A286852 Number of partitions of n into unitary prime divisors of n.

1, 0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 1, 0, 2, 0, 1, 1, 21, 1, 0, 2, 2, 2, 0, 1, 2, 2, 1, 1, 28, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 5, 1, 2, 1, 0, 2, 42, 1, 1, 2, 43, 1, 0, 1, 2, 1, 1, 2, 49, 1, 1, 0, 2, 1, 5, 2, 2, 2, 1, 1, 10, 2, 1, 2, 2, 2

Offset: 0

Author

Ilya Gutkovskiy, Aug 01 2017

Keywords

• nonn

Examples

			a(6) = 2 because 6 has 4 divisors {1, 2, 3, 6} among which 2 are unitary prime divisors {2, 3} therefore we have [3, 3] and [2, 2, 2].
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..2309
• Eric Weisstein's World of Mathematics, Unitary Divisor
• Index entries for related partition-counting sequences

Crossrefs

Cf. A018818, A055231, A056169, A066874, A066882, A225244, A284289.

Programs

Mathematica

Join[{1}, Table[d = Divisors[n]; Coefficient[Series[Product[1/(1 - Boole[GCD[n/d[[k]], d[[k]]] == 1 && PrimeQ[d[[k]]]] x^d[[k]]), {k, Length[d]}], {x, 0, n}], x, n], {n, 1, 95}]]

PARI

A055231(n) = {my(f=factor(n)); for (k=1, #f~, if (f[k, 2] > 1, f[k, 2] = 0); ); factorback(f); } \\ From A055231
unitary_prime_factors(n) = { my(ufs = factor(A055231(n))); ufs[,1]~; };
partitions_into(n,parts,from=1) = if(!n,1,my(k = #parts, s=0); for(i=from,k,if(parts[i]<=n, s += partitions_into(n-parts[i],parts,i))); (s));
A286852(n) = if(n<2,1-n,partitions_into(n,vecsort(unitary_prime_factors(n), , 4))); \\ Antti Karttunen, Jul 02 2018

Formula

a(n) = [x^n] Product_{p|n, p prime, gcd(p, n/p) = 1} 1/(1 - x^p).

a(n) = 0 if n is a powerful number (A001694).