

A262813


Number of ordered ways to write n as x^3 + y^2 + z*(z+1)/2 with x >= 0, y >=0 and z > 0.


44



1, 2, 2, 2, 2, 2, 3, 2, 1, 4, 5, 3, 2, 2, 5, 3, 2, 4, 4, 4, 1, 4, 4, 2, 3, 3, 5, 3, 5, 5, 4, 5, 3, 4, 1, 4, 9, 6, 4, 4, 3, 3, 3, 3, 7, 8, 4, 3, 3, 3, 3, 5, 7, 5, 5, 4, 4, 4, 4, 4, 3, 4, 3, 8, 6, 4, 8, 3, 4, 5, 8, 7, 5, 5, 5, 3, 2, 8, 8, 6, 4, 7, 8, 2, 5, 7, 4, 6, 2, 5, 7, 10, 6, 5, 7, 3, 5, 1, 6, 5
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OFFSET

1,2


COMMENTS

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 21, 35, 98, 152, 306.
This has been verified for all n = 1..2*10^7.
In contrast with the conjecture, in 2015 the author refined a result of Euler by proving that any positive integer can be written as the sum of two squares and a positive triangular number.
See also A262815, A262816 and A262941 for similar conjectures.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103113.
ZhiWei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275283.


EXAMPLE

a(1) = 1 since 1 = 0^3 + 0^2 + 1*2/2.
a(2) = 2 since 2 = 0^3 + 1^2 + 1*2/2 = 1^3 + 0^2 + 1*2/2.
a(6) = 2 since 6 = 0^3 + 0^2 + 3*4/2 = 1^3 + 2^2 + 1*2/2.
a(9) = 1 since 9 = 2^3 + 0^2 + 1*2/2.
a(21) = 1 since 21 = 0^3 + 0^2 + 6*7/2.
a(35) = 1 since 35 = 0^3 + 5^2 + 4*5/2.
a(98) = 1 since 98 = 3^3 + 4^2 + 10*11/2.
a(152) = 1 since 152 = 0^3 + 4^2 + 16*17/2.
a(306) = 1 since 306 = 1^3 + 13^2 + 16*17/2.


MATHEMATICA

TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[nx^3y^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[nx^3]}]; Print[n, " ", r]; Continue, {n, 1, 100}]


CROSSREFS

Cf. A000217, A000290, A000578, A254885, A262785, A262815, A262816, A262941.
Sequence in context: A054990 A046921 A262954 * A188794 A161966 A187188
Adjacent sequences: A262810 A262811 A262812 * A262814 A262815 A262816


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 03 2015


STATUS

approved



