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%I #8 Mar 19 2016 00:07:22
%S 1,1,2,2,2,2,3,1,3,4,2,5,2,3,4,2,3,4,5,1,4,3,3,4,3,4,5,5,3,6,5,3,3,6,
%T 2,4,6,3,9,4,2,3,4,3,7,6,3,6,2,4,2,6,5,7,6,4,5,3,6,4,11,1,5,9,3,6,5,3,
%U 8,8
%N Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.
%C Conjecture: (i) a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.
%C (ii) a(n) = 1 only for n = 1, 2, 8, 20, 62, 97, 296, 1493, 4283, 4346, 5433.
%C In contrast, the author conjectured in A262813 that any positive integer can be expressed as the sum of a nonnegative cube, a square and a positive triangular number.
%H Zhi-Wei Sun, <a href="/A270559/b270559.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.
%e a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.
%e a(2) = 1 since 2 = (-1)^4 + (-1)^3 + 1^2 + 1*2/2.
%e a(8) = 1 since 8 = 1^4 + 1^3 + 0^2 + 3*4/2.
%e a(20) = 1 since 20 = (-2)^4 + (-2)^3 + 3^2 + 2*3/2.
%e a(62) = 1 since 62 = (-2)^4 + (-2)^3 + 3^2 + 9*10/2.
%e a(97) = 1 since 97 = 1^4 + 1^3 + 2^2 + 13*14/2.
%e a(296) = 1 since 296 = (-4)^4 + (-4)^3 + 7^2 + 10*11/2.
%e a(1493) = 1 since 1493 = (-2)^4 + (-2)^3 + 0^2 + 54*55/2.
%e a(4283) = 1 since 4283 = (-6)^4 + (-6)^3 + 50^2 + 37*38/2.
%e a(4346) = 1 since 4346 = (-3)^4 + (-3)^3 + 49^2 + 61*62/2.
%e a(5433) = 1 since 5433 = (-8)^4 + (-8)^3 + 14^2 + 57*58/2.
%t TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
%t Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^4-x^3],r=r+1],{y,0,Sqrt[n]},{x,-1-Floor[(n-y^2)^(1/4)],(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,1,10000}]
%Y Cf. A000217, A000290, A000578, A000583, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.
%K nonn
%O 1,3
%A _Zhi-Wei Sun_, Mar 18 2016
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