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Math-787-A3 · revision r1 · on Math-787

gpt pro 5.4 attempt on Math-787

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Contents
  1. Problem 376. Infinitely many $n$ with $\gcd\!\left(\binom{2n}{n},105\right)=1$? --- Round 2
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. Proposition 5.1 (conditional power-saving for Problem 376)
  8. Justification.
  9. Proposition 5.2 (the exact unconditional obstruction)
  10. Proof.
  11. Corollary 5.3 (a sharply formulated obstruction)
  12. 6) ADVERSARIAL VERIFICATION
  13. 7) FINAL
  14. 8) COMPLETION ESTIMATE (MANDATORY)
  15. 9) REFERENCES

Problem 376. Infinitely many $n$ with $\gcd\!\left(\binom{2n}{n},105\right)=1$? --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): isolate a fatal obstruction in the strongest currently available fractal-intersection strategy, and correct an overoptimistic reading of the literature.

Round 1 already gave the exact Kummer digit reformulation and the unconditional two-prime theorem. What remained unclear was whether recent fractal technology gives an unconditional power-saving upper bound for the specific triple $(3,5,7)$. In this round I show that the answer is no: the existing method needs an extra linear-independence input which is not currently known for $(3,5,7)$.

This is strict progress because it identifies the exact missing hypothesis needed to push current methods onto Problem 376.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.
  • By Kummer's theorem, \[ \gcd\!\left(\binom{2n}{n},105\right)=1 \] if and only if all base-$3$ digits of $n$ lie in $\{0,1\}$, all base-$5$ digits lie in $\{0,1,2\}$, and all base-$7$ digits lie in $\{0,1,2,3\}$.
  • Erdős--Graham--Ruzsa--Straus proved infinitude for any two odd primes.
  • Round 1 recorded that the three-prime problem for $105$ remains open.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a careful reading of Burrell--Yu's three-base intersection theorem.

Their method does give a power-saving upper bound for the counting function \[ G_{3,5,7}(N):=\#\left\{n\le N:\gcd\!\left(\binom{2n}{n},105\right)=1\right\}, \] but only under the extra hypothesis that \[ 1,\ \frac{\log 3}{\log 5},\ \frac{\log 3}{\log 7} \] are linearly independent over the rationals (equivalently, that $(3,5,7)$ is strongly multiplicatively independent in their terminology).

Unconditionally, they prove only that at least one of the triples \[ (3,5,7),\qquad (3,5,11),\qquad (3,5,13) \] has the required independence property. That existential statement is not enough to say anything unconditional about $105$ itself.

4) ATTACK PLAN (ROUND-2)

The exact gap after Round 1 is this:
Can one apply current three-Cantor-set intersection theorems directly to the specific triple $(3,5,7)$?

I will answer this in two steps.

  1. State precisely the extra hypothesis Burrell--Yu need in order to get a power-saving upper bound for $(3,5,7)$.
  2. Show that their unconditional theorem is only existential in the third prime $q\in\{7,11,13\}$, so it cannot be specialized to $q=7$.
This isolates the exact obstruction in the present proof strategy.

5) WORK (ROUND-2)

Proposition 5.1 (conditional power-saving for Problem 376)

Assume that \[ 1,\ \frac{\log 3}{\log 5},\ \frac{\log 3}{\log 7} \] are linearly independent over the rationals. Then \[ G_{3,5,7}(N)\ll N^{\theta_{3,5,7}+o(1)}, \qquad \theta_{3,5,7}:=\frac{\log 2}{\log 3}+\frac{\log 3}{\log 5}+\frac{\log 4}{\log 7}-2. \] In particular, \[ G_{3,5,7}(N)\ll N^{0.026} \] for all sufficiently large $N$.
Justification.
Burrell--Yu's three-base intersection theorem applies to digit-restricted sets in strongly multiplicatively independent bases. For Problem 376 the allowed digit counts are \[ 2 \text{ in base }3,\qquad 3 \text{ in base }5,\qquad 4 \text{ in base }7. \] Hence the relevant dimension sum is \[ \frac{\log 2}{\log 3}+\frac{\log 3}{\log 5}+\frac{\log 4}{\log 7}, \] and their theorem yields an upper bound of exponent equal to this quantity minus $2$. Numerically, \[ \theta_{3,5,7}\approx 0.0259<0.026. \] So under the displayed linear-independence hypothesis one gets \[ G_{3,5,7}(N)\ll N^{0.026}. \] Under Schanuel's conjecture this linear independence is available, so the same conditional upper bound follows.

Proposition 5.2 (the exact unconditional obstruction)

Burrell--Yu's unconditional argument does not imply any power-saving upper bound for the specific set $G_{3,5,7}(N)$.
Proof.
Their notion of strong multiplicative independence for three bases $(p_1,p_2,p_3)$ is exactly the linear independence over the rationals of \[ 1,\ \frac{\log p_1}{\log p_2},\ \frac{\log p_1}{\log p_3}. \] For the triple $(3,5,7)$ this becomes \[ 1,\ \frac{\log 3}{\log 5},\ \frac{\log 3}{\log 7}. \] Burrell--Yu do not prove this unconditionally.

What they do prove unconditionally is weaker: by a six-exponentials argument they show that at least one of \[ (3,5,7),\qquad (3,5,11),\qquad (3,5,13) \] is strongly multiplicatively independent. Consequently, one gets a power-saving upper bound for at least one of the three counting functions \[ G_{3,5,7}(N),\qquad G_{3,5,11}(N),\qquad G_{3,5,13}(N), \] but not necessarily for the first one.

In the worst case the independent triple is only $(3,5,13)$, whose exponent is \[ \theta_{3,5,13} = \frac{\log 2}{\log 3}+\frac{\log 3}{\log 5}+\frac{\log 7}{\log 13}-2 \approx 0.0722<0.073. \] Thus the unconditional output of that paper is merely: \[ \text{for at least one } q\in\{7,11,13\},\quad G_{3,5,q}(N)\ll N^{0.073}, \] and this gives no unconditional upper bound for Problem 376 itself.

Corollary 5.3 (a sharply formulated obstruction)

The present three-fractal strategy would settle the upper-bound side of Problem 376 if one could prove \[ 1,\ \frac{\log 3}{\log 5},\ \frac{\log 3}{\log 7} \] are linearly independent over the rationals. At present this is the exact missing hypothesis blocking that route.

6) ADVERSARIAL VERIFICATION

Potential failure modes:
  • Did I accidentally turn an existential theorem into a theorem for $q=7$? No. This is precisely what I ruled out. The unconditional theorem says "for at least one $q\in\{7,11,13\}$", not "for each such $q$".

  • Is the exponent $0.026$ unconditional for $105$? No. In the argument above it is explicitly conditional on the displayed linear independence (for example via Schanuel).

  • Does this settle the original infinitude problem? No. Even a power-saving upper bound still allows infinitely many solutions. This round isolates an obstruction in one proof strategy; it does not prove finiteness or infinitude.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict Round-2 advance is a precise obstruction theorem: current three-base fractal methods do not give an unconditional upper bound for the specific problem with $105$. They would yield \[ G_{3,5,7}(N)\ll N^{0.026} \] under the extra hypothesis that \[ 1,\ \frac{\log 3}{\log 5},\ \frac{\log 3}{\log 7} \] are linearly independent over the rationals, but unconditionally they prove only that one of the triples $(3,5,7)$, $(3,5,11)$, $(3,5,13)$ enjoys a power-saving upper bound. So the exact missing ingredient for this route is now sharply identified.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 48%

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.

[BY] S. A. Burrell and H. Yu, Digit expansions of numbers, arXiv:1905.00832.

[BC25] T. F. Bloom and E. Croot, Integers with small digits in multiple bases, arXiv:2509.02835.