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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]

Index entries for sequences related to sums of cubes

Index entries for related partition-counting sequences

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: A185119 A280130 A304002 * A287457 A185118 A240332

Adjacent sequences:  A279757 A279758 A279759 * A279761 A279762 A279763

KEYWORD

nonn

AUTHOR

Ilya Gutkovskiy, Dec 18 2016

STATUS

approved

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Last modified October 20 17:24 EDT 2018. Contains 316392 sequences. (Running on oeis4.)