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.
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, 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, 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
Offset: 0
Keywords
Examples
a(14) = 1 since 14 = 1^2 + 2*3/2 + 4*5/2 with floor(sqrt(14)) == 1 (mod 2). a(44) = 1 since 44 = 4^2 + 0*1/2 + 7*8/2 with floor(sqrt(44)) == 4 (mod 2). a(63) = 1 since 63 = 5^2 + 4*5/2 + 7*8/2 with floor(sqrt(63)) == 5 (mod 2).
References
- L. E. Dickson, History of the Theory of Numbers, vol. II, AMS Chelsea Publ., 1999, p. 24.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
- Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
- 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.
Programs
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Mathematica
TQ[n_]:=IntegerQ[Sqrt[8n+1]] 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}]; Print[n," ",r];Label[aa];Continue,{n,0,100}]
Comments