A101907 Numbers n-1 such that the arithmetic mean of the first n Fibonacci numbers (beginning with F(0)) is an integer.

0, 3, 5, 8, 10, 18, 23, 28, 30, 33, 40, 45, 47, 58, 60, 70, 71, 78, 88, 93, 95, 99, 100, 105, 108, 119, 128, 130, 138, 143, 148, 150, 165, 178, 180, 190, 191, 198, 200, 210, 213, 215, 219, 225, 228, 238, 239, 240, 248, 250, 268, 270, 273, 280, 287, 310, 320, 330

Offset

1,2

Comments

The sum of the first n Fibonacci numbers is F(n+2)-1, sequence A000071.

Knott discusses the factorization of these numbers. - T. D. Noe, Oct 10 2005

Links

T. D. Noe, Table of n, a(n) for n = 1..1000

Y. Bugeaud, F. Luca, M. Mignotte and S. Siksek, On Fibonacci numbers with few prime divisors, Proc. Japan Acad., 81, Ser. A (2005), pp. 17-20.

H. R. Morton, Fibonacci-like sequences and greatest common divisors, The American Mathematical Monthly, Vol. 102, No. 8 (October 1995), pp. 731-734.

M. Ward, The prime divisors of Fibonacci numbers, Pacific J. Math., Vol. 11, No. 1 (1961), pp. 379-386.

Eric W. Weisstein's World of Mathematics, Arithmetic mean

Eric W. Weisstein's World of Mathematics, Fibonacci

Formula

Numbers n-1 such that (F(0)+ F(1)+ ... + F(n-1)) / n is an integer. F(i) is the i-th Fibonacci number.

a(n) = A219612(n) - 1. - Altug Alkan, Dec 29 2015

Example

n=4 : (F(0)+F(1)+F(2)+F(3))/4 = (0+1+1+2)/4 = 1. So n-1 = 4-1 = 3 is a term.
n=6 : (F(0)+F(1)+F(2)+F(3)+F(4)+F(5))/6 = (0+1+1+2+3+5)/6 = 2. So n-1 = 6-1 = 5 is a term.

Mathematica

Select[ Range[0, 500], Mod[Fibonacci[ # + 2] - 1, # + 1] == 0 &] (* Robert G. Wilson v *)

Prog

(PARI) is(n)=((Mod([1, 1; 1, 0], n+1))^(n+2))[1, 2]==1 \\ Charles R Greathouse IV, Feb 04 2013

Crossrefs

Cf. A000045, A000071. See A111035 for another version.

Cf. A219612. - Altug Alkan, Dec 29 2015

Sequence in context: A212987 A217919 A127700 * A242250 A117668 A184410

Adjacent sequences: A101904 A101905 A101906 * A101908 A101909 A101910

Keyword

easy,nonn

Author

Ctibor O. Zizka, Jul 27 2008

Extensions

Edited and extended by Robert G. Wilson v, Aug 03 2008

Definition corrected by Altug Alkan, Dec 29 2015

Status

approved