Erdős Problem 684 at Density One: Small-prime Parts of Binomial Coefficients and Gaussian Fluctuations

Abstract: For $0\leq k\leq n$, let $u(n,k)$ be the largest divisor of $\binom nk$ whose prime factors are at most $k$. Erdős Problem #684 concerns the special threshold $u(n,k)&gt;n<sup>2$ and asks how early this small-prime part can be forced to become large. We prove the density-one analogue for every fixed power threshold. If $f_c(n)$ is the least $k$ for which $u(n,k)&gt;n<sup>c$, then, for each fixed $c&gt;0$, [ f_c(n)=\left(\frac{c}{1-γ}+o(1)\right)\log n ] for almost all positive integers $n$. In particular, [ f_2(n)=\left(\frac{2}{1-γ}+o(1)\right)\log n =(4.730544237\ldots+o(1))\log n ] for the Erdős #684 threshold. This is a normal-order theorem, not a pointwise resolution of the corresponding worst-case problem. The constant $1-γ$ is arithmetic. Kummer's theorem rewrites $\log u(n,k)$ as a sum of carry indicators, and complete-residue averaging gives [ m(k)=k\sum_{p\leq k}\frac{\log p}{p-1}-\log k!=(1-γ)k+o(k). ] The cancellation in this formula moves the typical crossing from the naive scale $c\log n$ to $c(1-γ)^{-1}\log n$. We prove the required concentration uniformly for every $k\leq A\log X$ on one dyadic interval, after discarding a zero-density exceptional set caused by large powers of small primes dividing one of the nearby integers $n,n-1,\ldots$. We also prove Gaussian fluctuations in the logarithmic range. If $k=k(X)\to\infty$, $k\leq A\log X$, and $n$ is uniform in $[X,2X)\cap\mathbb Z$, then [ \frac{\log u(n,k)-m(k)}{\sqrt{V(k)}}\Rightarrow \mathcal N(0,1), \qquad V(k)\sim (2-\log(2π))k\log k. ] Higher prime powers are needed for the mean, but after centering their aggregate is $L<sup>2$-negligible on the Gaussian scale; the variance comes only from the prime levels.

• The paper proves that for every fixed c > 0, the first k with the small-prime part of \(\binom{n}{k}\) exceeding \(n^c\) satisfies \(f_c(n) = (c/(1-\gamma)+o(1))\log n\) for almost all n, giving \(f_2(n) \sim 4.730544237\log n\).
• The paper uses Kummer’s carry-counting theorem, truncation of large prime powers, and uniform fourth-moment bounds to establish concentration across all relevant \(k \leq A\log X\) without assuming monotonicity.
• The paper proves a central limit theorem after centering by the exact arithmetic mean, with variance \(V(k) \sim (2-\log(2\pi))k\log k\), while emphasizing that the result does not resolve the worst-case pointwise form of Erdős Problem 684.

Overview and main results

For $0 \le k \le n$, let $u(n,k)$ denote the largest divisor of $\binom{n}{k}$ whose prime factors are all at most $k$, and let $f_c(n) := \min\{k : u(n,k) > n^c\}$. The case $c=2$ is the density-one counterpart of Erdős Problem #684, which asks for bounds on the least $k$ forcing the small-prime part of $\binom{n}{k}$ to exceed $n^2$ (2606.08216). The paper proves two theorems. First, a normal-order theorem: for every fixed $c>0$,

$u(n,k)$0

for almost all positive integers $u(n,k)$1, where $u(n,k)$2 is the Euler–Mascheroni constant. For the Erdős threshold this gives $u(n,k)$3 on a set of natural density one. Second, a central limit theorem: if $u(n,k)$4 with $u(n,k)$5 and $u(n,k)$6 is uniform in $u(n,k)$7, then

$u(n,k)$8

The author is explicit that the first result is a normal-order theorem, not a pointwise resolution of the worst-case problem; recent work cited there gives polylogarithmic worst-case upper bounds and logarithmic lower-bound examples (Alexeev et al., 31 Mar 2026).

The arithmetic origin of the constant

Kummer's theorem is recast in residue form: $u(n,k)$9 equals the number of levels $\binom{n}{k}$0 with $\binom{n}{k}$1, so $\binom{n}{k}$2 is a sum of carry indicators weighted by $\binom{n}{k}$3. Averaging each indicator over complete residue systems modulo $\binom{n}{k}$4 yields the deterministic mean

$\binom{n}{k}$5

A naive first-order heuristic would predict crossing at $\binom{n}{k}$6, i.e., coefficient $\binom{n}{k}$7. The exact mean instead satisfies $\binom{n}{k}$8, because the two large terms $\binom{n}{k}$9 (from the Mertens–von Mangoldt sum) and $k$0 cancel to leading order, leaving a linear main term. This cancellation shifts the typical crossing to $k$1 and is the sole source of the constant $k$2 in the Erdős case.

Uniform concentration via fourth moments

The normal-order theorem requires concentration of $k$3 around $k$4 simultaneously for every integer $k$5 on one dyadic interval. Two mechanisms accomplish this:

• Truncation of large prime powers. Levels $k$6 contribute only when $k$7 divides one of $k$8; a union bound over primes $k$9 and logarithmically many shifts shows these events occur for $f_c(n) := \min\{k : u(n,k) > n^c\}$0 integers $f_c(n) := \min\{k : u(n,k) > n^c\}$1. Any fixed truncation exponent below $f_c(n) := \min\{k : u(n,k) > n^c\}$2 would suffice, since four moduli appear in the moment expansion and their lcm must be $f_c(n) := \min\{k : u(n,k) > n^c\}$3.
• Fourth-moment estimate. After truncation, lcm's of quadruples of moduli are at most $f_c(n) := \min\{k : u(n,k) > n^c\}$4, so interval averages may be replaced by complete-residue averages with summable error. Chinese-remainder factorization reduces the complete-residue fourth moment to independent centered "prime-tower" variables; nestedness of the high-level carry events gives bounded second and fourth moments per prime, yielding $f_c(n) := \min\{k : u(n,k) > n^c\}$5 uniformly in $f_c(n) := \min\{k : u(n,k) > n^c\}$6.

Markov's inequality plus a union over $f_c(n) := \min\{k : u(n,k) > n^c\}$7 values of $f_c(n) := \min\{k : u(n,k) > n^c\}$8 produces an exceptional set of size $f_c(n) := \min\{k : u(n,k) > n^c\}$9 — this is precisely why a fourth rather than second moment is needed. Combining the lower exclusion (no $c=2$0 below $c=2$1 works) with a single exhibited value $c=2$2 above it yields the theorem. Notably, no monotonicity of $c=2$3 in $c=2$4 is used or available, since both $c=2$5 and the permitted prime set vary with $c=2$6.

Gaussian fluctuations

The fluctuation analysis separates scales. Higher prime-power levels ($c=2$7) contribute deterministically to $c=2$8 and cannot be discarded before centering; after centering by the full mean, however, their aggregate $c=2$9 size is $k$0, proved by splitting at $k$1 and using a short-residue-window estimate that avoids losing a factor $k$2 when the admissible classes form an initial segment. On the Gaussian scale $k$3, only the prime levels survive.

The variance constant comes from

$k$4

evaluated via Stieltjes integration against $k$5 and the elementary integral $k$6. The prime-level sum satisfies a Lindeberg triangular-array CLT against independent Bernoulli variables $k$7 with success probability $k$8; fixed moments transfer from the model to the dyadic interval via periodic averaging, with errors $k$9, and Rosenthal's inequality supplies uniform integrability. Slutsky's theorem then upgrades to a fully standardized CLT around the interval mean and variance.

Limitations and open questions

The paper concedes its scope plainly. The density-one formulation discards exactly the integers $\binom{n}{k}$0 for which a large power of a small prime divides one of $\binom{n}{k}$1, $\binom{n}{k}$2; such congruence obstructions have zero natural density but can dominate individual worst-case inputs, so the result must not be quoted as resolving Erdős Problem #684 pointwise. The exceptional-set bound $\binom{n}{k}$3 is quantitative but weak, and the truncation exponent $\binom{n}{k}$4 is tied to the fourth-moment argument. The CLT is established only in the logarithmic range $\binom{n}{k}$5; behavior at larger $\binom{n}{k}$6, and any distributional statement for $\binom{n}{k}$7 itself beyond its normal order, remain open.

Conclusion

The paper determines the normal order of the first small-prime crossing of binomial coefficients for every fixed power threshold, with leading constant $\binom{n}{k}$8 arising from an exact cancellation between $\binom{n}{k}$9 and $n^2$0, and establishes Gaussian fluctuations with variance constant $n^2$1 driven entirely by prime-level carries. It thereby supplies the typical answer to Erdős Problem #684 while explicitly leaving the pointwise question open.