A294879 Number of proper divisors of n that are in Perrin sequence, A001608.

0, 0, 0, 1, 0, 2, 0, 1, 1, 2, 0, 2, 0, 2, 2, 1, 0, 2, 0, 3, 2, 1, 0, 3, 1, 1, 1, 2, 0, 4, 0, 1, 1, 2, 2, 3, 0, 1, 1, 3, 0, 3, 0, 2, 2, 1, 0, 3, 1, 3, 2, 1, 0, 2, 1, 2, 1, 2, 0, 5, 0, 1, 2, 1, 1, 3, 0, 2, 1, 4, 0, 3, 0, 1, 2, 1, 1, 3, 0, 3, 1, 1, 0, 4, 2, 1, 2, 2, 0, 4, 1, 1, 1, 1, 1, 3, 0, 2, 1, 3, 0, 4, 0, 1, 3, 1, 0, 3, 0, 4, 1, 2, 0, 2, 1, 2, 2, 1, 2, 5, 0

Offset: 1

Author

Antti Karttunen, Nov 10 2017

Keywords

• nonn

Examples

			For n = 22, with proper divisors [1, 2, 11], only 2 is in A001608, thus a(22) = 1.
For n = 121, with proper divisors [1, 11], neither of them is in A001608, thus a(121) = 0. Note that this is the first zero not in A008578.
For n = 644, with proper divisors [1, 2, 4, 7, 14, 23, 28, 46, 92, 161, 322, 644], both 2 and 7 are in A001608, thus a(644) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..24914

Crossrefs

Cf. A001608, A014981, A074788, A294878, A294880.

Cf. also A293433, A293435.

Programs

PARI

A001608(n) = if(n<0, 0, polsym(x^3-x-1, n)[n+1]);
A294878(n) = { my(k=1,v); while((v=A001608(k))A294879(n) = sumdiv(n,d,(dA294878(d));

Formula

a(n) = Sum_{d|n, dA294878(d).

a(n) = A294880(n) - A294878(n).