A268507 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w > 0, w >= x <= y <= z such that x^2*y^2 + y^2*z^2 + z^2*x^2 is a square, where w,x,y,z are nonnegative integers.

1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 2, 3, 2, 1, 4, 4, 2, 2, 3, 3, 1, 2, 3, 5, 4, 1, 5, 5, 1, 1, 5, 4, 4, 3, 2, 5, 1, 3, 7, 6, 3, 2, 5, 4, 1, 1, 5, 7, 6, 2, 5, 8, 1, 3, 4, 3, 5, 2, 5, 7, 4, 1, 8, 8, 3, 4, 6, 6, 1, 4, 6, 9, 5, 2, 6, 7, 1, 2

Offset: 1

Author

Zhi-Wei Sun, Apr 16 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k, 4^k*m (k = 0,1,2,... and m = 3, 7, 23, 31, 39, 47, 55, 71, 79, 151, 191, 551).

(ii) For each triple (a,b,c) = (1,4,4), (1,4,16), (1,4,26), (1,4,31), (1,4,34), (1,9,9), (1,9,11), (1,9,17), (1,9,21), (1,9,27), (1,9,33), (1,9,41), (1,18,24), (1,36,44), (3,4,8), (4,6,9), (4,8,19), (4,8,27), (4,9,36), (4,16,41), (4,19,29), (5,9,25), (7,9,33), (7,25,49), (9,10,45), (9,12,28), (9,16,36), (9,21,49), (9,24,37), (9,25,27), (9,25,45), (9,30,40), (9,32,64), (9,34,36), (9,44,61), (14,25,40), (16,17,36), (16,20,25), (24,36,39), (25,40,64), (25,45,51), (27,36,37), (28,44,49), (32,49,64), (36,43,45), (36,54,58), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that a*x^2*y^2 + b*y^2*z^2 + c*z^2*x^2 is a square.

See also A269400, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

The author has proved in arXiv:1604.06723 that a(n) > 0 for any positive integer n. - Zhi-Wei Sun, May 09 2016

Examples

			a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 > 0 = 0 < 1 and 0^2*0^2 + 0^2*1^2 + 1^2*0^2 = 0^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1 > 0 < 1 = 1 and 0^2*1^2 + 1^2*1^2 + 1^2*0^2 = 1^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 = 1 = 1 < 2 and 1^2*1^2 + 1^2*2^2 + 2^2*1^2 = 3^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 3 > 1 < 2 < 3 and 1^2*2^2 + 2^2*3^2 + 3^2*1^2 = 7^2.
a(31) = 1 since 31 = 5^2 + 1^2 + 1^2 + 2^2 with 5 > 1 = 1 < 2 and 1^2*1^2 + 1^2*2^2 + 2^2*1^2 = 3^2.
a(39) = 1 since 39 = 5^2 + 1^2 + 2^2 + 3^2 with 5 > 1 < 2 < 3 and 1^2*2^2 + 2^2*3^2 + 3^2*1^2 = 7^2.
a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + 5^2 with 3 > 2 < 3 < 5 and 2^2*3^2 + 3^2*5^2 + 5^2*2^2 = 19^2.
a(55) = 1 since 55 = 7^2 + 1^2 + 1^2 + 2^2 with 7 > 1 = 1 < 2 and 1^2*1^2 + 1^2*2^2 + 2^2*1^2 = 3^2.
a(71) = 1 since 71 = 3^2 + 1^2 + 5^2 + 6^2 with 3 > 1 < 5 < 6 and 1^2*5^2 + 5^2*6^2 + 6^2*1^2 = 31^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 3^2 + 6^2 with 5 > 3 = 3 < 6 and 3^2*3^2 + 3^2*6^2 + 6^2*3^2 = 27^2.
a(151) = 1 since 151 = 5^2 + 3^2 + 6^2 + 9^2 with 5 > 3 < 6 < 9 and 3^2*6^2 + 6^2*9^2 + 9^2*3^2 = 63^2.
a(191) = 1 since 191 = 3^2 + 1^2 + 9^2 + 10^2 with 3 > 1 < 9 < 10 and 1^2*9^2 + 9^2*10^2 + 10^2*1^2 = 91^2.
a(551) = 1 since 551 = 15^2 + 3^2 + 11^2 + 14^2 with 15 > 3 < 11 < 14 and 3^2*11^2 + 11^2*14^2 + 14^2*3^2 = 163^2.
		

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], 2016-2017.

Crossrefs

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

Programs

Mathematica

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