cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A076984 Number of Fibonacci numbers that are divisors of the n-th Fibonacci number.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 2, 5, 2, 3, 4, 4, 2, 5, 2, 5, 4, 3, 2, 7, 3, 3, 4, 5, 2, 7, 2, 5, 4, 3, 4, 8, 2, 3, 4, 7, 2, 7, 2, 5, 6, 3, 2, 9, 3, 5, 4, 5, 2, 7, 4, 7, 4, 3, 2, 11, 2, 3, 6, 6, 4, 7, 2, 5, 4, 7, 2, 11, 2, 3, 6, 5, 4, 7, 2, 9, 5, 3, 2, 11, 4, 3, 4, 7, 2, 11, 4, 5, 4, 3, 4, 11, 2, 5, 6, 8, 2, 7, 2
Offset: 1

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Author

Amarnath Murthy, Oct 25 2002

Keywords

Comments

a(A001605(n)) = 2; a(A105802(n)) = n.
It is well known that if k is a divisor of n then F(k) divides F(n). Hence if n has d divisors, one expects that a(n)=d. However because F(1)=F(2)=1, there is one fewer Fibonacci divisor when n is even. So for even n, a(n)=d-1. - T. D. Noe, Jan 18 2006

Examples

			n=12, A000045(12)=144: 5 of the 15 divisors of 144 are also Fibonacci numbers, a(12) = #{1, 2, 3, 8, 144} = 5.
		

Crossrefs

Programs

  • Maple
    with(combinat, fibonacci):a[1] := 1:for i from 2 to 229 do s := 0:for j from 2 to i do if((fibonacci(i) mod fibonacci(j))=0) then s := s+1:fi:od:a[i] := s:od:seq(a[l],l=2..229);
  • Mathematica
    Table[s=DivisorSigma[0, n]; If[OddQ[n], s, s-1], {n, 100}] (Noe)
  • PARI
    {a(n)=if(n<1, 0, numdiv(n)+n%2-1)} /* Michael Somos, Sep 03 2006 */
    
  • PARI
    {a(n)=if(n<1, 0, sumdiv(n,d, d!=2))} /* Michael Somos, Sep 03 2006 */

Formula

a(n) = A023645(n) + 1. - T. D. Noe, Jan 18 2006
a(n) = tau(n) - [n is even] = A000005(n) - A059841(n). Proof: gcd(Fib(m), Fib(n)) = Fib(gcd(m, n)) and Fib(2) = 1. - Olivier Wittenberg, following a conjecture of Ralf Stephan, Sep 28 2004
The number of divisors of n excluding 2.
a(2n) = A066660(n). a(2n-1) = A099774(n). - Michael Somos, Sep 03 2006
a(3*2^(Prime(n-1)-1)) = 2n + 1 for n > 3. a(3*2^A068499[n]) = 2n + 1, where A068499(n) = {1,2,3,4,6,10,12,16,18,...}. - Alexander Adamchuk, Sep 15 2006

Extensions

Corrected and extended by Sascha Kurz, Jan 26 2003
Edited by N. J. A. Sloane, Sep 14 2006. Some of the comments and formulas may need to be adjusted to reflect the new offset.