A260406 -id:A260406 - cp's OEIS frontend

A260072 Primes p such that (p-1)^2+1 divides 2^(p-1)-1.

17, 257, 8209, 65537, 649801

Offset: 1

Jaroslav Krizek, Jul 22 2015

a(6), if it exists, is larger than 1.7*10^12. - Giovanni Resta, Jul 23 2015

N = 1382401 is the smallest composite number such that (n-1)^2+1 divides 2^(n-1)-1, cf. A260407; see also A081762 and A260406. The sequence contains all Fermat primes 2^2^k+1 > 5 (A019434). - M. F. Hasler, Jul 24 2015

			17 is in this sequence because (17 - 1)^2 + 1 = 257 divides 2^(17 - 1) - 1 = 65535; 65535 / 257 = 255.

Cf. A081762 (primes p such that (p-1)^2 - 1 divides 2^(p-1) - 1).

[n: n in [1..2000000] | IsPrime(n) and (2^(n-1)-1) mod ((n-1)^2 + 1) eq 0];

fQ[n_] := PowerMod[2, n-1, (n-1)^2 + 1] == 1; p = 2; lst = {}; While[p < 10^9, If[ fQ@ p, AppendTo[lst, p]]; p = NextPrime@ p] (* Robert G. Wilson v, Jul 24 2015 *)

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

Original entry on oeis.org

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

Offset: 1

M. F. Hasler, Jul 24 2015

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

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

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

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

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

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

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A260072 Primes p such that (p-1)^2+1 divides 2^(p-1)-1.

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