A351219 Multiplicative with a(p^e) = Fibonacci(e+1).

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 8, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 5, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 13, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 5, 5, 1, 1, 2, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Feb 05 2022

Keywords

• nonn
• mult
• easy

Comments

These numbers were called Zetanacci numbers by Bruckman (1983).

The distinct values of the terms are in A065108.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Paul S. Bruckman, Problem H-359, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 21, No. 3 (1983), p. 238; Zetanacci, Solution to Problem H-359 by C. Georghiou, ibid., Vol. 23, No. 1 (1985), pp. 91-92.

Crossrefs

Cf. A000045, A005117, A065108, A065488.

Programs

Mathematica

f[p_, e_] := Fibonacci[e + 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = fibonacci(f[k,2]+1); f[k,2]=1); factorback(f); \\ Michel Marcus, Feb 05 2022

PARI

for(n=1, 100, print1(direuler(p=2, n, 1/(1 - X - X^2))[n], ", ")) \\ Vaclav Kotesovec, Feb 10 2022

Python

from math import prod
from sympy import factorint, fibonacci
def a(n): return prod(fibonacci(e+1) for p, e in factorint(n).items())
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Feb 05 2022

Formula

Dirichlet g.f.: Product_{p prime} 1/(1 - p^(-s) - p^(-2*s)).

a(n) = 1 if and only if n is a squarefree number (A005117).

Sum_{k=1..n} a(k) ~ c * n, where c = A065488 = Product_{p primes} (1 + 1/(p^2 - p - 1)) = 2.67411272557... - Vaclav Kotesovec, Feb 10 2022