A260407 Numbers n such that (n-1)^2+1 divides 2^(n-1)-1.

1, 17, 257, 8209, 65537, 649801, 1382401, 4294967297

Offset: 1

Author

M. F. Hasler, Jul 24 2015

Keywords

• nonn
• more

Comments

a(7) = 1382401 is the first composite number of this sequence (which makes it different from A260072).

The Fermat numbers 2^(2^k)+1 = A000215(k) with k>1 are a subsequence of this sequence. I conjecture that they are equal to the intersection of this and A260406 (except for the conventional 1).

Conjecture: also numbers n such that ((2^k)^(n-1)-1) == 0 mod ((n-1)^2+1) for all k >= 1. - Jaroslav Krizek, Jun 02 2016

Crossrefs

Cf. A000215, A081762, A247165, A260072, A260406.

Programs

Magma

[n: n in [1..10^6] | (2^(n-1)-1) mod ((n-1)^2+1) eq 0 ]; // Vincenzo Librandi, Jul 25 2015

Mathematica

Join [{1},Select[Range[43*10^8],PowerMod[2,#-1,(#-1)^2+1]==1&]] (* Harvey P. Dale, Sep 07 2018 *)

PARI

forstep(n=1,1e7,2,Mod(2,(n-1)^2+1)^(n-1)==1&&print1(n","))

Formula

a(n) = A247165(n)+1.