A113175 - OEIS

A113175

Replace each prime p in prime-factorization of n with p-th Fibonacci number.

1, 1, 2, 1, 5, 2, 13, 1, 4, 5, 89, 2, 233, 13, 10, 1, 1597, 4, 4181, 5, 26, 89, 28657, 2, 25, 233, 8, 13, 514229, 10, 1346269, 1, 178, 1597, 65, 4, 24157817, 4181, 466, 5, 165580141, 26, 433494437, 89, 20, 28657, 2971215073, 2, 169, 25, 3194, 233

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OFFSET

1,3

COMMENTS

If, for p = prime, p^(m_{n,p}) is highest power of p dividing n, m = nonnegative integer, then a(n) is product over all primes of F(p)^(m_{n,p}), where F(p) = p-th Fibonacci number (A000045).

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..4000

FORMULA

Totally multiplicative with a(p) = F(p). - Franklin T. Adams-Watters, Jun 05 2006

EXAMPLE

63 = 3^2 * 7^1. So a(63) = F(3)^2 * F(7)^1 = 4 * 13 = 52.

MATHEMATICA

Times@@@Table[Fibonacci[#[[1]]]^#[[2]]&/@FactorInteger[n], {n, 60}] (* Harvey P. Dale, Mar 30 2024 *)

PROG

(Sage) [1]+[prod([fibonacci(x[0])^x[1] for x in factor(n)]) for n in range(2, 53)] # Danny Rorabaugh, Apr 03 2015

CROSSREFS

Cf. A113176, A000045.