A268271 Primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all quadratic residues of p in the first (p-1)/2 iterations (for some b).

11, 19, 29, 31, 59, 71, 79, 89, 101, 131, 179, 181, 191, 229, 239, 251, 271, 311, 349, 359, 379, 401, 419, 431, 439, 479, 491, 499, 509, 571, 599, 631, 659, 719, 739, 751, 761, 839, 941, 971, 1019, 1021, 1039, 1051, 1061, 1091, 1109, 1171, 1229, 1249, 1259, 1319, 1361, 1399

Offset: 1

Author

Michel Marcus, Mar 02 2016

Keywords

• nonn

Examples

			p=11 is a term since, modulo 11, the sequence 1, 4, 5, 9, 3 satisfies 5=4+1, 9=5+4, 3=9+5, 1=9+3, ..., with a period of (11-1)/2 = 5.
		

Links

• Alexandru Gica, Quadratic Residues in Fibonacci Sequences, Fibonacci Quart. 46/47 (2008/2009), no. 1, 68-72.

Crossrefs

Subsequence of A045468.

Cf. A003147 (similar sequence for a different period).

Cf. A168429, A070373 (examples of such Fibonacci-type sequences).

Programs

PARI

findr(p) = {for (k=1, (p-1)/2, if ((k^2 % p) == 5, return(k)););}
isok(p) = {if ((p % 2) && isprime(p), pm = p % 5; if ((pm == 1) || (pm == 4), rf = findr(p);(znorder(Mod((1+rf)/2, p)) == (p-1)/2) || (znorder(Mod((1-rf)/2, p)) == (p-1)/2);););}