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Least prime p = 1 (mod n) which divides Fibonacci((p-1)/n).
3

%I #18 Feb 06 2025 22:02:20

%S 11,29,139,61,211,541,2269,89,199,281,859,661,911,2269,2221,2081,2789,

%T 2161,3041,421,2521,19009,21529,3001,9901,5981,2161,2269,26449,2221,

%U 31249,19681,17491,2789,3571,25309,30859,3041,6709,3001,9349,2521,13159,19009

%N Least prime p = 1 (mod n) which divides Fibonacci((p-1)/n).

%H Alois P. Heinz, <a href="/A168171/b168171.txt">Table of n, a(n) for n = 1..1000</a>

%e 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.

%e 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.

%e 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.

%t a[1] = 11;

%t a[n_] := For[p = 1, True, p = p + n, If[PrimeQ[p] && Divisible[Fibonacci[(p - 1)/n], p], Return[p]]];

%t a /@ Range[100] (* _Jean-François Alcover_, Oct 14 2019 *)

%o (PARI) for(n=1,99,forprime(p=1,oo,(p-1)%n & next; fibonacci((p-1)/n)%p || print1(p, ", ") || next(2)))

%Y 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]).

%K nonn

%O 1,1

%A _M. F. Hasler_, Nov 25 2009