A272977 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 3*x^2*y + z^2*w a square, where w is a nonzero integer and x,y,z are nonnegative integers with x >= z.

2, 4, 1, 3, 8, 1, 1, 4, 3, 11, 3, 1, 9, 5, 3, 3, 10, 7, 6, 9, 3, 6, 1, 1, 11, 15, 4, 2, 13, 2, 2, 4, 4, 16, 5, 4, 13, 5, 2, 10, 12, 6, 5, 1, 12, 6, 1, 3, 7, 19, 2, 10, 10, 6, 3, 1, 2, 12, 7, 3, 15, 7, 4, 3, 16, 8, 6, 9, 5, 6, 1, 7, 12, 19, 3, 3, 7, 2, 4, 9

Offset: 1

Author

Zhi-Wei Sun, May 11 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 23, 47, 71, 147, 199, 263, 439, 16^k*m (k = 0,1,2,... and m = 6, 12, 24, 44, 56, 140, 156, 174, 204, 284, 4652).

(ii) For each ordered pair (a,b) = (7,1), (8,1), (9,2), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with a*x^2*y + b*z^2*w a square, where x,y,z are nonnegative integers and w is a nonzero integer.

Compare this conjecture with the one in A270073.

See arXiv:1604.06723 for more refinements of Lagrange's four-square theorem.

Examples

			a(1) = 2 since 1 = 1^2 + 0^2 + 0^2 + 1^2 with 1 > 0 and 3*1^2*0 + 0^2*1 = 0^2, and also 1 = 1^2 + 0^2 + 0^2 + (-1)^2 with 1 > 0 and 3*1^2*0 + 0^2*(-1) = 0^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1 = 1 and 3*1^2*0 + 1^2*1 = 1^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with 2 > 1 and 3*2^2*0 + 1^2*1 = 1^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + (-2)^2 with 1 = 1 and 3*1^2*1 + 1^2*(-2) = 1^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + (-3)^2 with 1 = 1 and 3*1^2*1 + 1*(-3) = 0^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 3^2 + (-2)^2 with 3 = 3 and 3*3^2*1 + 3^2*(-2) = 3^2.
a(24) = 1 since 24 = 2^2 + 0^2 + 2^2 + 4^2 with 2 = 2 and 3*2^2*0 + 2^2*4 = 4^2.
a(44) = 1 since 44 = 3^2 + 5^2 + 3^2 + 1^2 with 3 = 3 and 3*3^2*5 + 3^2*1 = 12^2.
a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + (-5)^2 with 3 = 3 and 3*3^2*2 + 3^2*(-5) = 3^2.
a(56) = 1 since 56 = 6^2 + 0^2 + 2^2 + 4^2 with 6 > 2 and 3*6^2*0 + 2^2*4 = 4^2.
a(71) = 1 since 71 = 5^2 + 6^2 + 3^2 + (-1)^2 with 5 > 3 and 3*5^2*6 + 3^2*(-1) = 21^2.
a(140) = 1 since 140 = 5^2 + 3^2 + 5^2 + (-9)^2 with 5 = 5 and 3*5^2*3 + 5^2*(-9) = 0^2.
a(147) = 1 since 147 = 11^2 + 0^2 + 5^2 + 1^2 with 11 > 5 and 3*11^2*0 + 5^2*1 = 5^2.
a(156) = 1 since 156 = 7^2 + 3^2 + 7^2 + 7^2 with 7 = 7 and 3*7^2*3 + 7^2*7 = 28^2.
a(174) = 1 since 174 = 13^2 + 0^2 + 2^2 + 1^2 with 13 > 2 and 3*13^2*0 + 2^2*1 = 2^2.
a(199) = 1 since 199 = 9^2 + 1^2 + 9^2 + 6^2 with 9 = 9 and 3*9^2*1 + 9^2*6 = 27^2.
a(204) = 1 since 204 = 1^2 + 9^2 + 1^2 + (-11)^2 with 1 = 1 and 3*1^2*9 + 1^2*(-11) = 4^2.
a(263) = 1 since 263 = 3^2 + 14^2 + 3^2 + 7^2 with 3 = 3 and
3*3^2*14 + 3^2*7 = 21^2.
a(284) = 1 since 284 = 13^2 + 3^2 + 5^2 + (-9)^2 with 13 > 5 and 3*13^2*3 + 5^2*(-9) = 36^2.
a(439) = 1 since 439 = 13^2 + 5^2 + 7^2 + (-14)^2 with 13 > 7 and 3*13^2*5 + 7^2*(-14) = 43^2.
a(4652) = 1 since 4652 = 11^2 + 21^2 + 11^2 + (-63)^2 with 11 = 11 and 3*11^2*21 + 11^2*(-63) = 0^2.
		

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.GM], 2016.
• Zhi-Wei Sun, Refine Lagrange's four-square theorem, a message to Number Theory List, April 26, 2016.

Crossrefs

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888.

Programs

Mathematica

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