A331109 - OEIS

%I #16 Aug 06 2024 06:14:29

%S 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,

%T 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,

%U 4,8,2,8,2,4,4,4,4,8,2,8,4,4,2,8,4,4,4

%N 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).

%C Dual-Zeckendorf-infinitary divisors are analogous to infinitary divisors (A077609) with dual Zeckendorf expansion instead of binary expansion.

%C First differs from A286324 at n = 32.

%H Amiram Eldar, <a href="/A331109/b331109.txt">Table of n, a(n) for n = 1..10000</a>

%F Multiplicative with a(p^e) = 2^A112310(e).

%e 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.

%t 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];

%t 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]]]]];

%t f[p_, e_] := dualZeck[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%Y Cf. A037445, A038148, A074848, A077609, A104326, A112310, A286324, A318465.

%K nonn,mult

%O 1,2

%A _Amiram Eldar_, Jan 09 2020