A366029 - OEIS

A366029

Numbers k > 1 such that (k-1)^(2^k-2) == 1 (mod (2^k-1)*k^2).

2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 257, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 65537

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OFFSET

1,1

COMMENTS

Conjecture: for k > 1, (2^k-1)*k^2 divides (k-1)^(2^k-2)-1 if and only if 2^k-1 is (a Mersenne) prime or k = 2^(2^m)+1 (is a Fermat number).

If so, composite terms are Fermat composite numbers k = 2^(2^m)+1. The smallest (for m = 5) is k = 4294967297 = 641*6700417 and probably all larger (for m > 5).

If my conjecture is true, primes in this sequence that are not Mersenne exponents are Fermat prime numbers k = 2^(2^m)+1 for m > 2, namely k = 257, 65537 and there are probably no more.

LINKS

Table of n, a(n) for n=1..29.

MATHEMATICA

q[n_] := PowerMod[n - 1, 2^n - 2, n^2*(2^n - 1)] == 1; Select[Range[1000], q] (* Amiram Eldar, Oct 02 2023 *)

PROG

(Python)

def A366029_gen(startvalue=1): # generator of terms>= startvalue

return filter(lambda k:pow(k-1, (m:=(1<<k)-1)-1, m*k**2)==1, count(max(startvalue, 1)))

A366029_list = list(islice(A366029_gen(), 10)) # Chai Wah Wu, Oct 30 2023

CROSSREFS

Cf. A000043, A000215.

Sequence in context: A336721 A120857 A233516 * A000043 A109799 A152961