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A168171 Least prime p = 1 (mod n) which divides Fibonacci((p-1)/n). 3
11, 29, 139, 61, 211, 541, 2269, 89, 199, 281, 859, 661, 911, 2269, 2221, 2081, 2789, 2161, 3041, 421, 2521, 19009, 21529, 3001, 9901, 5981, 2161, 2269, 26449, 2221, 31249, 19681, 17491, 2789, 3571, 25309, 30859, 3041, 6709, 3001, 9349, 2521, 13159, 19009 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
LINKS
EXAMPLE
For n=1, all numbers p satisfy p=1 (mod n), but p=11 is the least prime that divides F((p-1)/1)=F(p-1)=F(10)=55.
For n=2, all odd numbers, thus all primes p>2, satisfy p=1 (mod n), but p=29 is the first one to divide F((p-1)/2) = F(14) = 377 = 13*29.
For n=5, a(n)=211 is the smallest Artiad, i.e. prime p=1 (mod 5) which divides F((p-1)/5) = F(42) = 211*1269736.
MATHEMATICA
a[1] = 11;
a[n_] := For[p = 1, True, p = p + n, If[PrimeQ[p] && Divisible[Fibonacci[(p - 1)/n], p], Return[p]]];
a /@ Range[100] (* Jean-François Alcover, Oct 14 2019 *)
PROG
(PARI) for(n=1, 99, forprime(p=1, default(primelimit), (p-1)%n & next; fibonacci((p-1)/n)%p | print1(p", ") | next(2)); error(n))
CROSSREFS
Cf. 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]).
Sequence in context: A099911 A118638 A088460 * A249993 A080083 A115972
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 25 2009
STATUS
approved

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Last modified April 25 15:53 EDT 2024. Contains 371989 sequences. (Running on oeis4.)