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%I #39 Feb 25 2018 20:16:06
%S 2,3,4,6,10,11,14,19,20,22,29,31,38,41,54,55,56,58,59,61,62,71,76,79,
%T 80,82,89,93,95,101,109,110,118,121,122,123,124,131,139,142,145,149,
%U 151,152,153,155,158,165,174,178,179,181,190,191,196,199,202,205,209,211,213
%N Numbers n for which Fibonacci(n) mod n equals some nonzero Fibonacci(k) and k divides n.
%C If Fibonacci(n) (mod n) = 1, then we assume that k=1 (even though Fibonacci(2) also equals 1). Subsequence of A182625.
%H Andrew Howroyd, <a href="/A187966/b187966.txt">Table of n, a(n) for n = 1..2000</a>
%e 14 is in this sequence because fib(14)=377 is congruent to 13 (mod 14), 13=fib(7), and 7 divides 14.
%t nn = 13; f = Table[Fibonacci[n], {n, nn}]; okQ[n_] := Module[{pos = Position[f, Mod[Fibonacci[n], n]]}, pos != {} && Mod[n, pos[[1, 1]]] == 0]; Select[Range[f[[-1]]], okQ] (* _T. D. Noe_, Apr 04 2011 *)
%o (PARI) ok(n)={my(m=fibonacci(n)%n); fordiv(n, k, my(t=fibonacci(k)); if(t>=m, return(t==m))); 0} \\ _Andrew Howroyd_, Feb 25 2018
%Y Cf. A182625.
%K nonn
%O 1,1
%A _Carmine Suriano_, Mar 30 2011
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