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

1,2

Comments

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

Links

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

Formula

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

Example

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

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 *)

Prog

(Python)
sum, prpr, prev = 0, 0, 1
for i in range(1, 1000):
  sum += prpr
  if sum % i == 0:  print i,
  prpr, prev = prev, prpr+prev
(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

Crossrefs

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

Sequence in context: A020935 A267149 A198696 * A258743 A292660 A190081

Adjacent sequences: A219609 A219610 A219611 * A219613 A219614 A219615

Keyword

nonn,easy

Author

Alex Ratushnyak, May 03 2013

Status

approved