%I #30 Jul 21 2023 09:15:19

%S 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,

%T 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,

%U 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

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

%C Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 21, 35, 98, 152, 306.

%C This has been verified for all n = 1..2*10^7.

%C Conjecture verified up to 10^11. - _Mauro Fiorentini_, Jul 18 2023

%C If z >= 0, a(n) = 1 only for n = 21, 35, 98, 306. - _Mauro Fiorentini_, Jul 20 2023

%C 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.

%C See also A262815, A262816 and A262941 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A262813/b262813.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127(2007), 103-113.

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.07.024">On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d)</a>, J. Number Theory 171(2017), 275-283.

%e a(1) = 1 since 1 = 0^3 + 0^2 + 1*2/2.

%e a(2) = 2 since 2 = 0^3 + 1^2 + 1*2/2 = 1^3 + 0^2 + 1*2/2.

%e a(6) = 2 since 6 = 0^3 + 0^2 + 3*4/2 = 1^3 + 2^2 + 1*2/2.

%e a(9) = 1 since 9 = 2^3 + 0^2 + 1*2/2.

%e a(21) = 1 since 21 = 0^3 + 0^2 + 6*7/2.

%e a(35) = 1 since 35 = 0^3 + 5^2 + 4*5/2.

%e a(98) = 1 since 98 = 3^3 + 4^2 + 10*11/2.

%e a(152) = 1 since 152 = 0^3 + 4^2 + 16*17/2.

%e a(306) = 1 since 306 = 1^3 + 13^2 + 16*17/2.

%t TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]

%t Do[r=0;Do[If[TQ[n-x^3-y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000217, A000290, A000578, A254885, A262785, A262815, A262816, A262941.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Oct 03 2015