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%I #37 Feb 04 2023 05:29:59
%S 1,4,6,9,11,19,24,29,31,34,41,46,48,59,61,71,72,79,89,94,96,100,101,
%T 106,109,120,129,131,139,144,149,151,166,179,181,191,192,199,201,211,
%U 214,216,220,226,229,239,240,241,249,251,269,271,274,281,288,311
%N Numbers k that divide the sum of the first k Fibonacci numbers (beginning with F(0)).
%C Numbers k such that A000045(k+1) == 1 (mod k). - _Robert Israel_, Oct 13 2015
%H Robert Israel, <a href="/A219612/b219612.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A101907(n) + 1. - _Altug Alkan_, Dec 29 2015
%e Sum of first 6 Fibonacci numbers is 0+1+1+2+3+5 = 12. Because 6 divides 12, 6 is in the sequence.
%p fmod:= proc(a, b) local A, n, f1, f2, f;
%p uses LinearAlgebra[Modular];
%p A:= Mod(b, <<1, 1>|<1, 0>>, integer[8]);
%p MatrixPower(b, M, a)[1, 2];
%p end proc:
%p 1, op(select(t -> fmod(t+1,t) = 1, [$2..10^4])); # _Robert Israel_, Oct 13 2015
%t okQ[n_] := n == 1 || Mod[Fibonacci[n+1], n] == 1;
%t Select[Range[1000], okQ] (* _Jean-François Alcover_, Feb 04 2023 *)
%o (Python)
%o sum, prpr, prev = 0, 0, 1
%o for i in range(1, 1000):
%o sum += prpr
%o if sum % i == 0: print i,
%o prpr, prev = prev, prpr+prev
%o (PARI) lista(nn) = {sf = 0; for (n=0, nn, sf += fibonacci(n); if (sf % (n+1) == 0, print1(n+1, ", ")););} \\ _Michel Marcus_, Jun 05 2013
%Y Cf. A000045, A000071, A023172, A045345, A101907.
%K nonn,easy
%O 1,2
%A _Alex Ratushnyak_, May 03 2013
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