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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).

%I A268271 #35 Jan 05 2025 19:51:40
%S A268271 11,19,29,31,59,71,79,89,101,131,179,181,191,229,239,251,271,311,349,
%T A268271 359,379,401,419,431,439,479,491,499,509,571,599,631,659,719,739,751,
%U A268271 761,839,941,971,1019,1021,1039,1051,1061,1091,1109,1171,1229,1249,1259,1319,1361,1399
%N 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).
%H A268271 Alexandru Gica, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/46_47-1/Gica_11-08.pdf">Quadratic Residues in Fibonacci Sequences</a>, Fibonacci Quart. 46/47 (2008/2009), no. 1, 68-72.
%e A268271 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.
%o A268271 (PARI) findr(p) = {for (k=1, (p-1)/2, if ((k^2 % p) == 5, return(k)););}
%o A268271 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);););}
%Y A268271 Subsequence of A045468.
%Y A268271 Cf. A003147 (similar sequence for a different period).
%Y A268271 Cf. A168429, A070373 (examples of such Fibonacci-type sequences).
%K A268271 nonn
%O A268271 1,1
%A A268271 _Michel Marcus_, Mar 02 2016