%I #47 Mar 31 2021 19:00:44
%S 3,7,13,17,23,37,43,53,67,73,83,97,103,127,137,157,163,167,173,193,
%T 197,223,227,257,277,283,293,313,317,337,367,373,383,397,433,443,457,
%U 463,467,487,503,523,547,577,587,593,607,613,617,643,647,653,673,683,727
%N Related to Pisano periods: integers k such that the period of Fibonacci numbers mod k equals 2*(k+1).
%C Terms are primes with final digit 3 or 7.
%C Apparently these are the primes given in A003631 without 2 and A216067. - _Klaus Purath_, Dec 11 2020
%C If k is a term, then for m=5*k the period of Fibonacci numbers mod m equals 2*(m+5). - _Matthew Goers_, Jan 13 2021
%H Michael De Vlieger, <a href="/A071774/b071774.txt">Table of n, a(n) for n = 1..3969</a>
%H Bob Bastasz, <a href="https://www.fq.math.ca/Papers1/58-5/bastasz.pdf">Lyndon words of a second-order recurrence</a>, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.
%t Select[Prime@ Range[129], Function[n, Mod[Last@ NestWhile[{Mod[#2, n], Mod[#1 + #2, n], #3 + 1} & @@ # &, {1, 1, 1}, #[[1 ;; 2]] != {0, 1} &], n] == Mod[2 (n + 1), n] ]] (* _Michael De Vlieger_, Mar 31 2021, after _Leo C. Stein_ at A001175 *)
%o (PARI) for(n=2,5000,t=2*(n+1);good=1;if(fibonacci(t)%n==0, for(s=0,t,if(fibonacci(t+s)%n!=fibonacci(s)%n,good=0;break); if(s>1&&s<t-1&&fibonacci(s)%n==0,cur=s;good2=1;for(ss=0,s, if(fibonacci(ss+s)%n!=fibonacci(ss)%n,good2=0;break)); if(good2,good=0;break););); if(good,print1(n,",")))) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Dec 21 2004
%o (PARI) forprime(p=3,3000,if(p%5==2||p%5==3,a=1;b=0;c=1;while(a!=0||b!=1,c++;d=a;a=b;a=(a+d)%p;b=d%p);if(c==(2*(p+1)),print1(p",")))) /* _V. Raman_, Nov 22 2012 */
%Y Cf. A001175, A060305, A071776.
%Y Cf. A003631, A216067.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Jun 04 2002
%E More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Dec 21 2004