A318465 The number of Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the Zeckendorf expansion (A014417) of each s(i) contains only terms that are in the Zeckendorf expansion of r(i).

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 2, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4, 4, 2, 8, 4, 4, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 8

Offset: 1

Author

Antti Karttunen, Aug 30 2018

Keywords

• nonn
• mult

Comments

Zeckendorf-infinitary divisors are analogous to infinitary divisors (A077609) with Zeckendorf expansion instead of binary expansion. - Amiram Eldar, Jan 09 2020

Examples

			a(16) = 4 since 16 = 2^4 and the Zeckendorf expansion of 4 is 101, i.e., its Zeckendorf representation is a set with 2 terms: {1, 3}. There are 4 possible exponents of 2: 0, 1, 3 and 4, corresponding to the subsets {}, {1}, {3} and {1, 3}. Thus 16 has 4 Zeckendorf-infinitary divisors: 2^0 = 1, 2^1 = 2, 2^3 = 8, and 2^4 = 16.
		

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, A318464, A318469.

Cf. A014417, A037445, A038148, A074848, A077609.

Cf. also A318472, A318474.

Programs

Mathematica

fb[n_] := Block[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; Fibonacci[1 + Position[Reverse@fr, ?(# == 1 &)]]]; f[p, e_] := 2^Length@fb[e]; a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n])); Array[a, 100] (* Amiram Eldar, Jan 09 2020 after Robert G. Wilson v at A014417 *)

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); }
A318465(n) = factorback(apply(e -> 2^A007895(e),factor(n)[,2]));

Formula

Multiplicative with a(p^e) = 2^A007895(e), where A007895(n) gives the number of terms in the Zeckendorf representation of n.

a(n) = 2^A318464(n).

Extensions

Name edited and interpretation in terms of divisors added by Amiram Eldar, Jan 09 2020