A081320 Largest 3-smooth divisor of n-th Fibonacci number.

1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 144, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 288, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 432, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 576, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 144, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 864, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 144, 1, 1, 2, 3, 1, 8, 1, 3, 2, 1, 1, 1152, 1

Offset: 1

Author

Reinhard Zumkeller, May 20 2003

Keywords

• nonn

Comments

Conjecture: for n>12 and n>0 modulo 12: a(n)=a(n-12) and a(12*k)=A065331(k)*144.

The first part of the conjecture follows from the fact that the Fibonacci numbers are a strong divisibility sequence. - Charles R Greathouse IV, Sep 24 2012

Examples

			Fibonacci(36) = 14930352 = 2^4 * 3^3 * 17 * 19 * 107, therefore a(36) = 2^4 * 3^3 = 432.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003586, A065331, A000045.

Programs

Mathematica

a[n_] := Times @@ ({2, 3}^IntegerExponent[Fibonacci[n], {2, 3}]);
Table[a[n], {n, 1, 1000}] (* Jean-François Alcover, Oct 15 2021 *)

PARI

fibord(n,p)=if(n==0, return(oo)); my(u=3,t); while((t=((Mod([1,1;1,0],p^u))^n)[1,2])==0, u*=2); valuation(t,p)
a(n)=if(n%12, return(gcd(fibonacci(n%12),24))); 3^fibord(n,3)<Charles R Greathouse IV, Nov 13 2015

Formula

a(n) = A065331(A000045(n)).