A351219 - OEIS

%I #18 Feb 10 2022 06:01:40

%S 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,

%T 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,

%U 1,1,1,6,1,1,2,2,1,1,1,5,5,1,1,2,1,1,1

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

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

%C The distinct values of the terms are in A065108.

%H Amiram Eldar, <a href="/A351219/b351219.txt">Table of n, a(n) for n = 1..10000</a>

%H Paul S. Bruckman, <a href="https://fq.math.ca/Scanned/21-3/advanced21-3.pdf">Problem H-359</a>, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 21, No. 3 (1983), p. 238; <a href="https://www.fq.math.ca/Scanned/23-1/advanced23-1.pdf">Zetanacci</a>, Solution to Problem H-359 by C. Georghiou, ibid., Vol. 23, No. 1 (1985), pp. 91-92.

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

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

%F 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

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

%o (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

%o (Python)

%o from math import prod

%o from sympy import factorint, fibonacci

%o def a(n): return prod(fibonacci(e+1) for p, e in factorint(n).items())

%o print([a(n) for n in range(1, 88)]) # _Michael S. Branicky_, Feb 05 2022

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, 1/(1 - X - X^2))[n], ", ")) \\ _Vaclav Kotesovec_, Feb 10 2022

%Y Cf. A000045, A005117, A065108, A065488.

%K nonn,mult,easy

%O 1,4

%A _Amiram Eldar_, Feb 05 2022