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%I A255191 #11 Feb 16 2015 11:20:14
%S A255191 1,1,1,1,2,1,3,2,1,1,2,2,2,2,1,2,4,2,2,2,4,2,4,1,2,4,2,2,2,3,3,4,3,1,
%T A255191 2,2,4,4,4,2,4,2,5,4,1,3,6,4,2,3,3,3,5,2,2,5,5,3,3,3,3,4,3,1,5,5,4,6,
%U A255191 2,2,6,4,6,5,4,2,6,3,3,3,5,5,6,3,2,7,2,4,4,2,5,6,6,3,4,5,2,6,3,2,6
%N A255191 Number of solutions to the equation n = x^2 + y*(y+1)/2 + z*(z+1)/2 (x,y,z=0,1,...) with x == Floor[sqrt(n)] (mod 2) and y <= z.
%C A255191 By Dickson's book in the reference, E. Lionnet observed in 1872 that any nonnegative integer can be written as the sum of a square and two triangular numbers.
%C A255191 We have a(n) > 0 for all n. This follows from Theorem 1(i) and Lemma 2 of Sun's reference in 2007. Note that if n = m*(m+1) with m a nonnegative integer then floor(sqrt(n)) = m.
%D A255191 L. E. Dickson, History of the Theory of Numbers, vol. II, AMS Chelsea Publ., 1999, p. 24.
%H A255191 Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
%H A255191 Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
%H A255191 Zhi-Wei Sun, On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z), arXiv:1502.03056 [math.NT], 2015.
%e A255191 a(14) = 1 since 14 = 1^2 + 2*3/2 + 4*5/2 with floor(sqrt(14)) == 1 (mod 2).
%e A255191 a(44) = 1 since 44 = 4^2 + 0*1/2 + 7*8/2 with floor(sqrt(44)) == 4 (mod 2).
%e A255191 a(63) = 1 since 63 = 5^2 + 4*5/2 + 7*8/2 with floor(sqrt(63)) == 5 (mod 2).
%t A255191 TQ[n_]:=IntegerQ[Sqrt[8n+1]]
%t A255191 Do[r=0;Do[If[Mod[Floor[Sqrt[n]]-x,2]==0&&TQ[n-x^2-y*(y+1)/2], r=r+1],{x,0,Sqrt[n]},{y,0,(Sqrt[4(n-x^2)+1]-1)/2}];
%t A255191 Print[n," ",r];Label[aa];Continue,{n,0,100}]
%Y A255191 Cf. A000217, A000290, A254885.
%K A255191 nonn
%O A255191 0,5
%A A255191 _Zhi-Wei Sun_, Feb 16 2015
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