A247220 Numbers k such that k^2 + 1 divides 2^k + 1.

0, 2, 4, 386, 20136, 59140, 373164544

Offset

1,2

Comments

a(8) > 2*10^9. - Hiroaki Yamanouchi, Nov 29 2014

a(8) > 2*10^10. - Chai Wah Wu, Dec 07 2014

All terms of the sequence are even. a(5), a(6) and a(7) are of the form 2*p + 2 where p is a prime and mod(p, 14) = 1. - Farideh Firoozbakht, Dec 07 2014

From Jianing Song, Jan 13 2019: (Start)

Among the known terms only a(3) and a(4) are of the form 2*p where p is a prime.

a(n)^2 + 1 is prime for 2 <= n <= 7. Among these primes, the multiplicative order of 2 modulo a(n)^2 + 1 is 2*a(n) except for n = 5, in which case it is 2*a(n)/3. (End)

a(8) > 10^12. - Giovanni Resta, May 05 2020

Links

Table of n, a(n) for n=1..7.

Example

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

Prog

(PARI) for(n=0, 10^5, if(Mod(2, n^2+1)^n==-1, print1(n, ", "))); \\ Joerg Arndt, Nov 30 2014
(Python)
from gmpy2 import powmod
A247220_list = [i for i in range(10**7) if powmod(2, i, i*i+1) == i*i]
# Chai Wah Wu, Dec 03 2014

Crossrefs

Sequence in context: A012562 A263959 A216024 * A012753 A012440 A058172

Adjacent sequences: A247217 A247218 A247219 * A247221 A247222 A247223

Keyword

nonn,more

Author

Juri-Stepan Gerasimov, Nov 26 2014

Extensions

a(7) from Hiroaki Yamanouchi, Nov 29 2014

Status

approved