

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
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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

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97120.


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[nx^32*y^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[(nx^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

ZhiWei Sun, Jul 17 2016


STATUS

approved



