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

1,1

Links

Table of n, a(n) for n=1..54.

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

Example

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.

Prog

(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); ); ); }

Crossrefs

Subsequence of A045468.

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

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

Sequence in context: A123976 A045468 A196095 * A053032 A342961 A277123

Adjacent sequences: A268268 A268269 A268270 * A268272 A268273 A268274

Keyword

nonn

Author

Michel Marcus, Mar 02 2016

Status

approved