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%I A279760 #12 Aug 31 2017 04:36:16
%S A279760 1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,
%T A279760 0,1,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,
%U A279760 0,0,1,0,1,0,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1
%N A279760 Expansion of Product_{k>=1} 1/(1 - x^(prime(k)^3)).
%C A279760 Number of partitions of n into cubes of primes (A030078).
%H A279760 Antti Karttunen, <a href="/A279760/b279760.txt">Table of n, a(n) for n = 0..2055</a>
%H A279760 M. Bernstein and N. J. A. Sloane, <a href="http://arXiv.org/abs/math.CO/0205301">Some canonical sequences of integers</a>, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
%H A279760 M. Bernstein and N. J. A. Sloane, <a href="/A003633/a003633_1.pdf">Some canonical sequences of integers</a>, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
%H A279760 <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%H A279760 <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>
%F A279760 G.f.: Product_{k>=1} 1/(1 - x^(prime(k)^3)).
%e A279760 a(35) = 1 because we have [27, 8].
%e A279760 For n = 152, there are two solutions: 152 = 5^3 + 3^3 = 19 * 2^3, thus a(152) = 2. This is also the first point where the sequence obtains value larger than one. - _Antti Karttunen_, Aug 31 2017
%t A279760 nmax = 120; CoefficientList[Series[Product[1/(1 - x^(Prime[k]^3)), {k, 1, nmax}], {x, 0, nmax}], x]
%o A279760 (PARI) A279760(n,m=8) = { my(s=0,p); if(!n,1,for(c=m,n,if((ispower(c,3,&p)&&isprime(p)), s+=A279760(n-c,c))); (s)); }; \\ _Antti Karttunen_, Aug 31 2017
%Y A279760 Cf. A000607, A003108, A030078, A078128, A090677.
%K A279760 nonn
%O A279760 0
%A A279760 _Ilya Gutkovskiy_, Dec 18 2016

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