A272336 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x^4 + y^3*z a square, where x,y,z,w are nonnegative integers with z > 0.

1, 3, 2, 1, 4, 5, 1, 3, 5, 5, 5, 3, 4, 7, 3, 1, 10, 10, 4, 5, 8, 5, 1, 5, 6, 12, 7, 1, 10, 7, 1, 3, 11, 8, 6, 6, 4, 11, 2, 5, 15, 11, 4, 5, 12, 6, 2, 3, 8, 15, 8, 4, 13, 16, 2, 7, 11, 6, 10, 4, 11, 13, 4, 1, 15, 20, 5, 10, 15, 9, 1, 10, 10, 18, 11, 4, 15, 9, 1, 5

Offset: 1

Author

Zhi-Wei Sun, Apr 26 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 23, 31, 71, 191, 271, 391, 503, 943, 1591, 2351, 2791, 4791, 8863, 9983, 4^k*m (k = 0,1,2,... and m = 1, 7, 79).

(ii) For each ordered pair (a,b) = (1,1), (1,15), (1,20), (1,36), (1,60), (9,260), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and w > 0 (or z > 0) such that a*x^4 + b*y^3*z is a square.

See also A280831 for a similar conjecture.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0^4 + 0^3*1 = 0^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1^4 + 2^3*1 = 3^2.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 1^4 + 2^3*3 = 5^2.
a(31) = 1 since 31 = 1^2 + 2^2 + 1^2 + 5^2 with 1^4 + 2^3*1 = 3^2.
a(71) = 1 since 71 = 5^2 + 6^2 + 1^2 + 3^2 with 5^4 + 6^3*1 = 29^2.
a(79) = 1 since 79 = 3^2 + 6^2 + 3^2 + 5^2 with 3^4 + 6^3*3 = 27^2.
a(191) = 1 since 191 = 7^2 + 6^2 + 5^2 + 9^2 with 7^4 + 6^3*5 = 59^2.
a(271) = 1 since 271 = 5^2 + 10^2 + 5^2 + 11^2 with 5^4 + 10^3*5 = 75^2.
a(391) = 1 since 391 = 9^2 + 6^2 + 15^2 + 7^2 with 9^4 + 6^3*15 = 99^2.
a(503) = 1 since 503 = 5^2 + 6^2 + 1^2 + 21^2 with
5^4 + 6^3*1 = 29^2.
a(943) = 1 since 943 = 6^2 + 3^2 + 27^2 + 13^2 with 6^4 + 3^3*27 = 45^2.
a(1591) = 1 since 1591 = 18^2 + 27^2 + 3^2 + 23^2 with 18^4 + 27^3*3 = 405^2.
a(2351) = 1 since 2351 = 6^2 + 45^2 + 13^2 + 11^2 with 6^4 + 45^3*13 = 1089^2.
a(2791) = 1 since 2791 = 19^2 + 38^2 + 19^2 + 25^2 with 19^4 + 38^3*19 = 1083^2.
a(4791) = 1 since 4791 = 9^2 + 2^2 + 41^2 + 55^2 with 9^4 + 2^3*41 = 83^2.
a(8863) = 1 since 8863 = 27^2 + 54^2 + 27^2 + 67^2 with 27^4 + 54^3*27 = 2187^2.
a(9983) = 1 since 9983 = 63^2 + 54^2 + 17^2 + 53^2 with 63^4 + 54^3*17 = 4293^2.
		

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.NT], 2016.
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190. (Cf. Conjecture 4.10(iv).)

Crossrefs

Cf. A000118, A000290, A262357, A268507, A269400, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x^4+y^3*z],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,1,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

Extensions

Typo in example fixed by Zak Seidov, Apr 26 2016