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 A281826 Number of ways to write n as x^3 + 2*y^3 + 3*z^3 + p(k), where x,y,z are nonnegative integers, k is a positive integer, and p(.) is the partition function given by A000041. 1
 1, 2, 3, 4, 5, 5, 5, 5, 4, 4, 5, 4, 5, 4, 5, 4, 4, 6, 4, 4, 5, 5, 4, 4, 6, 6, 7, 8, 6, 8, 8, 9, 7, 8, 9, 4, 5, 5, 6, 3, 6, 7, 6, 4, 6, 6, 7, 5, 7, 4, 4, 4, 4, 7, 6, 8, 7, 7, 8, 6, 5, 8, 4, 5, 5, 7, 6, 8, 11, 7, 5, 7, 6, 5, 3, 6, 4, 4, 4, 9 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: (i) a(n) > 0 for all n > 0. (ii) Any positive integer n can be written as x^3 + 2*y^3 + 4*z^3 + p(k) with x,y,z nonnegative integers and k a positive integer. (iii) For each c = 3, 4, any positive integer n can be written as x^3 + 2*y^3 + c*z^3 + A000009(k) with x,y,z nonnegative and k a positive integer. We have verified the conjecture for n up to 1.3*10^6. On the author's request, Prof. Qing-Hu Hou at Tianjin Univ. has verified all the three parts of the conjecture for n up to 10^9. - Zhi-Wei Sun, Feb 06 2017 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(1) = 1 since 1 = 0^3 + 2*0^3 + 3*0^3 + p(1). a(2) = 2 since 2 = 1^3 + 2*0^3 + 3*0^3 + p(1) = 0^3 + 2*0^3 + 3*0^3 + p(2). a(75) = 3 since 75 = 4^3 + 2*0^3 + 3*0^3 + p(6) = 3^3 + 2*1^3 + 3*2^3 + p(8) = 0^3 + 2*2^3 + 3*1^3 + p(11). MATHEMATICA CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)] p[n_]:=p[n]=PartitionsP[n] Do[r=0; Do[If[p[k]>n, Goto[bb]]; Do[If[CQ[n-p[k]-3x^3-2y^3], r=r+1], {x, 0, ((n-p[k])/3)^(1/3)}, {y, 0, ((n-p[k]-3x^3)/2)^(1/3)}]; Continue, {k, 1, n}]; Label[bb]; Print[n, " ", r]; Continue, {n, 1, 80}] CROSSREFS Cf. A000009, A000041, A000578, A280455. Sequence in context: A062186 A085763 A273264 * A062985 A168092 A210032 Adjacent sequences:  A281823 A281824 A281825 * A281827 A281828 A281829 KEYWORD nonn AUTHOR Zhi-Wei Sun, Jan 31 2017 STATUS approved

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Last modified May 11 13:41 EDT 2021. Contains 343791 sequences. (Running on oeis4.)