A318464 Additive with a(p^e) = A007895(e), where A007895(n) gives the number of terms in Zeckendorf representation of n.

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 3

Offset: 1

Author

Antti Karttunen, Aug 30 2018

Keywords

• nonn

Comments

From Amiram Eldar, Aug 09 2024: (Start)

The number of factors of n of the form p^Fibonacci(k), where p is a prime and k >= 2, when the factorization is uniquely done using the Zeckendorf representation of the exponents in the prime factorization of n.

Equivalently, the number of Zeckendorf-infinitary divisors of n (defined in A318465) that are prime powers (A246655). (End)

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Cf. A007895, A077761, A246655, A318465, A318469.

Programs

Mathematica

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; a[n_] := Total[z /@ FactorInteger[n][[;; , 2]]]; a[1] = 0; Array[a, 100] (* Amiram Eldar, May 15 2023 *)

PARI

A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
A007895(n) = { my(s=0); while(n>0, s++; n -= fibonacci(1+A072649(n))); (s); }
A318464(n) = vecsum(apply(e -> A007895(e),factor(n)[,2]));

Formula

a(n) = A007814(A318465(n)).

a(n) = A001222(A318469(n)).

Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{k>=2} (A007895(k)-A007895(k-1)) * P(k) = 0.05631817952062180045..., where P(s) is the prime zeta function. - Amiram Eldar, Oct 09 2023