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 A279760 Expansion of Product_{k>=1} 1/(1 - x^(prime(k)^3)). 4
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0 COMMENTS Number of partitions of n into cubes of primes (A030078). LINKS Antti Karttunen, Table of n, a(n) for n = 0..2055 M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version] M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures] FORMULA G.f.: Product_{k>=1} 1/(1 - x^(prime(k)^3)). EXAMPLE a(35) = 1 because we have [27, 8]. 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 MATHEMATICA nmax = 120; CoefficientList[Series[Product[1/(1 - x^(Prime[k]^3)), {k, 1, nmax}], {x, 0, nmax}], x] PROG (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 CROSSREFS Cf. A000607, A003108, A030078, A078128, A090677. Sequence in context: A015241 A253513 A014025 * A287457 A285898 A185118 Adjacent sequences:  A279757 A279758 A279759 * A279761 A279762 A279763 KEYWORD nonn AUTHOR Ilya Gutkovskiy, Dec 18 2016 STATUS approved

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Last modified January 20 03:32 EST 2019. Contains 319323 sequences. (Running on oeis4.)