A293435 a(n) is the number of the proper divisors of n that are Fibonacci numbers (A000045).
0, 1, 1, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 2, 3, 3, 1, 3, 1, 3, 2, 2, 1, 4, 2, 3, 2, 2, 1, 4, 1, 3, 2, 2, 2, 3, 1, 2, 3, 4, 1, 4, 1, 2, 3, 2, 1, 4, 1, 3, 2, 3, 1, 3, 2, 3, 2, 2, 1, 4, 1, 2, 3, 3, 3, 3, 1, 3, 2, 3, 1, 4, 1, 2, 3, 2, 1, 4, 1, 4, 2, 2, 1, 4, 2, 2, 2, 3, 1, 4, 2, 2, 2, 2, 2, 4, 1, 2, 2, 3, 1, 4, 1, 4, 4
Offset: 1
Keywords
Examples
For n = 55, its proper divisors are [1, 5, 11], of which only two, namely 1 and 5 are in A000045, thus a(55) = 2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10946
Programs
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Mathematica
With[{s = Fibonacci@ Range[2, 40]}, Table[DivisorSum[n, 1 &, And[MemberQ[s, #], # != n] &], {n, 105}]] (* Michael De Vlieger, Oct 09 2017 *)
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PARI
A010056(n) = { my(k=n^2); k+=(k+1)<<2; (issquare(k) || (n>0 && issquare(k-8))) }; \\ This function from Charles R Greathouse IV, Jul 30 2012 A293435(n) = sumdiv(n,d,(d
A010056(d));
Formula
a(n) = Sum_{d|n, dA010056(d).
G.f.: Sum_{k>=2} x^(2*Fibonacci(k)) / (1 - x^Fibonacci(k)). - Ilya Gutkovskiy, Apr 14 2021
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A079586 - 1 = 2.359885... . - Amiram Eldar, Jul 05 2025