A247165 Numbers m such that m^2 + 1 divides 2^m - 1.

0, 16, 256, 8208, 65536, 649800, 1382400, 4294967296

Offset: 1

Author

Juri-Stepan Gerasimov, Nov 30 2014

Keywords

• nonn
• more

Comments

Contains 2^(2^k) = A001146(k) for k >= 2. - Robert Israel, Dec 02 2014

a(9) > 10^12. - Hiroaki Yamanouchi, Sep 16 2018

For each n, a(n)^2 + 1 belongs to A176997, and thus a(n) belongs to either A005574 or A135590. - Max Alekseyev, Feb 08 2024

Examples

			0 is in this sequence because 0^2 + 1 = 1 divides 2^0 - 1 = 1.
		

Crossrefs

Cf. A247219 (n^2 - 1 divides 2^n - 1), A247220 (n^2 + 1 divides 2^n + 1).

Cf. A005574, A135590, A176997.

Programs

Magma

[n: n in [1..100000] | Denominator((2^n-1)/(n^2+1)) eq 1];

Maple

select(n -> (2 &^ n - 1) mod (n^2 + 1) = 0, [$1..10^6]); # Robert Israel, Dec 02 2014

Mathematica

a247165[n_Integer] := Select[Range[0, n], Divisible[2^# - 1, #^2 + 1] &]; a247165[1500000] (* Michael De Vlieger, Nov 30 2014 *)

PARI

for(n=0,10^9,if(Mod(2,n^2+1)^n==+1,print1(n,", "))); \\ Joerg Arndt, Nov 30 2014

Python

A247165_list = [n for n in range(10**6) if n == 0 or pow(2,n,n*n+1) == 1]
# Chai Wah Wu, Dec 04 2014

Extensions

a(8) from Chai Wah Wu, Dec 04 2014

Edited by Jon E. Schoenfield, Dec 06 2014