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A318465
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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).
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6
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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
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OFFSET
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1,2
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COMMENTS
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Zeckendorf-infinitary divisors are analogous to infinitary divisors (A077609) with Zeckendorf expansion instead of binary expansion. - Amiram Eldar, Jan 09 2020
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LINKS
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FORMULA
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Multiplicative with a(p^e) = 2^A007895(e), where A007895(n) gives the number of terms in the Zeckendorf representation of n.
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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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,mult
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AUTHOR
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EXTENSIONS
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Name edited and interpretation in terms of divisors added by Amiram Eldar, Jan 09 2020
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STATUS
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approved
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