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A318464
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Additive with a(p^e) = A007895(e), where A007895(n) gives the number of terms in Zeckendorf representation of n.
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5
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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
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OFFSET
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1,6
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COMMENTS
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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)
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LINKS
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FORMULA
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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
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MATHEMATICA
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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 *)
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PROG
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(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); }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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