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Math-787-A2 · revision r1 · on Math-787

gpt pro 5.2 attempt on Math-787

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Contents
  1. Problem 376. Infinitely many $n$ with $\binom{2n}{n}$ coprime to $105$?
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE / CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. 1. Digit formulation via Kummer
  7. 2. What is known unconditionally
  8. 3. Partial progress toward $r\ge 3$
  9. VERIFICATION
  10. FINAL
  11. COMPLETION ESTIMATE

Problem 376. Infinitely many $n$ with $\binom{2n}{n}$ coprime to $105$?

FORMAL RESTATEMENT

Determine whether the set \[ \mathcal{S}:=\{n\in\mathbb{N}: \gcd\!\left(\binom{2n}{n},105\right)=1\} \] is infinite.

QUICK LITERATURE / CONTEXT CHECK

This problem is stated as open in Erdős--Graham $1980$ and remains open in the Erdős problems database.

A key equivalence comes from Kummer's theorem: for a prime $p$, $p\nmid \binom{2n}{n}$ iff there are no carries when adding $n+n$ in base $p$, which is equivalent to every base-$p$ digit of $n$ being $<p/2$.

Erdős--Graham--Ruzsa--Straus $1975$ prove that for any two odd primes $p,q$ there are infinitely many $n$ with $\binom{2n}{n}$ coprime to $pq$.

For the specific triple $(3,5,7)$, many examples are known computationally (OEIS A030979 begins $0,1,10,756,757,3160,3186,3187,3250,\dots$), but infinitude is unknown. Bloom--Croot $2025$ give a "weak" result for $\ge 3$ primes when the primes are sufficiently large: they produce infinitely many $n$ for which almost all base-$p$ digits are $<p/2$, implying $\binom{2n}{n}$ is coprime to the product of those primes up to a factor $\le n^\varepsilon$.

Conditional and quantitative information can be found in work of Han Yu $2022$: under Schanuel's conjecture and an additional conjecture on thickness, the set $\mathcal{S}$ is infinite; under Schanuel's conjecture one has a power-saving upper bound $\#(\mathcal{S}\cap[1,N])\ll N^{0.026\dots}$.

ATTACK PLAN

  1. Use Kummer/Lucas to rewrite the gcd condition as a digit restriction in bases $3,5,7$.
  2. Interpret each digit-restricted set as a Cantor-like set; the problem becomes whether three multiplicatively independent Cantor sets have infinite intersection.
  3. Current approaches in the literature include: (i) "carry-poor" constructions (EGRS for $r=2$; Bloom--Croot for $r\ge 3$ but with exceptions), and (ii) fractal-geometry methods (Newhouse gap lemma, thickness) giving conditional results and upper bounds.
  4. A full resolution for $(3,5,7)$ would likely require a new mechanism forcing simultaneous small-digit expansions in three small bases (or proving this cannot happen beyond finitely many $n$).

WORK

1. Digit formulation via Kummer

Let $p$ be a prime and write the base-$p$ expansion \[ n=\sum_{j\ge 0} a_j p^j,\qquad 0\le a_j\le p-1. \] Kummer's theorem states that the $p$-adic valuation $\nu_p\!\left(\binom{2n}{n}\right)$ equals the number of carries when adding $n+n$ in base $p$. In particular, \[ p\nmid \binom{2n}{n} \iff \text{no carries occur in }n+n\text{ (base }p) \iff \forall j,\ 2a_j\le p-1 \iff \forall j,\ a_j\le \frac{p-1}{2}. \] Thus \[ \gcd\!\left(\binom{2n}{n},105\right)=1 \iff \begin{cases} \text{all base-}3\text{ digits of }n\text{ lie in }\{0,1\},\\ \text{all base-}5\text{ digits of }n\text{ lie in }\{0,1,2\},\\ \text{all base-}7\text{ digits of }n\text{ lie in }\{0,1,2,3\}. \end{cases} \] The example $756=(1001000)_3=(11011)_5=(2130)_7$ is one explicit solution.

2. What is known unconditionally

  • (Two primes.) EGRS $1975$ prove that for any two odd primes $p,q$ there are infinitely many $n$ such that $\binom{2n}{n}$ is coprime to $pq$. This corresponds to the existence of infinitely many integers whose base-$p$ digits are $<p/2$ and base-$q$ digits are $<q/2$.
  • (Three primes, small.) For $(3,5,7)$ the problem is open. Direct computation gives at least the following 12 solutions up to $20000$: \[ 0,1,10,756,757,3160,3186,3187,3250,7560,7561,7651. \] These agree with the initial segment of OEIS A030979.

3. Partial progress toward $r\ge 3$

Bloom--Croot $2025$ show that for $r\ge 1$, if $g_1,\dots,g_r$ are distinct coprime integers that are sufficiently large (depending only on $r$), then for every $\varepsilon>0$ there are infinitely many integers $n$ such that, simultaneously for all $i$, all but at most $\varepsilon\log n$ of the base-$g_i$ digits of $n$ are $\le g_i/2$. Their Corollary 1 yields (for $r\ge 3$ and sufficiently large primes $p_i$) a factorisation \[ \binom{2n}{n}=n_1 n_2, \qquad (n_1,p_1\cdots p_r)=1,\quad n_2\le n^{\varepsilon}, \] which is a "weak" version of the desired coprimality.

Han Yu $2022$ gives conditional results in the exact direction of Problem 376: under Schanuel's conjecture and an additional conjecture on thickness of the digit sets, the intersection for bases $(3,5,7)$ is infinite; under Schanuel one also gets a strong power-saving upper bound on the counting function.

VERIFICATION

The equivalence between $p\nmid\binom{2n}{n}$ and the digit condition is standard via Kummer's carry interpretation; the digit condition for $105$ is just the conjunction for $p\in\{3,5,7\}$. The numerical list up to $20000$ was verified by a brute-force digit check.

FINAL

LABEL: UNRESOLVED.
SUB-LABEL: N/A.
The problem for $105=3\cdot 5\cdot 7$ is open. We recorded the exact digit reformulation, the unconditional $r=2$ theorem of Erdős--Graham--Ruzsa--Straus, and partial $r\ge 3$ results (Bloom--Croot; Yu), plus computational evidence.

COMPLETION ESTIMATE

A full solution would require either (i) a new construction producing infinitely many integers with simultaneously small digits in bases $3,5,7$ (no carries in all three bases), or (ii) a proof that the intersection is finite. Existing methods handle $r=2$ strongly, and $r\ge 3$ only in a weakened "almost all digits" sense or under strong transcendence/fractal hypotheses.