A331109 The number of dual-Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the dual Zeckendorf expansion (A104326) of each s(i) contains only terms that are in the dual Zeckendorf expansion of r(i).
1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 8, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4
Offset: 1
Examples
a(32) = 4 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.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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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 == {}, 1, 2^Total[v[[i[[1, 1]] ;; -1]]]]]; f[p_, e_] := dualZeck[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
Formula
Multiplicative with a(p^e) = 2^A112310(e).
Comments