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 A078128 Number of ways to write n as sum of cubes>1. 13
 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, 2, 0, 0, 1, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 1, 0, 2, 1, 0, 1, 0, 0, 1, 0, 2, 1, 0, 2, 0, 0, 1, 0, 2, 1, 0, 2, 0, 0, 1, 0, 2, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,64 COMMENTS a(A078129(n))=0; a(A078130(n))=1; a(A078131(n))>0; Conjecture (lower bound): for all k exists b(k) such that a(n)>k for n>b(k); see b(0)=A078129(83)=154 and b(1)=A078130(63)=218. LINKS Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 Eric Weisstein's World of Mathematics, Cubic Number. FORMULA a(n) = 1/n*Sum_{k=1..n} (b(k)-1)*a(n-k), a(0) = 1, where b(k) is sum of cube divisors of k. - Vladeta Jovovic, Nov 20 2002 From Vaclav Kotesovec, Jan 05 2017: (Start) a(n) = A003108(n) - A003108(n-1). a(n) ~ exp(4*(Gamma(1/3) * Zeta(4/3))^(3/4) * n^(1/4) / 3^(3/2)) * (Gamma(1/3) * Zeta(4/3))^(3/2) / (8 * 3^(5/2) * Pi^2 * n^2). (End) EXAMPLE a(160)=4: 160 = 20*2^3 = 4^3+12*2^3 = 2*4^3+4*2^3 = 5^3+3^3+2^3. MATHEMATICA nmax = 105; CoefficientList[Series[Product[1/(1 - x^(k^3)), {k, 1, nmax}], {x, 0, nmax}], x] // Differences (* Jean-François Alcover, Mar 01 2019, after Vaclav Kotesovec *) CROSSREFS Cf. A000578, A003108, A001235, A078132, A078133, A078134. Sequence in context: A291437 A302231 A228101 * A191269 A216602 A236233 Adjacent sequences:  A078125 A078126 A078127 * A078129 A078130 A078131 KEYWORD nonn AUTHOR Reinhard Zumkeller, Nov 19 2002 STATUS approved

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Last modified June 14 19:28 EDT 2021. Contains 345038 sequences. (Running on oeis4.)