OFFSET
1,2
COMMENTS
First differs from A188999 at n = 32.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
Multiplicative with a(p^e) = Product_{i} (p^s(i) + 1), where s(i) are the terms in the dual Zeckendorf representation of e (A104326).
EXAMPLE
a(32) = 45 since 32 = 2^5 and the dual Zeckendorf expansion of 5 is 110, i.e., its dual Zeckendorf representation is a set with 2 terms: {2, 3}. There are 4 possible exponents of 2: 0, 2, 3 and 5, corresponding to the subsets {}, {2}, {3} and {2, 3}. Thus 32 has 4 dual-Zeckendorf-infinitary divisors: 2^0 = 1, 2^2 = 4, 2^3 = 8, and 2^5 = 32, and their sum is 1 + 4 + 8 + 32 = 45.
MATHEMATICA
fibTerms[n_] := Module[{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--]; fr];
dualZeck[n_] := Module[{v = fibTerms[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i]] == 1 && v[[i + 1]] == 0 && v[[i + 2]] == 0, v[[i]] = 0; v[[i + 1]] = 1; v[[i + 2]] = 1; If[i > 2, i -= 3]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, {}, v[[i[[1, 1]] ;; -1]]]];
f[p_, e_] := p^Fibonacci[1 + Position[Reverse@dualZeck[e], _?(# == 1 &)]];
a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
Amiram Eldar, Jan 09 2020
STATUS
approved