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Rough-integer densities and the Buchstab identity
This page reconstructs Lemma 7.4 of the
weighted-dilation manuscript
for the
extraction of primes 2u-1,
as part of the
elementary estimates of the smooth-successor route.
The formulation and use of the estimates follow OpenAI; the densities and
minimum-coordinate identity are classical Buchstab rough-number estimates
(A. Buchstab, Mat. Sbornik 2(44) (1937), 1239–1246). The derivation and
the range checks below are Plasma AI's. The
E3 examination
has a distinct PASS grade, examined on 2026-10-16 UTC, accepting
refutation-failed at its stated classical-premise and parameter scopes.
The grade retains the original-source
reading limits and, for the weighted E3 consumer consequence, its Type II/P
hypothesis.
Notation is that of the
proxy-discrepancy page:
cells C=(Y,eΔY] with Δ=L−S, R(C) the number of W-rough
integers in C, and cI(C) the proxy constant of a test I.
The densities
For w>γ>0 put Dγ(w)=∑l≥1Tl(γ,w), where
T1(γ,w)=1/w and, for l≥2, with s(t)=t1+⋯+tl−1,
This is the manuscript's (7.15). Tl vanishes unless lγ≤w, so the
sum is finite. Writing tl=w−s(t), the l-th term is 1/l! times the
integral of ∏i≤lti−1 over
{t∈Rl:ti≥γ,∑ti=w} against dt1⋯dtl−1:
ordered lists of l prime logarithms, scaled by L, with sum w.
As a cross-check that the arguments below do not use, $D_\gamma(w)=
\omega(w/\gamma)/\gamma$ for Buchstab's function ω; for
2γ≤w≤3γ both sides equal (1+log(w/γ−1))/w.
Continuity. For fixed b,ε>0 and wmax, D is jointly
continuous on the compact set
$K(b,\varepsilon,w_{\max})={(\gamma,w):\gamma\ge b,
\gamma+\varepsilon\le w\le w_{\max}}$, thresholds w=lγ included.
Indeed only l≤wmax/b occur. On Ωl(γ,w) the integrand is
at most b−l, since ti≥γ and w−s(t)≥γ, and
Ωl(γ,w)⊂[b,wmax]l−1. As (γ′,w′)→(γ,w)
the integrand on Ωl(γ′,w′) converges to that on
Ωl(γ,w) at every point off the hyperplanes ti=γ and
s(t)=w−γ, a null set, so dominated convergence gives continuity of each
Tl; at w=lγ the domain shrinks to a point and Tl→0.
Counts in short cells
Fix the precision S≥1, as in every consumed application. The prime
number theorem with error
OA(y(logy)−A), used with A=S+2, gives uniformly for y with
logy≥c0L, c0>0 fixed,
π(eΔy)−π(y)=logyΔy(1+O(L−1)).
The balanced prime proxy (7.16). Let 0<κ<1/50. A cell meeting
Bκ=[x1/2−2κ,x1/2+2κ] has
w=logY/L∈[0.46−o(1),0.54], so by the display and item 1 of the
proxy-discrepancy proposition
cpr(C)=R(C)π(eΔY)−π(Y)=wLV(W)1+o(1)≤LV(W)3
for x≥x0(κ,S). The constant 3 is absolute: it does not depend on
κ or S, which affect only x0. This is the uniformity that the
balanced constant Cbal of the extraction requires.
Rough counts (7.20). Fix b,ε>0 and wmax. Uniformly for
cells C=(Y,eΔY] and exponents γ with
(γ,logY/L)∈K(b,ε,wmax),
Proof. The integers of C divisible by p2 for some prime p>xγ
number at most ∑p>xγeΔY/p2≤3Yx−b=o(ΔY/L). A
squarefree m∈C with P−(m)>xγ is a product of $l\le l_{\max}=
\lfloor(w_{\max}+1)/b\rfloor$ distinct primes above xγ, and its l!
orderings give it total weight one when each ordered list carries weight
1/l!. Dropping distinctness only adds lists, so the squarefree count is at
most ∑l≤lmaxNl/l!, where Nl counts ordered
l-tuples of primes above xγ with product in C.
For l=1, every prime of C exceeds xγ because
w≥γ+ε, so N1=(1+O(L−1))ΔY/(wL), the term
T1. For l≥2 fix p1,…,pl−1 with product Q and
ti=logpi/L. A last prime exists
only if eΔY/Q>xγ, that is s(t)<w−γ+Δ/L; then
log(Y/Q)=L(w−s(t))≥bL/2, and the last prime lies in (Y/Q,eΔY/Q].
Dropping its lower cutoff and using the display,
It remains to compare μ⊗(l−1) with ν⊗(l−1), where
ν=dt/t. By Mertens' estimate
∑p≤y1/p=loglogy+M+O(1/logy), every interval satisfies
μ((a,a′])=ν((a,a′])+O(1/(bL)) for γ≤a<a′≤wmax. Fix
η>0 and cover (γ,wmax]l−1 by the cubes
∏i(γ+kiη,γ+(ki+1)η], truncating the last interval
in each coordinate at wmax. There are at most
Nη=(wmax/η+1)l−1 of them. Each cube has
μ⊗-measure equal to its ν⊗-measure up to
Oη,b,l(1/L). Replace φ, which is at most 2/b, by its supremum
φ+ on each cube. Then
∫φdμ⊗≤∫φ+dν⊗+Oη,b,l,wmax(L−1).
On cubes inside {s<w−γ+Δ/L} the oscillation of 1/(w−s) is at
most 4(l−1)η/b2. The cubes meeting the hyperplane
s=w−γ+Δ/L lie in a slab of width 2(l−1)η, whose
ν⊗-measure is Ob,l,wmax(η). The faces ti=γ are
cube faces, so they cause no boundary error. Finally the strip
w−γ≤s<w−γ+Δ/L has ν⊗-measure
Ob,l,wmax(Δ/L). Hence
∫φdμ⊗(l−1)≤l!Tl(γ,w)+O(η)+oη(1),
uniformly on K(b,ε,wmax). Here the constant in O(η)
depends only on b,l,wmax, and the oη(1) tends to zero uniformly
once η and S have been fixed. For any desired error first choose a
fixed η, then let x grow; this proves a uniform o(1) without
interchanging the choices of S and x. □
The cofactor bound (7.17). Let b1≤γ<γ′≤1/2−κ be a bin.
Suppose C contains an integer m=N/p with x≤N≤5x and p a prime in
(xγ,xγ′]. With a=logp/L∈(γ,γ′],
for all sufficiently large x. Thus (γ,w) lies in the compact set
K(b1,κ,1), where the rough-count bound holds uniformly and D is
uniformly continuous. Dividing by
R(C)=(1+o(1))ΔYV(W),
uniformly over all such cells. This is (7.17) exactly at the consumed range.
The only property of N=2u−1 used is x≤N≤5x, which also holds for
2u+1; the change of sign moves w by O(1/L) and does not change the
rough-integer density or its prime-factor cutoff. No density at w=γ is
used: on the consumed range w−γ≥2κ+o(1), with b1 and κ
fixed before x.
The minimum-coordinate identity (7.18)
Identity. For 0<b1<1/2,
∫b11/2Dα(1−α)αdα=Db1(1)−1,
with the integrand at α=1/2 read as its left limit 2; for
1/3<α<1/2 only T1 survives and Dα(1−α)=1/(1−α).
Proof.T1(b1,1)=1. For l≥2 write Tl(b1,1) on the simplex
Σl={t∈Rl:ti≥b1,∑ti=1} with the measure
l!1∏i≤lti−1dt1⋯dtl−1. This measure is
invariant under all permutations of the l coordinates: the density is
symmetric, a transposition of two of t1,…,tl−1 has Jacobian −1, and
the transposition of tj with tl=1−∑i<lti replaces tj by
1−∑i<lti, a linear change with determinant −1. Off a null set
exactly one coordinate is minimal; by symmetry,
Put α=t1. Two coordinates are at least α and sum to at most one,
so b1≤α≤1/2. For fixed α, the remaining
t2,…,tl−1 range, up to null boundaries, over
Ωl−1(α,1−α), with integrand
(1−α−∑2≤i<lti)−1∏2≤i<lti−1; for l=2 the
inner integral is the value (1−α)−1. Since l/l!=1/(l−1)!, Tonelli's
theorem gives
Tl(b1,1)=∫b11/2Tl−1(α,1−α)αdα.
Summing the finitely many l≥2 proves the identity. □
The term 1 is the one-prime contribution to Db1(1); the integral is
the composite part, organized by the least prime factor.
The bound (7.19)
Bound. There are absolute b0,c′>0 such that for 0<b1≤b0,
Db1(1)≤b1e−γE(1+c′e−1/(20b1)).
Lower count. Let Φ(x)=#{x<m≤2x:P−(m)>xb1}. Every m with
Ω(m)=l has at most l! ordered prime factorizations, so ordered lists of
primes above xb1 with product in (x,2x], each with weight 1/l!, give
every integer total weight at most one, repetitions included. Fix
η′>0. The case l=1 gives π(2x)−π(x)=(1+o(1))x/L. For l≥2
restrict the first l−1 primes to ti>b1 and s(t)<1−b1−η′ and sum the
last prime over (x/Q,2x/Q]; it is then automatically above
xb1+η′, and the prime number theorem gives
(1+O(L−1))x/(QL(1−s(t))) of them. The same cube comparison, with infima on
cubes in place of suprema, gives
Upper count. The fourth instance of the
block sieve, at
z=xb1 with g(p)=1/p and h=2⌊1/(40b1)⌋, gives
Φ(x)≤xV(xb1)(1+c1e2e−1/(20b1))+x1/5+2b1 once h≥h0,
which fixes b0. By Mertens, LV(xb1)→e−γE/b1, and the
remainder is o(x/L).
Combining, $D_{b_1}(1)\le\liminf L\Phi(x)/x\le\limsup L\Phi(x)/x\le
(e^{-\gamma_E}/b_1)(1+c'e^{-1/(20b_1)})$ with c′=c1e2. □
The bound is a statement about a fixed number, obtained from two ordinary
counts; no unrestricted pointwise density theorem is imported.
The mesh step
The extraction sums the bin constants
supγ≤a≤γ′Dγ(1−a)log(γ′/γ). On
{(γ,a):b1≤γ≤a≤1/2−κ} the map
(γ,a)↦Dγ(1−a) is uniformly continuous, since there
1−a−γ≥2κ; let ω be its modulus. For a partition of
[b1,1/2−κ] of mesh at most δ0, every α and a in the
bin [γ,γ′] satisfy
Dγ(1−a)≤Dα(1−α)+ω(2δ0). Integrating against
dα/α over each bin and summing,
A fixed mesh makes the last term at most 1/10, as the extraction requires.
The bins then remove at most
LSXA{Db1(1)−1+101+o(1)} by the identity, while
the preliminary sieve retains at least
LSXA{b1e−γE(1−O(e−1/(20b1)))+o(1)}.
By (7.19) the difference is at least
LSXA{1−101−O(b1−1e−1/(20b1))+o(1)}, as on
the extraction page.