A219612 Numbers k that divide the sum of the first k Fibonacci numbers (beginning with F(0)).

1, 4, 6, 9, 11, 19, 24, 29, 31, 34, 41, 46, 48, 59, 61, 71, 72, 79, 89, 94, 96, 100, 101, 106, 109, 120, 129, 131, 139, 144, 149, 151, 166, 179, 181, 191, 192, 199, 201, 211, 214, 216, 220, 226, 229, 239, 240, 241, 249, 251, 269, 271, 274, 281, 288, 311

Offset: 1

Author

Alex Ratushnyak, May 03 2013

Keywords

• nonn
• easy

Comments

Numbers k such that A000045(k+1) == 1 (mod k). - Robert Israel, Oct 13 2015

Examples

			Sum of first 6 Fibonacci numbers is 0+1+1+2+3+5 = 12. Because 6 divides 12, 6 is in the sequence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A000071, A023172, A045345, A101907.

Programs

Maple

fmod:= proc(a, b) local A, n, f1, f2, f;
  uses LinearAlgebra[Modular];
  A:= Mod(b, <<1, 1>|<1, 0>>, integer[8]);
  MatrixPower(b, M, a)[1, 2];
end proc:
1, op(select(t -> fmod(t+1,t) = 1, [$2..10^4])); # Robert Israel, Oct 13 2015

Mathematica

okQ[n_] := n == 1 || Mod[Fibonacci[n+1], n] == 1;
Select[Range[1000], okQ] (* Jean-François Alcover, Feb 04 2023 *)

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

Python

sum_, prpr, prev = 0, 0, 1
for i in range(1, 1000):
  sum_ += prpr
  if sum_ % i == 0:  print(i, end=', ')
  prpr, prev = prev, prpr+prev

Formula

a(n) = A101907(n) + 1. - Altug Alkan, Dec 29 2015