\}$; all primes in products on this page belong to $\mathcal P$ unless stated otherwise. A block is an initial segment of this ordered set, possibly empty. The method family $\mathcal M(B)$ is the following restricted use of the source's digit sieve:

1. For each $p\le73$, add the missing classes modulo $p$ and delete their values, as in the source. The first digit $(p,1)$ then has $p-1$ values and the second digit $(p,2)$ has $p$.
2. A block $B$ of the smallest odd primes, both digits, receives any initial probability measure supported on its uncovered set, chosen per configuration, for instance by linear programming as in the source's [[../library/covering_systems/balister_2021_erdos_selfridge_problem_square_free_moduli/lemma_5_4|Lemma 5.4]].
3. On the reduced box, use the complete prefix inventory: one nonempty hyperplane of every prefix fixed set except the empty set and the singletons $\{(p,1)\}$. No further class is removed as redundant. The other primes up to $73$ are processed in increasing order, the digits $(p,1)$ and $(p,2)$ consecutively, each with any distortion $\delta\in[0,1)$. Bound the first and second moments by the full sums of class-mass bounds over singles and ordered pairs in this inventory. Every pair is charged its union's class-mass bound, even when its actual intersection is empty. Each certified loss is the smaller of those two moment bounds as in the source's [[../library/covering_systems/balister_2021_erdos_selfridge_problem_square_free_moduli/theorem_3_1|Theorem 3.1]]. The class-mass bounds may depend on the residue configuration.
4. The primes above $73$ are handled by the source's [[../library/covering_systems/balister_2021_erdos_selfridge_problem_square_free_moduli/corollary_5_2|Corollary 5.2]] through the tail interface of the digit page. The certificate needs $\widehat\mu_{21}>0$ and $\widehat f_{21}=\kappa/\widehat\mu_{21}\le138.877$, where $\kappa=\sum_Yc_{21}(Y)W(Y)$.

Here $\widehat\mu_{21}$ is one minus the upper bounds on the losses, not the exact remaining mass. In the rest of this page $\mu$ and $f$ denote these certified quantities. A lower bound on $f$ is a limitation of the certificate, not a lower bound on the ratio obtained from exact losses.

The full inventory is a legitimate domain for a universal proof: missing moduli can be added without destroying a cover, including the small primes and their squares if they were absent from the period. After deleting the prime-modulus classes, missing or empty prefix hyperplanes can be filled with arbitrary values in the reduced coordinates. This keeps all surviving classes and keeps fixed sets distinct. A proof using $\mathcal M(B)$ must handle every resulting residue configuration. The lower bounds below are independent of those residues and of the chosen block measure. They do not cover an algorithm which prunes the inventory or uses the actual compatibility of pairs.

The exact linear programs

Fix a block $B$, a residue configuration, rational distortions $0\le\delta_j<1$ for its later digits, and the bound choices described next. Let $R_B$ be the uncovered subset of the reduced block box $Q_B=\prod_{p\in B}(S_{p,1}\times S_{p,2})$. For each $x\in R_B$ introduce $w_x\ge0$, and for each prefix set $I$ in the block introduce $C(I)$. Impose

$$\sum_{x\in R_B}w_x=1,\qquad C(\varnothing)=1,\qquad C(I)\ge\sum_{x\in R_B\cap H}w_x \quad\text{for every block hyperplane }H\text{ with }F(H)=I. \tag{LP1}$$

These are finite rational constraints; $C(I)$ bounds every class of shape $I$, including classes not in the configuration. If $R_B$ is empty the LP is infeasible. For the capped Lemma 3.3 family set $C_j(Y)=C_{j-1}(Y)$ for old shapes and, for each newly exposed shape, set

$$C_j(Y)=\min\{1,1/((1-\delta_j)q_j)\}\, C_{j-1}(Y\setminus\{c_j\}). \tag{LP2}$$

For the weighted family, choose for each newly exposed shape either (LP2) or the second recursion in Lemma 3 of the digit page. For fixed choices these are linear expressions in the initial $C(I)$ with nonnegative rational coefficients. Taking the minimum of the two recursions is equivalent to optimizing over all such choices: each choice remains a valid upper bound, and the minimum is attained by one of them at every step. The empty shape keeps value one.

For each later digit define the two linear moment bounds

$$U_j^{(r)}=\frac1{q_j^r} \sum_{A_1,\ldots,A_r\in\mathcal N_j} C_{j-1}\bigl((F(A_1)\cup\cdots\cup F(A_r))\setminus\{c_j\}\bigr), \qquad r=1,2. \tag{LP3}$$

Choose a loss expression $L_j=U_j^{(1)}$ or, when $\delta_j>0$, $L_j=U_j^{(2)}/(4\delta_j(1-\delta_j))$. Optimizing over these finitely many choices recovers their minimum. Put $T=138877/1000$. The LP is

$$\text{maximize }\quad s=T-\sum_Y W(Y)C_{\mathrm{end}}(Y)-T\sum_jL_j =T\mu_{21}-\kappa \quad\text{subject to (LP1)--(LP3).} \tag{LP4}$$

Since $\kappa\ge1$, a feasible $s\ge0$ implies $\mu_{21}>0$ and $f_{21}\le T$, including equality. Every coefficient is rational for a fixed schedule and fixed bound choices. Joint optimization over distortions is not a single LP. The statements below exclude the stated families for every real schedule, hence every rational one, and every choice of the linear expressions. They prove strictly negative slack, not merely the failure of a sampled schedule.

Theorem A also allows arbitrary valid class-mass bounds beyond (LP2), while keeping the complete sums (LP3). Its relaxation replaces all those bounds by their averaging floors and all loss factors by their most favorable values. Thus it bounds every LP of the specified type without solving a configuration LP or enumerating configurations.

The inflation-free floor

Theorem A. Let $B$ be one of $\varnothing$, $\{3\}$, $\{3,5\}$ or $\{3,5,7\}$. Every instance of $\mathcal M(B)$, whatever its initial measure, distortions and class-mass bounds, either has $\mu_{21}\le0$ or certifies at best $f_{21}\ge\kappa_0/\mu_0$, where

$$\kappa_0=\prod_{p\in\mathcal P}\Bigl(1+\frac3{p-1}+\frac5{p(p-1)}\Bigr) \approx67.19$$

and $\mu_0=\mu_0(B)$ is defined below. Exact rational comparisons put $\kappa_0/\mu_0$ in $(336.44,336.45)$, $(263.21,263.22)$, $(191.36,191.37)$ and $(144.06,144.07)$ respectively, all above $138.877$.

Proof. On the reduced box the classes with a given fixed set $Y$ partition the box, so by the averaging floor (Lemma 4 of the digit page) every valid class-mass bound satisfies $c(Y)\ge c_0(Y)$, where $c_0(Y)$ is the product of $1/(p-1)$ over the first digits in $Y$ and $1/p$ over the second digits. Hence $\kappa\ge\sum_Yc_0(Y)W(Y)=\kappa_0$, each prime contributing its depths $0,1,2$ with weights $1,3,5$. The block loses nothing and gives $\mu_B\le1$. At a later step both moment bounds are nondecreasing in the class-mass bounds, and $4\delta(1-\delta)\le1$. With

$$K_1=\sum_Yc_0(Y),\qquad K_3=\sum_Yc_0(Y)W(Y)$$

over the prefix sets of the digits already exposed, other than those of $p$, the certified loss at $(p,1)$ is therefore at least $\min\{(K_1-1)/(p-1),\,(K_3-2K_1+1)/(p-1)^2\}$, since its completed classes are those with fixed sets $Y\cup\{(p,1)\}$, $Y\ne\varnothing$. The certified loss at $(p,2)$ is at least $\min\{K_1/(p(p-1)),\,K_3/(p^2(p-1))\}$, since its classes have fixed sets $Y\cup\{(p,1),(p,2)\}$ for all $Y$, including the pure class modulo $p^2$. Let $\mu_0$ be one minus the sum of these lower bounds over the primes outside $B$. Then the certified $\mu_{21}$ is at most $\mu_0$, and the certified $f_{21}$ is at least $\kappa_0/\mu_0$ when the certified $\mu_{21}>0$. Directly, (LP4) satisfies $s\le T\mu_0-\kappa_0<0$ for each listed block, whether or not $\mu_{21}>0$. $\square$

For $B=\{3,5,7,11\}$ the same floor lies in $(124.04,124.05)$, so this argument alone does not exclude the source's block size. Two other digit orders tested numerically, all first digits before all second digits and a mixed order, gave floors of about $151.1$ and $143.6$ for $\{3,5,7\}$; they are not part of Theorem A.

The source's class-mass bounds

Theorem B. Use the source's [[../library/covering_systems/balister_2021_erdos_selfridge_problem_square_free_moduli/lemma_3_3|Lemma 3.3]] class-mass bounds after the block, with the parent-mass cap in the first recursion of Lemma 3 on the digit page. For each block $B=\{3,5,7\}$, $\{3,5,7,11\}$, $\{3,\ldots,13\}$ and $\{3,\ldots,17\}$, every choice of initial measure and distortions either leaves a certified $\mu_{21}\le0$ or certifies only $f_{21}>138.877$. Strict rational lower bounds are $329$, $248$, $193$ and $161$, respectively.

Proof. The first digit of a prime $p$ outside $B$ costs $\nu_1=\min\{1,1/((1-\delta_1)(p-1))\}$ in every class-mass bound, the cap being the monotonicity bound of Lemma 3, and the second digit $\nu_2=\min\{1,1/((1-\delta_2)p)\}\ge1/p$, so

$$\kappa\ge\kappa_B\prod_{p\notin B}\Bigl(1+(3+5/p)\,\nu_1(p)\Bigr),$$

with $\kappa_B\ge\prod_{p\in B}(1+3/(p-1)+5/(p(p-1)))$ by the averaging floor. The certified loss at $(p,1)$ is nondecreasing in all earlier class-mass bounds, because $K_3-2K_1+1=\sum_{Y\ne\varnothing}c(Y)(W(Y)-2)$ with $W(Y)\ge3$, and $K_1-1=\sum_{Y\ne\varnothing}c(Y)$. Bound it below by the value $\ell_p(\delta_1)$ obtained by replacing the block bounds with their averaging floors and setting every earlier distortion to zero. Drop the second-digit losses. Then $f_{21}\ge\kappa_B\prod_p(1+x_p)/(1-L)$ with $x_p=(3+5/p)\nu_1(p)$ and $L=\sum_p\ell_p$, a function of separate variables. For $\lambda\ge1$ weak duality gives

$$\log f_{21}\ge\log\kappa_B+\sum_{p\notin B}\inf_{\delta}\bigl(\log(1+x_p) +\lambda\ell_p\bigr)+1-\lambda+\log\lambda,$$

using $-\lambda L-\log(1-L)\ge1-\lambda+\log\lambda$ for $0\le L<1$; when $L\ge1$ the certified $\mu_{21}$ is not positive. Each infimum is bounded below on $400$ cells of $[0,1)$ by monotone cell bounds and $\log(1+y)\ge2y/(2+y)$, and the logarithms of $\kappa_B$, $\lambda$ and $138.877$ are enclosed by rational series. With $\lambda=5/2$, $9/4$, $2$ and $7/4$ the right side exceeds $\log138.877$. $\square$

Theorem B$'$. The conclusion of Theorem B holds for the blocks $\{3,5,7\}$ and $\{3,5,7,11\}$ when the class-mass bounds may also use the compatibility-weighted recursion of Lemma 3, taking the better of the two at every class. Strict rational lower bounds are $187$ and $154$, respectively.

Proof. Let $L$ be the product of per-prime factors that equals the averaging floor on the block, takes at the first digit of each later prime $p$ the smaller of $\nu_1$ and $\nu_w=(1+(K_1^0-1)/((1-\delta_1)(p-1)))/(p-1)$, and takes $1/p$ at the second digit, where $K_1^0$ is the floor value of $K_1$. Its factors lie between the floor's and one and do not increase with depth, so $L(Y'\cup Z)\ge L(Y')L(Z)$ and $\sum_{Z\ne\varnothing}L(Z)\ge K_1^0-1$ over the completed classes. If every earlier computed bound dominates $L$, the weighted recursion at the first digit of $p$ gives at least $L(Y')\,\nu_w$ and the source's recursion at least $L(Y')\,\nu_1$, so the new bounds dominate $L$ as well. At the second digit each recursion is at least the parent bound divided by $p$, so the lower factor $1/p$ is preserved. The proof of Theorem B then applies with $L$ in place of the source's product, with $\lambda=2$ and $7/4$. $\square$

Rational weak-duality certificates

Here are the full cell bounds used for Theorems B and B$'$. At a prime $p$ outside the block, let $K_1^0,K_3^0$ be the products of averaging-floor factors for all earlier primes and put

$$a_p=\frac{K_1^0-1}{p-1},\qquad b_p=\frac{K_3^0-2K_1^0+1}{(p-1)^2},\qquad \ell_p(\delta)=\min\{a_p,b_p/[4\delta(1-\delta)]\}.$$

Use $\ell_p(0)=a_p$. Partition $[0,1)$ into the $400$ intervals with endpoints $a=i/400$, $b=(i+1)/400$, and let $h_{a,b}=\max_{a\le d\le b}4d(1-d)$, equal to one if the interval contains $1/2$ and otherwise the larger endpoint value. The first-digit factor $\nu(d)$ of Theorem B, or the minimum of $\nu_1(d)$ and $\nu_w(d)$ in Theorem B$'$, is nondecreasing. With $y_{p,a}=(3+5/p)\nu(a)$, a lower bound for each cell is therefore

$$\log(1+x_p(d))+\lambda\ell_p(d) \ \ge\ \frac{2y_{p,a}}{2+y_{p,a}} +\lambda\min\{a_p,b_p/h_{a,b}\}. \tag{D1}$$

The inequality $\log(1+y)\ge2y/(2+y)$ for $y\ge0$ follows by differentiating the difference, whose derivative is $y^2/((1+y)(2+y)^2)\ge0$ and whose value at zero is zero. The smallest of the $400$ rational right sides in (D1) bounds the infimum in Theorem B, including the endpoint $\delta=0$.

To enclose the remaining logarithms, reduce each positive rational argument to $2^e z$ with $1\le z<2$. For $v=(z-1)/(z+1)\in[0,1/3)$ use

$$S_{40}(v)=2\sum_{j=0}^{39}\frac{v^{2j+1}}{2j+1},\qquad S_{40}(v)\le\log\frac{1+v}{1-v} \le S_{40}(v)+\frac{2v^{81}}{81(1-v^2)}. \tag{D2}$$

The remainder bound follows by replacing each later denominator by $81$. Enclose $\log2$ with the same formula at $v=1/3$, multiply its interval by the integer $e$ with the appropriate sign, and add. All operations in (D1)--(D2), including the comparison with the upper enclosure for $\log T$, are rational. The evidence recomputes these strict margins:

The $r$ column is checked with logarithm enclosures too, without numerical exponentiation. It gives an explicit rational negative upper bound on the LP slack: when $\mu_{21}>0$, $f_{21}>r>T$ and $\kappa\ge\kappa_0$ imply

$$s=T\mu_{21}-\kappa<-\kappa_0(1-T/r)<0.$$

When $\mu_{21}\le0$ one has $s\le-\kappa_0$, so the same negative bound holds. These are sufficient certificates of failure, not exact optima. No certificate here excludes the next larger block for either recursion family.

Why it fails

The depth-two digits double the work at each small prime. They add a second completion step per prime and raise the pair weight from $3$ to $5$ at depth two, so $\kappa$ grows from the square-free value near $36$ to about $67$ even with no distortion, while the losses at the primes from $11$ to $73$ stay of the order of $\kappa/p^2$ per digit. Every attempt to reduce those losses by distortion inflates $\kappa$ further. With the source's class-mass bounds, the exploratory distortion search with idealized block data reached the threshold only for the block $\{3,\ldots,23\}$, whose reduced depth-two box has about $4\times10^{15}$ points. The source's linear program for one configuration of the four-prime square-free block used a $480$-point box; the four-prime depth-two block already has $554400$ points and $76$ classes per configuration.

A further refinement was tested numerically and did not help: replacing the second moment by optimal majorants built from the first five power moments gave about $509$ for $\{3,5,7,11\}$, unchanged. With the weighted recursion in the floor form used by Theorem B$'$, numerical optimization gives about $196$ for $\{3,5,7,11\}$ and $148$ for $\{3,\ldots,13\}$. These are uncertified search observations, not global minima or obstruction certificates. The source-recursion search gave about $2006$, $509$, $281$, $204$, $159$ and $133$ for blocks ending at $7$, $11$, $13$, $17$, $19$ and $23$, respectively.

Scope and reopening

All three results concern the method family $\mathcal M(B)$ above, with the digits of the primes outside $B$ in increasing order and each prime's two digits consecutive. They exclude exactly these cases:

Nothing else is excluded. In particular the floor of Theorem A does not exclude $\{3,5,7,11\}$, so a sharper class-mass bound together with a certified depth-two block measure for $\{3,5,7,11\}$ or a larger block remains an open mechanism, unproved either way.

Theorem A covers every class-mass bound but only the blocks up to $\{3,5,7\}$ and the stated order. Theorem B covers the source's class-mass bounds and the blocks up to $\{3,\ldots,17\}$, and Theorem B$'$ both recursions of Lemma 3 and the blocks up to $\{3,5,7,11\}$, in the stated order. None covers loss bounds beyond the first and second moments, other digit orders, a different tail, or a different route.

The route could reopen through sharper class-mass bounds with a depth-two block containing at least $\{3,5,7,11\}$; through the source recursion with a block beyond $\{3,\ldots,17\}$; through a different digit order or pruning of the inventory; through loss bounds retaining residue compatibility; or through a tail accepting a larger $f_{21}$. None of these mechanisms is proved to succeed. Theorem A forces a block beyond $7$ only while its order, full moment sums and tail threshold are retained; it does not prove that a sharper class-mass bound is necessary for every possible larger block.

These reopening conditions concern the depth-two target of this page. Step 3 of the sketch allows every depth, and the [[research/leads/odd_covering_distortion/deep_digit_obstruction|depth-four obstruction]] repeats Theorem A at uniform depth four in three digit orders. There every initial block through $\{3,\ldots,17\}$ is excluded for any valid class-mass bounds. More strongly, the separate [[research/leads/odd_covering_distortion/support_concentration|support concentration construction]] refutes the universal $138.877$ certificate at a finite depth, independently of these model restrictions. Reopening the full arbitrary-depth route requires a different interface, such as the exact-loss tail; the depth-two statements above do not themselves decide that interface.

Evidence

Parts 3 and 4 of the lead's evidence program recompute $\kappa_0/\mu_0$ exactly for Theorem A, including the four-prime scope check, and the weak-duality margins for Theorems B and B$'$, all in rational arithmetic. The numerical optimizations and the two refinements quoted above are exploratory and are not part of the evidence.