A272351 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x^4 + 8*y*z*(y^2+z^2) a fourth power, where w,x,y,z are nonnegative integers with w >= x and y > z.
1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 2, 3, 2, 1, 4, 5, 2, 2, 3, 3, 2, 2, 3, 5, 4, 1, 5, 6, 1, 1, 5, 4, 5, 3, 2, 5, 2, 3, 7, 7, 3, 2, 5, 4, 2, 1, 5, 8, 7, 2, 5, 9, 1, 3, 4, 4, 5, 2, 5, 8, 6, 1, 8, 8, 4, 4, 6, 5, 1, 5, 5, 10, 6, 2, 6, 8, 1, 2
Offset: 1
Keywords
Examples
a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 = 0, 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4. a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4. a(3) = 1 since 3 = 1^2 + 1^2 + 1^2 + 0^2 with 1 = 1, 1 > 0 and 1^4 + 8*1*0*(1^2+0^2) = 1^4. a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1 = 1, 3 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4. a(31) = 1 since 31 = 5^2 + 1^2 + 2^2 + 1^2 with 5 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4. a(55) = 1 since 55 = 7^2 + 1^2 + 2^2 + 1^2 with 7 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4. a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 3 > 1, 6 > 5 and 1^4 + 8*6*5*(6^2+5^2) = 11^4. a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5 > 3, 6 > 3 and 3^4 + 8*6*3*(6^2+3^2) = 9^4. a(151) = 1 since 151 = 5^2 + 3^2 + 9^2 + 6^2 with 5 > 3, 9 > 6 and 3^4 + 8*9*6*(9^2+6^2) = 15^4. a(191) = 1 since 191 = 3^2 + 1^2 + 10^2 + 9^2 with 3 > 1, 10 > 9 and 1^4 + 8*10*9*(10^2+9^2) = 19^4.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
- Zhi-Wei Sun, Refine Lagrange's four-square theorem, a message to Number Theory List, April 26, 2016.
Crossrefs
Programs
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Mathematica
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]] QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)] Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&QQ[x^4+8y*z*(y^2+z^2)],r=r+1],{z,0,(Sqrt[2n-1]-1)/2},{y,z+1,Sqrt[n-z^2]},{x,0,Sqrt[(n-y^2-z^2)/2]}];Print[n," ",r];Continue,{n,1,80}]
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