A283239 Number of ways to write n as x^2 + y^2 + z*(3*z-1)/2 with x,y,z integers such that x + 2*y is a square.
1, 2, 3, 2, 2, 5, 5, 5, 1, 2, 4, 2, 5, 3, 3, 4, 3, 7, 5, 3, 8, 3, 5, 3, 2, 6, 3, 7, 5, 3, 4, 7, 5, 3, 4, 6, 5, 3, 3, 7, 5, 5, 5, 1, 6, 7, 6, 4, 3, 2, 5, 5, 9, 6, 3, 7, 5, 7, 5, 4, 8, 5, 6, 4, 6, 5, 5, 8, 5, 6, 6, 7, 5, 5, 5, 7, 5, 6, 2, 4, 12
Offset: 0
Keywords
Examples
a(8) = 1 since 8 = 1^2 + 0^2 + (-2)*(3*(-2)-1)/2 with 1 + 2*0 = 1^2. a(43) = 1 since 43 = 1^2 + 4^2 + (-4)*(3*(-4)-1)/2 with 1 + 2*4 = 3^2. a(84) = 1 since 84 = 7^2 + (-3)^2 + (-4)*(3*(-4)-1)/2 with 7 + 2*(-3) = 1^2. a(133) = 1 since 133 = 4^2 + 0^2 + 9*(3*9-1)/2 with 4 + 2*0 = 2^2. a(253) = 1 since 253 = (-13)^2 + 7^2 + 5*(3*5-1)/2 with (-13) + 2*7 = 1^2. a(399) = 1 since 399 = 18^2 + (-7)^2 + (-4)*(3*(-4)-1)/2 with 18 + 2*(-7) = 2^2. a(488) = 1 since 488 = 9^2 + 20^2 + (-2)*(3*(-2)-1)/2 with 9 + 2*20 = 7^2. a(523) = 1 since 523 = 9^2 + 0^2 + (-17)*(3*(-17)-1)/2 with 9 + 2*0 = 3^2. a(803) = 1 since 803 = (-17)^2 + 13^2 + (-15)*(3*(-15)-1)/2 with (-17) + 2*13 = 3^2. a(7369) = 1 since 7369 = 0^2 + 72^2 + (-38)*(3*(-38)-1)/2 with 0 + 2*72 = 12^2.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
- Ju. V. Linnik, An asymptotic formula in an additive problem of Hardy-Littlewood, Izv. Akad. Nauk SSSR Ser. Mat., 24 (1960), 629-706 (Russian).
- Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), no. 7, 1367-1396.
- Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190.
- Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
Programs
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Mathematica
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; PenQ[n_]:=PenQ[n]=SQ[24n+1]; Do[r=0;Do[If[PenQ[n-x^2-y^2],Do[If[SQ[(-1)^i*x+2(-1)^j*y],r=r+1],{i,0,Min[x,1]},{j,0,Min[y,1]}]],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]}];Print[n," ",r];Continue,{n,0,80}]
Comments