Smooth integers in every short block, exponent by exponent

---

Target and conventions

The target is Problem 962, whose assessment records both of its questions as open; this research does not change that status. Let $k(n)$ be the largest $k$ for which some integer $m$ with $0\le m\le n$ has each of $m+1,\ldots,m+k$ divisible by a prime greater than $k$. In Erdős's 1976 notation, $n_k$ is the least $n\ge0$ such that each of $n+1,\ldots,n+k$ has a prime factor greater than $k$, and $k(n)=\max\{k:n_k\le n\}$. Allowing $m=0$ or requiring $m\ge1$ changes no bound below: every upper bound here is proved for all $m\ge0$, and the lower bound of L252 uses a positive $m$.

A block $m+1,\ldots,m+K$ whose terms all have a prime factor greater than $K$ is a block with no $K$-smooth member, and its first $K'$ terms form such a block for every $K'\le K$. Hence, for every integer $K\ge1$,

$$k(n)<K\iff\text{every block }(m,m+K]\text{ with }0\le m\le n \text{ contains a }K\text{-smooth integer.}$$

This is an assertion about every block, not almost every block: a theorem that holds outside an exceptional set of $m$ says nothing about $k(n)$, because the one bad block may be exceptional.

Two targets are kept apart.

• Waypoints: $k(n)\ll n^\theta$ with $\theta<1/2$, that is, $n_k\gg k^{1/\theta}$.
• The displayed question: $\log k(n)\le(\log n)^{1/2+o(1)}$, Erdős's conjecture (7) in inverse form. No power waypoint proves it; it needs $K=n^{o(1)}$.

Standing

Here $n_2(p)$ is the least quadratic nonresidue modulo $p$. The lower bound $\log k(n)\ge(1/\sqrt2-o(1))\sqrt{\log n\log\log n}$ and its sources are on the problem page. Erdős and Sárközy's theorem is P. Erdős and A. Sárközy, On the prime factors of $\binom nk$ and of consecutive integers, Utilitas Math. 16 (1979), 197--215 (public scan in the Rényi Institute's Erdős archive); it proves the bound $n_k>k^{2+\varepsilon}$ that Erdős wrote in 1976 he could not show, in the strong form $n_k>k^{5/2}/16$. The literature comparison matches it and the other every-interval theorems at their exact parameters.

The argument

Held link 1: the near-square adjustment. If $c,q\ge1$ and $m\ge0$ are integers with $0<cq^2-m\le E$, then subtracting the largest $ct^2$ below $cq^2-m$ gives $N=c(q-t)(q+t)$ with $0<N-m\le2\sqrt{cE}+c$ and $t<q$ (Lemma 1 in the proofs of L250 and L251). So a $K$-smooth member of $(m,m+K]$ exists as soon as some pair $(c,q)$ with $c\le K$ and $q\lesssim K$ has $cq^2$ in a window $(m,m+E]$ with $E\asymp K^2/c$. The rest of the argument finds such a pair.

Held link 2: one multiplier (exponent $2/5$). With $c\asymp m/K^2$ chosen from $m$, the next square $q^2$ above $m/c\le K^2/16$ lies within $2q-1\le K/2+1$ of $m/c$, so $E\approx cK/2$ and the adjustment lands within about $c\sqrt{2K}$ of $m$. That is at most $K$ when $c\lesssim\sqrt K$, that is, $m\lesssim K^{5/2}$. L250 proves this with explicit constants for every K >= 64; it is Erdős and Sárközy's construction.

Held link 3: an averaged multiplier (exponent $11/28$). Let $c$ range over $(C,2C]$ with $C=\lceil16m/K^2\rceil$ and put $Y=\sqrt{m/C}\asymp K$. The pairs $(c,q)$ with $m<cq^2\le m+E$ number $\sum_c(\sqrt{(m+E)/c}-\sqrt{m/c})-S_C(m+E)+S_C(m)$, where

$$S_C(X)=\sum_{C<c\le2C}\psi\Bigl(\sqrt{X/c}\Bigr),\qquad \psi(u)=u-\lfloor u\rfloor-\tfrac12 .$$

The main term is at least $E/((2+\sqrt2)Y)\asymp K/C$. Vaaler's inequality turns $S_C$ into exponential sums $\sum_ce(r\sqrt{X/c})$, whose phase has third derivative of size $rY/C^3$, and the explicit third-derivative test of Granville and Ramaré gives

$$|S_C(X)|\le60\,C^{4/7}Y_X^{1/7}+40\,C^{3/4}(\log C)^{3/2} \qquad(C^{3/2}\le Y_X\le C^3/120),$$

with $Y_X=\sqrt{X/C}$ (Lemma 3 of the L251 proof). This is below the main term while $C\le K^{6/11}/1024$, so pairs exist for all $m$ up to $2^{-15}K^{28/11}$. L251 proves the resulting bound $k(n)<\max(64,(2^{15}n)^{11/28})$ completely; its analytic inputs are the held statements of Lemmas 7.1 and 8.6 of Granville and Ramaré (1996), consumed as stated, except that the proof corrects the placement of one misprinted sign in Lemma 7.1 and derives the corrected inequality.

Held link 4: Bourgain's exponent pair. Theorem 6 of Bourgain's Decoupling, exponential sums and the Riemann zeta function, in arXiv:1408.5794v2, gives $(\kappa,\lambda)=(13/84+\eta,55/84+\eta)$ for every $\eta>0$. The phase $-r\sqrt X\,c^{-1/2}$ has first derivative exactly $(r\sqrt X/2)c^{-3/2}$, so it lies in the monomial class specified in Section 5, including the range $C>\sqrt{rY_X}$. Its sum is bounded by $\ll_\eta C^{1/2}(rY_X)^{13/84+\eta}$. Vaaler then gives $|S_C(X)|\ll_\eta C^{\lambda/(1+\kappa)}Y_X^{\kappa/(1+\kappa)}$. For every fixed $\delta>0$ sufficiently small this is $o(K/C)$ uniformly when $\sqrt K/8<C\le32K^{21/38-\delta}$ and $Y_X\asymp K$. L253 writes out the source interface, frequency truncation and uniform endpoint construction, giving $k(n)\ll_\varepsilon n^{38/97+\varepsilon}$.

The composition is complete at tier 0. L250 covers the small endpoints; L251 proves an explicit bound for every $K\ge64$; L253 proves its stronger asymptotic bound for every sufficiently large $K$, with the threshold depending only on $\delta$. The claimed bounds use no unproved mathematical premise. The cited analytic theorems are external source inputs, and none of the project deductions has independent acceptance.

The exponent as a function of the multiplier-sum bound

The proof of L251 uses the third-derivative test only through the bound on $S_C$. Suppose instead that $|S_C(X)|\ll C^aY_X^b$ with $a\ge0$ and $0\le b<1$, uniformly for $C\le K$ and $Y_X\asymp K$. Then the same parameter choice works while $K/C\gg C^aK^b$, that is, for $C\ll K^{(1-b)/(1+a)}$, and gives

$$\theta=\frac{1+a}{3+2a-b},\qquad n_k\gg k^{2+(1-b)/(1+a)} .$$

This formula is an author derivation, not written out with constants; each use must recheck the side conditions (the ranges of $C$ and $Y_X$ and the secondary terms). An exponent pair $(\kappa,\lambda)$ used with Vaaler's inequality, in the regime $rY_X/C\ge1$ that holds here, gives $(a,b)=(\lambda/(1+\kappa),\kappa/(1+\kappa))$ and hence

$$\theta=\frac{1+\kappa+\lambda}{3+2\kappa+2\lambda} =\frac12-\frac{1}{2(3+2\kappa+2\lambda)} ,$$

so only $\kappa+\lambda$ matters, and smaller is better.

The second row explains why one-variable second-derivative attempts stall: at the critical size the second-derivative test is no better than the trivial bound. The fourth row is proved with its $\varepsilon$ slack in L253; taking the infimum does not remove that slack from its conclusion. Bounds for sawtooth sums that use the average over the frequency $r$ as well may do better than a separate bound at each frequency. A theorem of that kind still needs its exact hypotheses matched to this range.

Ceilings and obstructions

Scaling a fixed input (proved, elementary). Suppose every interval $(X,X+AX^\beta]$ contains an $AX^\alpha$-smooth integer, and one scales by a $K$-smooth multiplier $c$, using the input at $X=m/c$. Containing the whole supplied interval $c\cdot(X,X+AX^\beta]$ in the block requires $AcX^\beta\le K$, and $cX^\beta=c^{1-\beta}m^\beta\ge m^\beta$ for $c\ge1$ and $\beta\le1$. So the diagonal exponent obtained is at least $\beta$ whatever $\alpha$ is. More precisely, take $c\asymp m^{1-s}$, $0<s\le1$, using a power of $2$ so that $c$ is $K$-smooth for $K\ge2$. The two exponent requirements are $\theta\ge s\alpha$ and $\theta\ge1-s+s\beta$. Their maximum is minimized at $s=1/(1+\alpha-\beta)$ when $\beta\le\alpha$, giving $\alpha/(1+\alpha-\beta)$; when $\beta>\alpha$, its minimum is $\beta$ at $s=1$. All constant-factor changes are absorbed in the interval and smoothness constants. No input with interval length $x^{1/2}$ or longer can give $\theta<1/2$ this way. This covers the square-root results matched on the literature page; Harman's 1999 Theorem 2 requires interval exponent above $3/7$ even under the exponent-pair conjecture (its original known-pair discussion reaches between $0.45$ and $0.46$); Friedlander and Lagarias's $f(1/2)\le1/4$ gives $\frac{1/2}{1+1/2-1/4}=2/5$, the Erdős--Sárközy exponent again, and their $f(1/3)\le5/12$ gives $5/12$.

A window of products with one multiplier (author derivation; exponent pair conjecture still gives only $2/5$). A different count fixes $c\asymp m/K^2$ and looks for $ab$ with $a,b\le K$ in the window $(m/c,(m+K)/c]$ of length $K/c\asymp K^3/m$ near $T=m/c\asymp K^2$. The error term is a sawtooth sum $\sum_{a\asymp K}\psi(T/a)$ of divisor-problem type. A pointwise bound $T^\delta$ for it gives $\theta=1/(3-2\delta)$: the second-derivative test ($\delta=1/3$) gives $3/7$, and even square-root cancellation over the $K\asymp T^{1/2}$ terms ($\delta=1/4$) gives only $2/5$. This route therefore stays at or above the fixed-multiplier exponent. A related one-variable count with the reciprocal phase $m/a$ and the second-derivative test is retired at its stated scope: its requirements, as stated for that route (their derivation is not reproduced here), are $H>U^2/K$, $\sqrt{Hm}<K$ and $U^2\ge m/K$, for window length $H$ and factor size $U$. The first and third give $H>m/K^2$, so $\sqrt{Hm}>m/K$, and the second then forces $m<K^2$: no gain over the square-root bound. Its sum is a different one from the multiplier sum $S_C$.

Limits of this construction and estimate (exact scope). A product of three positive integers at most $K$ is at most $K^3$. For $m\ge K^3$ it therefore cannot lie in $(m,m+K]$, whatever method selects its factors. Thus the three-factor conclusion of L250, L251 and L253 cannot provide a diagonal power exponent below $1/3$. This is an exact obstruction for that output specification; a $K$-smooth integer can have more than three such factors.

There is a separate obstruction to improving the displayed exponent-pair calculation. Every exponent pair in its usual range has $\kappa\ge0$ and $\lambda\ge1/2$, so $s=\kappa+\lambda\ge1/2$. The formula $\theta=(1+s)/(3+2s)$ is increasing in $s$, and hence gives $\theta\ge3/8$. Even the exponent-pair conjecture reaches only the infimum $3/8$ through this calculation. Likewise, simply inserting a discrepancy estimate of size $C^{1/2}$ gives the sufficient condition $K/C\gg C^{1/2}$, or $C\ll K^{2/3}$. Neither observation rules out a smaller discrepancy, direct control of $S_C(m+E)-S_C(m)$, or a different counting method. They are limits of these estimates, not lower bounds on the true discrepancy.

More generally, a witness formed from $r$ positive factors at most $K$ cannot exceed $K^r$. A witness above $m$ requires $r>\log m/\log K$ when $K>1$. At the conjectured scale $\log K=(\log m)^{1/2+o(1)}$, the number of factors must therefore grow at least as $(\log m)^{1/2-o(1)}$. No fixed-factor construction can establish the displayed question.

The nonresidue barrier (proved). L252 shows k(p) >= n_2(p) - 1 for primes p = 3 (mod 4), by the argument Granville gives after his Conjecture 4.4.2. So a proof of $k(n)\ll n^\theta$ with $\theta<1/(4\sqrt e)=0.1516\ldots$ would improve the Burgess--Vinogradov exponent for the least quadratic nonresidue of these primes, and the displayed question would give $\log n_2(p)\le(\log p)^{1/2+o(1)}$ for them. This measures difficulty and refutes nothing; the generalized Riemann hypothesis gives $n_2(p)=O((\log p)^2)$.

Gaps and next tests

1. A stronger uniform pair count. The next discriminating test is a bound for the one-sided count of pairs $(c,q)$ with $C<c\le2C$ and $m<cq^2\le m+E$, where $Y=\sqrt{m/C}\asymp K$, $E\asymp K^2/C$ and $C=K^\nu$ for some $\nu>21/38$. The main term is $\asymp K/C$; an error $o(K/C)$ would improve L253. A bound for the difference $S_C(m+E)-S_C(m)$ may exploit the short window more efficiently than bounding the two sums separately. At $C\asymp K^{2/3}$, an adequate estimate would reach the infimum $3/8$. Ordinary exponent-pair substitution cannot pass that infimum.
2. Growing factor count. Passing below $1/3$ requires changing the three-factor output. The Friedlander--Lagarias construction for a fixed $r$ has interval exponent $1-2/r+1/(r2^{r-1})$; it is already above $1/2$ for $r\ge4$, so merely increasing $r$ in that construction does not supply the needed short intervals. Moreover its fixed-$r$ theorem gives no uniform control when $r$ grows. A useful new input must produce many bounded factors while retaining a gap at most $K$; no such unbounded-factor construction is supplied here.
3. The displayed question. It requires $K=n^{o(1)}$, hence smooth members with unboundedly many factors, in every block. By L252 any unconditional exponent below $1/(4\sqrt e)$ already passes through a new bound for quadratic nonresidues; the displayed question is far beyond that. No route toward it is proposed here.

The conclusions currently supported are the explicit bound $k(n)<\max(64,(2^{15}n)^{11/28})$ and the stronger asymptotic bound $k(n)\ll_\varepsilon n^{38/97+\varepsilon}$ for every $\varepsilon>0$, both author-recorded at tier 0. The displayed question remains open; no bound with a fixed exponent at most $38/97$ is proved here.