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Numbers k such that binomial(k, 2) divides binomial(2^k-2, 2).
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%I #35 Jan 24 2022 18:21:44

%S 2,3,7,11,19,31,43,127,163,211,271,311,331,379,487,571,631,811,883,

%T 991,1459,1471,1747,2311,2531,2647,2791,2971,3079,3631,3943,4091,5171,

%U 5419,6571,7591,8863,8911,9199,9791,9931,10891,11827,11971,13591,14407,15391,16759,17011,18523,19531,21871,22111,23431,24967

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

%C 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).

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

%H Charles R Greathouse IV, <a href="/A350402/b350402.txt">Table of n, a(n) for n = 1..10000</a>

%t Select[Range[2, 25000], Divisible[Binomial[2^# - 2, 2], Binomial[#,2]] &] (* _Amiram Eldar_, Dec 29 2021 *)

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

%o (PARI) isok(n) = (n>1) && ((binomial(2^n-2, 2) % binomial(n, 2)) == 0); \\ _Michel Marcus_, Jan 04 2022

%o (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

%Y Supersequence of A069051.

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

%K nonn

%O 1,1

%A _Juri-Stepan Gerasimov_, Dec 29 2021