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Least prime p == -1 (mod n) that divides Fibonacci((p+1)/n), or 0 if no such prime exists.
1

%I #11 Jan 02 2023 12:30:47

%S 2,13,47,0,0,113,307,0,233,0,967,0,2417,797,0,0,1087,233,5737,0,5417,

%T 5653,1103,0,0,2417,4373,0,6263,0,25357,0,3167,42533,0,0,4513,5737,

%U 2417,0,61417,5417,32507,0,0,36017,1597,0,97607,0,27947,0,42293,4373,0,0

%N Least prime p == -1 (mod n) that divides Fibonacci((p+1)/n), or 0 if no such prime exists.

%C Max Alekseyev has proved (cf. link) that a(n)=0 if n is a multiple of 4 or 5; for all other n, a prime a(n) with the required property seems to exist.

%H Max Alekseyev, <a href="http://list.seqfan.eu/oldermail/seqfan/2009-November/003046.html">Re: Primes p = nk-1 dividing Fibonacci( k )</a>, SeqFan mailing list, Nov. 2009.

%o (PARI) A168172(n) = n%4 && n%5 && forstep(p=n-1,1e9,n, isprime(p) || next; fibonacci((p+1)/n)%p || return(p))

%Y Cf. A168171 (least p | F[(p-1)/n]), A122487 (p | F[(p+1)/2]), A047652 (p | F[(p-1)/3]), A001583 (Artiads: p | F[(p-1)/5], A125252 (p | F[(p+1)/7]), A125253 (p | F[(p-1)/7]).

%K nonn

%O 1,1

%A _M. F. Hasler_, Nov 28 2009