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 A275150 Number of ordered ways to write n as x^3 + 2*y^2 + k*z^2, where x,y,z are nonnegative integers, k is 1 or 5, and k = 1 if z = 0. 3
 1, 2, 2, 2, 2, 2, 2, 2, 3, 4, 3, 2, 3, 3, 2, 1, 2, 4, 3, 4, 3, 2, 2, 3, 3, 3, 3, 4, 5, 2, 3, 2, 3, 5, 4, 4, 5, 3, 4, 3, 2, 3, 2, 2, 5, 5, 4, 2, 2, 5, 3, 5, 5, 3, 5, 5, 2, 3, 3, 4, 4, 2, 2, 4, 4, 6, 3, 5, 4, 2, 3, 4, 5, 5, 4, 4, 5, 5, 5, 1, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 15, 79, 120, 218, 399, 454, 622, 725, 3240. We have verified that a(n) > 0 for all n = 0..10^7. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..10000 Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. EXAMPLE a(0) = 1 since 0 = 0^3 + 2*0^2 + 0^2. a(15) = 1 since 15 = 2^3 + 2*1^2 + 5*1^2. a(79) = 1 since 79 = 3^3 + 2*4^2 + 5*2^2. a(120) = 1 since 120 = 2^3 + 2*4^2 + 5*4^2. a(218) = 1 since 218 = 6^3 + 2*1^2 + 0^2. a(399) = 1 since 399 = 5^3 + 2*3^2 + 16^2. a(454) = 1 since 454 = 0^3 + 2*15^2 + 2^2. a(622) = 1 since 622 = 2^3 + 2*17^2 + 6^2. a(725) = 1 since 725 = 5^3 + 2*10^2 + 20^2. a(3240) = 1 since 3240 = 7^3 + 2*38^2 + 3^2. MATHEMATICA SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]] TQ[n_]:=TQ[n]=SQ[n]||SQ[n/5] Do[r=0; Do[If[TQ[n-x^3-2*y^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[(n-x^3)/2]}]; Print[n, " ", r]; Continue, {n, 0, 80}] CROSSREFS Cf. A000290, A000578,  A262813, A262941, A262954, A270488, A274274, A275083. Sequence in context: A292027 A025845 A029393 * A303904 A173021 A109703 Adjacent sequences:  A275147 A275148 A275149 * A275151 A275152 A275153 KEYWORD nonn AUTHOR Zhi-Wei Sun, Jul 17 2016 STATUS approved

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Last modified July 28 10:51 EDT 2021. Contains 346326 sequences. (Running on oeis4.)