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A219612 Numbers n such that n divides sum of first n Fibonacci numbers (beginning with F(0)). 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Numbers n such that A000045(n+1) == 1 (mod n). - 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

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.

Cf. A101907. - Altug Alkan, Dec 29 2015

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

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Last modified April 23 18:34 EDT 2019. Contains 322387 sequences. (Running on oeis4.)