A319040 Numbers k > 1 such that Pell(k) == 1 (mod k).

7, 17, 23, 31, 35, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127, 137, 151, 167, 169, 191, 193, 199, 223, 233, 239, 241, 257, 263, 271, 281, 311, 313, 337, 353, 359, 367, 383, 385, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487, 503, 521, 569, 577, 593, 599

Offset: 1

Author

Jon E. Schoenfield, Sep 08 2018

Keywords

• nonn

Comments

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

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

Examples

			k = 7 is in the sequence since Pell(7) = 169 = 7 * 24 + 1 == 1 (mod 7).
k = 11 is not in the sequence: Pell(11) = 5741 = 11 * 522 - 1 !== 1 (mod 11).
k = 35 is in the sequence: Pell(35) = 8822750406821 = 35 * 252078583052 + 1 == 1 (mod 35).
		

Crossrefs

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

Programs

Maple

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]):
A319040List(600); # Peter Luschny, Sep 09 2018

Mathematica

Select[Range[500], Mod[Fibonacci[#, 2], #] == 1 &] (* Alonso del Arte, Sep 08 2018 *)