

A247220


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


6




OFFSET

1,2


COMMENTS

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
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)


LINKS



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]


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



