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A279329
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Number of partitions of n into distinct cubes.
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34
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1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
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OFFSET
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0
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COMMENTS
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In general, if m > 0 and g.f. = Product_{k>=1} (1 + x^(k^m)), then a(n) ~ exp((m+1) * ((2^(1/m)-1) * Gamma(1/m) * Zeta(1+1/m) / m^2)^(m/(m+1)) * (n/2)^(1/(m+1))) * ((2^(1/m)-1) * Gamma(1/m) * Zeta(1+1/m))^(m/(2*(m+1))) / (sqrt((m+1)*Pi) * 2^((2*m+3)/(2*(m+1))) * m^((m-1)/(2*(m+1))) * n^((2*m+1)/(2*(m+1)))).
a(12758) = 0 is the last zero in this sequence. - Antti Karttunen, Aug 30 2017
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LINKS
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FORMULA
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G.f.: Product_{k>=1} (1 + x^(k^3)).
a(n) ~ exp(2^(7/4) * 3^(-3/2) * ((2^(1/3)-1) * Gamma(1/3) * Zeta(4/3))^(3/4) * n^(1/4)) * ((2^(1/3)-1) * Gamma(1/3) * Zeta(4/3))^(3/8) / (2^(17/8) * 3^(1/4) * sqrt(Pi) * n^(7/8)).
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EXAMPLE
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a(9) = 1 because we have one solution, [8, 1].
a(216) = 2 because we have two solutions: 216 = 6^3 = 5^3 + 4^3 + 3^3. This is also the first point where the sequence obtains value larger than one. - Antti Karttunen, Aug 30 2017
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MATHEMATICA
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nmax = 10; CoefficientList[Series[Product[(1+x^(k^3)), {k, 1, nmax}], {x, 0, nmax^3}], x]
nmax = 10; poly = ConstantArray[0, nmax^3 + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[Do[poly[[j + 1]] += poly[[j - k^3 + 1]], {j, nmax^3, k^3, -1}]; , {k, 2, nmax}]; poly
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PROG
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(PARI) V279329=Vecsmall(prod(k=1, sqrtnint(#l=1+O(x^N=39800), 3), l+x^k^3)-1); A279329(n)=V279329[n+!n] \\ Needs stack of N*201 byte (allocatemem) to compute the series, only (N+1)*8 byte to store the vector. - M. F. Hasler, Jan 05 2020
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CROSSREFS
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Cf. A001476 (positions of zeros), A003997 (positions of nonzeros after a(0)).
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KEYWORD
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AUTHOR
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STATUS
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approved
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