cp's OEIS Frontend

A294333 Number of partitions of n into cubes dividing n.

%I A294333 #11 Jul 22 2018 08:48:24
%S A294333 1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,4,1,1,2,1,1,1,1,5,1,
%T A294333 1,1,1,1,1,1,6,1,1,1,1,1,1,1,7,1,1,1,1,1,3,1,8,1,1,1,1,1,1,1,10,1,1,1,
%U A294333 1,1,1,1,10,1,1,1,1,1,1,1,11,4,1,1,1,1,1,1,12,1,1,1,1,1,1,1,13,1,1,1,1,1,1,1,14,1
%N A294333 Number of partitions of n into cubes dividing n.
%H A294333 Antti Karttunen, <a href="/A294333/b294333.txt">Table of n, a(n) for n = 0..6640</a>
%H A294333 <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%H A294333 <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%F A294333 a(n) = 1 if n is a cubefree.
%F A294333 a(n) = 2 if n is a cube of prime.
%e A294333 a(8) = 2 because 8 has 4 divisors {1, 2, 4, 8} among which 2 are cubes {1, 8} therefore we have [8] and [1, 1, 1, 1, 1, 1, 1, 1].
%t A294333 Table[SeriesCoefficient[Product[1/(1 - Boole[Mod[n, k] == 0 && IntegerQ[k^(1/3)]] x^k), {k, 1, n}], {x, 0, n}], {n, 0, 105}]
%o A294333 (PARI)
%o A294333 cubes_dividing(n) = select(d -> ispower(d,3),divisors(n));
%o A294333 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 A294333 A294333(n) = if(n<2,1,partitions_into(n,vecsort(cubes_dividing(n), , 4))); \\ _Antti Karttunen_, Jul 21 2018
%Y A294333 Cf. A018818, A003108, A004709, A030078, A061704, A284345.
%K A294333 nonn
%O A294333 0,9
%A A294333 _Ilya Gutkovskiy_, Oct 28 2017