A282494 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y*(y+240*z) a positive square, where x,y,z,w are nonnegative integers.
1, 2, 1, 1, 4, 4, 1, 1, 3, 5, 4, 1, 3, 6, 3, 1, 6, 7, 3, 5, 9, 5, 1, 2, 6, 11, 7, 1, 7, 9, 2, 2, 6, 5, 5, 7, 7, 4, 1, 4, 10, 11, 3, 1, 9, 8, 2, 1, 5, 10, 8, 7, 10, 10, 4, 6, 8, 5, 4, 3, 9, 11, 4, 1, 11, 12, 4, 7, 13, 10, 2, 5, 5, 7, 7, 3, 10, 9, 1, 4
Offset: 1
Keywords
Examples
a(3) = 1 since 3 = 1^4 + 1^2 + 0^2 + 1^2 with 1*(1+240*0) = 1^2. a(4) = 1 since 4 = 0^4 + 2^2 + 0^2 + 0^2 with 2*(2+240*0) = 2^2. a(39) = 1 since 39 = 1^4 + 2^2 + 3^2 + 5^2 with 2*(2+240*3) = 38^2. a(188) = 1 since 188 = 3^4 + 5^2 + 1^2 + 9^2 with 5*(5+240*1) = 35^2. a(399) = 1 since 399 = 3^4 + 10^2 + 7^2 + 13^2 with 10*(10+240*7) = 130^2. a(428) = 1 since 428 = 0^4 + 10^2 + 2^2 + 18^2 with 10*(10+240*2) = 70^2. a(439) = 1 since 439 = 1^4 + 10^2 + 7^2 + 17^2 with 10*(10+240*7) = 130^2. a(508) = 1 since 508 = 1^4 + 5^2 + 11^2 + 19^2 with 5*(5+240*11) = 115^2. a(748) = 1 since 748 = 3^4 + 1^2 + 21^2 + 15^2 with 1*(1+240*21) = 71^2. a(1468) = 1 since 1468 = 2^4 + 10^2 + 26^2 + 26^2 with 10*(10+240*26) = 250^2. a(2828) = 1 since 2828 = 3^4 + 5^2 + 11^2 + 51^2 with 5*(5+240*11) = 115^2.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- 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.
Crossrefs
Programs
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Mathematica
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y*(y+240*z)],r=r+1],{x,0,(n-1)^(1/4)},{y,1,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,1,80}]
Comments