

A260407


Numbers n such that (n1)^2+1 divides 2^(n1)1.


2




OFFSET

1,2


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)^(n1)1) == 0 mod ((n1)^2+1) for all k >= 1.  Jaroslav Krizek, Jun 02 2016


LINKS

Table of n, a(n) for n=1..8.


FORMULA

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


MATHEMATICA

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


PROG

(PARI) forstep(n=1, 1e7, 2, Mod(2, (n1)^2+1)^(n1)==1&&print1(n", "))
(MAGMA) [n: n in [1..10^6]  (2^(n1)1) mod ((n1)^2+1) eq 0 ]; // Vincenzo Librandi, Jul 25 2015


CROSSREFS

Cf. A000215, A081762, A247165, A260072, A260406.
Sequence in context: A090457 A174408 A260072 * A193329 A256499 A217796
Adjacent sequences: A260404 A260405 A260406 * A260408 A260409 A260410


KEYWORD

nonn,more


AUTHOR

M. F. Hasler, Jul 24 2015


STATUS

approved



