%I #36 Sep 16 2018 04:42:04

%S 7,17,23,31,35,41,47,71,73,79,89,97,103,113,127,137,151,167,169,191,

%T 193,199,223,233,239,241,257,263,271,281,311,313,337,353,359,367,383,

%U 385,401,409,431,433,439,449,457,463,479,487,503,521,569,577,593,599

%N Numbers k > 1 such that Pell(k) == 1 (mod k).

%C It appears that most of the terms of this sequence are primes. The composite terms are 35, 169, 385, 899, 961, 1121, ... (A319042).

%C The primes in the sequence give A001132 (primes == +-1 (mod 8)), since for primes p we have Pell(p) == (2/p) (mod p) where (2/p) is the Legendre symbol. - _Jianing Song_, Sep 10 2018

%e k = 7 is in the sequence since Pell(7) = 169 = 7 * 24 + 1 == 1 (mod 7).

%e k = 11 is not in the sequence: Pell(11) = 5741 = 11 * 522 - 1 !== 1 (mod 11).

%e k = 35 is in the sequence: Pell(35) = 8822750406821 = 35 * 252078583052 + 1 == 1 (mod 35).

%p isA319040 := k -> simplify(2^(k-1)*hypergeom([1-k/2,(1-k)/2],[1-k],-1)) mod k = 1: A319040List := b -> select(isA319040, [$1..b]):

%p A319040List(600); # _Peter Luschny_, Sep 09 2018

%t Select[Range[500], Mod[Fibonacci[#, 2], #] == 1 &] (* _Alonso del Arte_, Sep 08 2018 *)

%Y Cf. A000129 (Pell numbers), A001132, A023173, A319041, A319042, A319043.

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Sep 08 2018