A275344 Number of ordered ways to write n as x^2 + y^2 + z^2 + 2*w^2 with x + 2*y + 3*z a square, where x,y,z,w are nonnegative integers.
1, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 2, 2, 2, 1, 1, 1, 3, 5, 3, 4, 3, 3, 2, 4, 1, 4, 3, 3, 4, 1, 4, 3, 1, 4, 3, 3, 8, 3, 2, 3, 2, 3, 2, 3, 3, 3, 4, 2, 2, 9, 3, 8, 7, 5, 5, 4, 2, 6, 4, 4, 9, 4, 4, 5, 4, 3, 8, 6, 5, 6, 5, 5, 5, 4, 2, 5, 5, 4, 6, 4
Offset: 0
Keywords
Examples
a(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 2*0^2 with 0 + 2*0 + 3*0 = 0^2. a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 2*0^2 with 1 + 2*0 + 3*0 = 1^2. a(3) = 1 since 3 = 1^2 + 0^2 + 0^2 + 2*1^2 with 1 + 2*0 + 3*0 = 1^2. a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 2*0^2 with 2 + 2*1 + 3*0 = 2^2. a(7) = 1 since 7 = 2^2 + 1^2 + 0^2 + 2*1^2 with 2 + 2*1 + 3*0 = 2^2. a(14) = 1 since 14 = 1^2 + 1^2 + 2^2 + 2*2^2 with 1 + 2*1 + 3*2 = 3^2. a(15) = 1 since 15 = 3^2 + 0^2 + 2^2 + 2*1^2 with 3 + 2*0 + 3*2 = 3^2. a(16) = 1 since 16 = 4^2 + 0^2 + 0^2 + 2*0^2 with 4 + 2*0 + 3*0 = 2^2. a(25) = 1 since 25 = 1^2 + 4^2 + 0^2 + 2*2^2 with 1 + 2*4 + 3*0 = 3^2. a(30) = 1 since 30 = 3^2 + 2^2 + 3^2 + 2*2^2 with 3 + 2*2 + 3*3 = 4^2. a(33) = 1 since 33 = 1^2 + 0^2 + 0^2 + 2*4^2 with 1 + 2*0 + 3*0 = 1^2. a(84) = 1 since 84 = 4^2 + 6^2 + 0^2 + 2*4^2 with 4 + 2*6 + 3*0 = 4^2. a(169) = 1 since 169 = 10^2 + 6^2 + 1^2 + 2*4^2 with 10 + 2*6 + 3*1 = 5^2. a(225) = 1 since 225 = 10^2 + 6^2 + 9^2 + 2*2^2 with 10 + 2*6 + 3*9 = 7^2.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
- Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
Programs
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Mathematica
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]] Do[r=0;Do[If[SQ[n-2*w^2-x^2-y^2]&&SQ[x+2*y+3*Sqrt[n-2w^2-x^2-y^2]],r=r+1],{w,0,Sqrt[n/2]},{x,0,Sqrt[n-2*w^2]},{y,0,Sqrt[n-2*w^2-x^2]}];Print[n," ",r];Continue,{n,0,80}]
Comments