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

%I A286852 #12 Feb 16 2025 08:33:45
%S A286852 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,
%T A286852 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,
%U A286852 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
%N A286852 Number of partitions of n into unitary prime divisors of n.
%H A286852 Antti Karttunen, <a href="/A286852/b286852.txt">Table of n, a(n) for n = 0..2309</a>
%H A286852 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/UnitaryDivisor.html">Unitary Divisor</a>
%H A286852 <a href="/index/Par#part">Index entries for related partition-counting sequences</a>
%F A286852 a(n) = [x^n] Product_{p|n, p prime, gcd(p, n/p) = 1} 1/(1 - x^p).
%F A286852 a(n) = 0 if n is a powerful number (A001694).
%e A286852 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].
%t A286852 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}]]
%o A286852 (PARI)
%o A286852 A055231(n) = {my(f=factor(n)); for (k=1, #f~, if (f[k, 2] > 1, f[k, 2] = 0); ); factorback(f); } \\ From A055231
%o A286852 unitary_prime_factors(n) = { my(ufs = factor(A055231(n))); ufs[,1]~; };
%o A286852 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));
%o A286852 A286852(n) = if(n<2,1-n,partitions_into(n,vecsort(unitary_prime_factors(n), , 4))); \\ _Antti Karttunen_, Jul 02 2018
%Y A286852 Cf. A018818, A055231, A056169, A066874, A066882, A225244, A284289.
%K A286852 nonn
%O A286852 0,7
%A A286852 _Ilya Gutkovskiy_, Aug 01 2017