A350402 Numbers k such that binomial(k, 2) divides binomial(2^k-2, 2).

2, 3, 7, 11, 19, 31, 43, 127, 163, 211, 271, 311, 331, 379, 487, 571, 631, 811, 883, 991, 1459, 1471, 1747, 2311, 2531, 2647, 2791, 2971, 3079, 3631, 3943, 4091, 5171, 5419, 6571, 7591, 8863, 8911, 9199, 9791, 9931, 10891, 11827, 11971, 13591, 14407, 15391, 16759, 17011, 18523, 19531, 21871, 22111, 23431, 24967

Offset: 1

Author

Juri-Stepan Gerasimov, Dec 29 2021

Keywords

• nonn

Comments

Conjecture: aside from the first term, this is a subsequence of A094179 (numbers congruent to 3 mod 4 which are divisible only by primes congruent to 3 mod 4).

The conjecture is false: a(2295) = 508606771 = 19531 * 26041 is not in A094179, nor even A004614. - Charles R Greathouse IV, Jan 22 2022

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Supersequence of A069051.

Cf. A069051 (binomial(k,2) divides binomial(2^k-1, 2)?), A094179, A350176.

Programs

Magma

[n: n in [2..25000] |  IsZero(Binomial(2^n-2, 2) mod Binomial(n, 2))];

Mathematica

Select[Range[2, 25000], Divisible[Binomial[2^# - 2, 2], Binomial[#,2]] &] (* Amiram Eldar, Dec 29 2021 *)

PARI

isok(n) = (n>1) && ((binomial(2^n-2, 2) % binomial(n, 2)) == 0); \\ Michel Marcus, Jan 04 2022

PARI

is(n)=my(m=n^2-n,t=Mod(2,m)^n-2); t*(t-1)==0 \\ Charles R Greathouse IV, Jan 20 2022