gpt pro 5.4 attempt on Math-788

Problem 377. Is $\displaystyle f(n)=\sum_{p\le n,\ p\nmid \binom{2n}{n}}\frac1p$ bounded? --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): strict progress by proving a new theorem that localizes any possible unboundedness to arbitrarily small primes.

Round 1 recorded the average-order results of Erdős--Graham--Ruzsa--Straus and left open whether $f(n)$ is bounded. In this round I prove a new uniform upper bound for the contribution of all primes in the range $n^{1/(m+1)}<p\le n$, for every fixed $m$. In particular, primes $\sqrt{n}<p\le n$ can contribute at most $\log 2+o(1)$. So if $f(n)$ is unbounded, the entire growth must come from smaller and smaller primes.

2) Round-1 FOUNDATION USED

I use only the Round-1 Kummer reformulation: for a prime $p$, \[ p\nmid \binom{2n}{n} \] if and only if every base-$p$ digit of $n$ is at most $(p-1)/2$. No other Round-1 proof is reproved.

3) NEW INSIGHT / TOOL (ROUND-2)

The new observation is a size-digit lemma:

if \[ n^{1/(k+1)}<p\le n^{1/k} \] and $p\nmid \binom{2n}{n}$, then the base-$p$ expansion of $n$ has exactly $k+1$ digits, each at most $(p-1)/2$, hence \[ n\le \frac{p^{k+1}-1}{2}. \] This forces \[ p\ge (2n+1)^{1/(k+1)}. \] So in each dyadic-like prime window $(n^{1/(k+1)},\,n^{1/k}]$, the "good" primes can only occupy the much shorter terminal interval \[ [(2n+1)^{1/(k+1)},\,n^{1/k}]. \] Summing reciprocal primes over these terminal intervals gives a new explicit bound.

4) ATTACK PLAN (ROUND-2)

The exact gap after Round 1 was that there was no structural theorem explaining where a possible blow-up of $f(n)$ could come from.

I now prove: \[ \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid \binom{2n}{n}}}\frac1p \le \log(m+1)+o(1) \qquad (m \text{ fixed}). \] This will imply that any unboundedness of $f(n)$ must be caused by primes \[ p\le n^{1/(m+1)} \] for every fixed $m$, since the complementary range $n^{1/(m+1)}<p\le n$ has uniformly bounded total weight. That is a genuine narrowing of the problem.

5) WORK (ROUND-2)

Write \[ C_n:=\binom{2n}{n}. \]

Lemma 5.1 (window lemma)

Fix an integer $k\ge 1$. For each $n$, define \[ I_k(n):=\sum_{\substack{n^{1/(k+1)}<p\le n^{1/k}\\ p\nmid C_n}}\frac1p. \] Then, as $n\to\infty$ with $k$ fixed, \[ I_k(n)\le \log\!\left(\frac{k+1}{k}\right)+o(1). \]

Proof.

Let \[ n^{1/(k+1)}<p\le n^{1/k} \] and suppose $p\nmid C_n$. By Kummer's theorem, every base-$p$ digit of $n$ is at most $(p-1)/2$.

Because $p\le n^{1/k}$, we have $p^k\le n$, so the base-$p$ expansion of $n$ has at least $k+1$ digits. Because $p>n^{1/(k+1)}$, we have $p^{k+1}>n$, so the expansion has at most $k+1$ digits. Hence $n$ has exactly $k+1$ base-$p$ digits: \[ n=a_0+a_1p+\cdots + a_kp^k, \qquad 0\le a_j\le \frac{p-1}{2}. \] Therefore \[ n\le \frac{p-1}{2}\sum_{j=0}^k p^j = \frac{p-1}{2}\cdot \frac{p^{k+1}-1}{p-1} = \frac{p^{k+1}-1}{2}. \] Rearranging gives \[ p\ge (2n+1)^{1/(k+1)}. \] So every prime counted by $I_k(n)$ lies in the interval \[ (2n+1)^{1/(k+1)}\le p\le n^{1/k}. \] Hence \[ I_k(n)\le \sum_{(2n+1)^{1/(k+1)}\le p\le n^{1/k}}\frac1p. \] Now use the classical Mertens asymptotic \[ \sum_{p\le x}\frac1p=\log\log x + B_1 + o(1). \] Taking the difference at the two endpoints yields \[ I_k(n)\le \log\log(n^{1/k})-\log\log((2n+1)^{1/(k+1)})+o(1). \] Since \[ \log\log(n^{1/k})=\log\log n - \log k \] and \[ \log\log((2n+1)^{1/(k+1)}) = \log\log(2n+1)-\log(k+1), \] we obtain \[ I_k(n)\le \log\!\left(\frac{k+1}{k}\right) + \log\!\left(\frac{\log n}{\log(2n+1)}\right) +o(1) = \log\!\left(\frac{k+1}{k}\right)+o(1). \] This proves the lemma. $\square$

Theorem 5.2 (large-prime contribution is uniformly bounded)

Fix $m\ge 1$. Then \[ \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid C_n}}\frac1p \le \log(m+1)+o(1) \qquad (n\to\infty). \]

Proof.

For primes $p\le n$, the condition $n^{1/(m+1)}<p\le n$ partitions uniquely into the windows \[ n^{1/(k+1)}<p\le n^{1/k}, \qquad k=1,2,\dots,m. \] Therefore \[ \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid C_n}}\frac1p = \sum_{k=1}^m I_k(n). \] Applying Lemma 5.1 to each fixed $k$ and summing, \[ \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid C_n}}\frac1p \le \sum_{k=1}^m \left(\log\!\left(\frac{k+1}{k}\right)+o(1)\right). \] Because $m$ is fixed, \[ \sum_{k=1}^m o(1)=o(1), \] and the logarithms telescope: \[ \sum_{k=1}^m \log\!\left(\frac{k+1}{k}\right)=\log(m+1). \] Hence \[ \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid C_n}}\frac1p \le \log(m+1)+o(1). \] This proves the theorem. $\square$

Corollary 5.3 (all possible unboundedness comes from very small primes)

Assume that $f(n)$ is unbounded. Then for every fixed $m\ge 1$ there exists a sequence $n_j\to\infty$ such that \[ \sum_{\substack{p\le n_j^{1/(m+1)}\\ p\nmid C_{n_j}}}\frac1p \to \infty. \]

Proof.

By definition, \[ f(n) = \sum_{\substack{p\le n^{1/(m+1)}\\ p\nmid C_n}}\frac1p + \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid C_n}}\frac1p. \] The second sum is at most $\log(m+1)+o(1)$ by Theorem 5.2. So if $f(n)$ is unbounded, the first sum must be unbounded along some sequence. $\square$

Special case

Taking $m=1$ gives \[ \sum_{\substack{\sqrt{n}<p\le n\\ p\nmid C_n}}\frac1p\le \log 2+o(1). \] Thus primes in the range $\sqrt{n}<p\le n$ can never be responsible for an unbounded growth of $f(n)$.

6) ADVERSARIAL VERIFICATION

Checks on the new argument:

• Number of digits. For $n^{1/(k+1)}<p\le n^{1/k}$, indeed $p^k\le n<p^{k+1}$, so the base-$p$ expansion has exactly $k+1$ digits. No edge case is missed.

• Could the top digit be smaller than $(p-1)/2$ and ruin the bound? No. Replacing the top digit by the maximal allowed value only enlarges the right-hand side, so the inequality \[ n\le \frac{p^{k+1}-1}{2} \] remains valid.

• Dependence on $m$. The theorem is for each fixed $m$. The $o(1)$ term may depend on $m$, which is harmless.

• Does this prove boundedness of $f(n)$? No. It only proves that the contribution from primes in the range $n^{1/(m+1)}<p\le n$ is uniformly bounded for each fixed $m$. The unresolved part is the cumulative contribution of much smaller primes.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The new theorem is: for every fixed $m\ge 1$, \[ \sum_{\substack{n^{1/(m+1)}<p\le n\\ p\nmid \binom{2n}{n}}}\frac1p \le \log(m+1)+o(1). \] In particular, \[ \sum_{\substack{\sqrt{n}<p\le n\\ p\nmid \binom{2n}{n}}}\frac1p \le \log 2+o(1). \] Therefore any possible unboundedness of $f(n)$ must come from primes \[ p\le n^{1/(m+1)} \] for every fixed $m$. This sharply localizes the remaining difficulty.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 57%

9) REFERENCES

[EGRS75] P. Erdős, R. L. Graham, I. Z. Ruzsa, and E. G. Straus, On the prime factors of $\binom{2n}{n}$, Math. Comp. 29 $1975$, 83--92.

[RS62] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 $1962$, 64--94.