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%I #31 Jan 06 2020 19:35:57
%S 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,
%T 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,
%U 0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0
%N Number of partitions of n into distinct cubes.
%C 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)))).
%C a(12758) = 0 is the last zero in this sequence. - _Antti Karttunen_, Aug 30 2017
%H Vaclav Kotesovec, <a href="/A279329/b279329.txt">Table of n, a(n) for n = 0..100000</a>
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%H <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>
%F G.f.: Product_{k>=1} (1 + x^(k^3)).
%F 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)).
%F For n >= 1, a(n) = A280130(n-1) + A280130(n). - _Antti Karttunen_, Aug 30 2017, after _Vaclav Kotesovec_'s Dec 26 2016 formula in the latter sequence.
%e a(9) = 1 because we have one solution, [8, 1].
%e 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
%t nmax = 10; CoefficientList[Series[Product[(1+x^(k^3)), {k, 1, nmax}], {x, 0, nmax^3}], x]
%t 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
%o (PARI) A279329(n,m=1) = { my(s=0); if(!n,1,for(c=m,n,if(ispower(c,3), s+=A279329(n-c,c+1))); (s)); }; \\ _Antti Karttunen_, Aug 30 2017
%o (PARI) apply( A279329(n,m=1)={if(n, sum(c=m,sqrtnint(n,3), A279329(n-c^3,c+1)), 1)}, [0..100]) \\ _M. F. Hasler_, Jan 05 2020
%o (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
%Y Cf. A000009, A003108, A033461, A259792, A279226, A280130, A280263.
%Y Cf. A001476 (positions of zeros), A003997 (positions of nonzeros after a(0)).
%Y Cf. A030272 (a(n^3)).
%K nonn,look
%O 0
%A _Vaclav Kotesovec_, Dec 10 2016
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