A271513 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 3*x^2 + 4*y^2 + 9*z^2 a square, where w, x, y and z are nonnegative integers.

1, 3, 2, 1, 4, 6, 3, 2, 2, 5, 6, 1, 2, 5, 4, 2, 4, 4, 3, 2, 6, 5, 1, 1, 3, 8, 6, 2, 4, 6, 6, 4, 2, 3, 8, 3, 7, 7, 1, 6, 6, 8, 6, 1, 2, 11, 7, 1, 2, 12, 8, 2, 7, 5, 9, 4, 4, 4, 7, 2, 4, 9, 4, 7, 4, 11, 6, 1, 5, 8, 7

Offset: 0

Author

Zhi-Wei Sun, Apr 09 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 11, 23, 43, 47, 67, 83, 107, 155, 323, 683, 803, 4^k*m (k = 0,1,2,... and m = 22, 38). [Conjecture verified for all natural numbers up to 10^9. - Mauro Fiorentini, Jul 04 2024]

(ii) Any natural number can be written as w^2 + x^2 + y^2 + z^2 with x, y, z integers and a*x^2 + b*y^2 + c*z^2 a square, whenever (a,b,c) is among the following triples: (1,3,12), (1,3,18), (1,3,21), (1,3,60), (1,5,15), (1,8,24), (1,12,15), (1,24,56), (1,24,72), (1,48,72), (1,48,168), (1,120,180), (1,192,288), (1,280,560), (3,9,13), (4,5,12), (4,5,60), (4,9,60), (4,12,21), (4,12,45), (4,12,69), (4,12,93), (4,12,237), (4,21,24), (4,21,36), (4,21,504), (4,24,93), (4,28,77), (4,45,120), (4,45,540), (4,45,600), (5,36,40), (7,9,126), (7,9,588), (8,16,73), (8,16,97), (8,49,112), (9,13,27), (9,16,24), (9,19,36), (9,21,91), (9,24,232), (9,28,63), (9,40,45), (9,40,56), (9,40,120), (9,45,115),(9,45,235), (12,13,24), (12,13,36), (12,36,37), (12,36,133), (13,36,72), (13,36,108), (15,24,25), (15,49,105), (16,17,48), (16,20,45), (16,21,84), (16,33,72), (16,33,176), (16,45,180), (16,48,57), (16,48,105), (16,48,233), (16,48,249), (19,45,57), (19,45,180), (21,25,35), (21,25,75), (21,28,36), (21,28,60), (21,43,105), (21,100,105),(24,25,72), (24,25,120), (24,48,97), (24,81,184), (24,120,145), (25,36,75), (25,40,56), (25,45,51), (25,45,99), (25,48,96), (25,48,144), (25,54,90), (25,75,81), (25,80,184), (25,96,120), (25,200,216), (28,33,36), (28,36,77), (28,72,189), (32,64,73), (33,36,220), (33,48,144), (33,72,256), (33,88,144), (36,45,100), (36,45,172), (37,81,243), (40,81,120), (40,81,240), (41,64,256), (45,48,76), (48,144,177), (49,56,64), (49,63,72), (55,141,165), (57,64,192), (60,105,196), (64,65,160), (72,73,144), (81,160,240), (85,140,196), (105,112,144), (112,144,153), (136,144,153), (144,145,240), (144,160,225),(148,189,252), (175,189,225). [Conjecture verified for all triples and all natural numbers up to 10^9. - Mauro Fiorentini, Jul 04 2024]

(iii) If a, b and c are positive integers such that any natural number can be written as w^2 + x^2 + y^2 + z^2 with x, y, z integers and a*x^2 + b*y^2 + c*z^2 a square, then a, b and c cannot be pairwise coprime.

This conjecture is stronger than Lagrange's four-square theorem. Moreover, there are many other suitable triples (a,b,c) for our purpose not listed in part (ii) of the conjecture. If a, b and c are positive integers such that any natural number can be written as w^2 + x^2 + y^2 + z^2 with x, y, z integers and a*x^2 + b*y^2 + c*z^2 a square, then one of a+b+c, 4*a+b+c, a+4*b+c and a+b+4*c must be a square since 2^2 + 1^2 + 1^2 + 1^2 is the unique way to express 7 as a sum of four squares.

Obviously, a(m^2*n) >= a(n) for all m,n = 1,2,3,....

See also A271510 and A271518 for related conjectures.

Examples

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^2 + 1^2 with 3*1^2 + 4*1^2 + 9*1^2 = 4^2.
a(11) = 1 since 11 = 1^2 + 3^2 + 0^2 + 1^2 with 3*3^2 + 4*0^2 + 9*1^2 = 6^2.
a(22) = 1 since 22 = 4^2 + 2^2 + 1^2 + 1^2 with 3*2^2 + 4*1^2 + 9*1^2 = 5^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 3*1^2 + 4*2^2 + 9*3^2 = 10^2.
a(38) = 1 since 38 = 0^2 + 6^2 + 1^2 + 1^2 with 3*6^2 + 4*1^2 + 9*1^2 = 11^2.
a(43) = 1 since 43 = 4^2 + 3^2 + 3^2 + 3^2 with 3*3^2 + 4*3^2 + 9*3^2 = 12^2.
a(47) = 1 since 47 = 3^2 + 6^2 + 1^2 + 1^2 with 3*6^2 + 4*1^2 + 9*1^2 = 11^2.
a(67) = 1 since 67 = 8^2 + 1^2 + 1^2 + 1^2 with 3*1^2 + 4*1^2 + 9*1^2 = 4^2.
a(83) = 1 since 83 = 0^2 + 9^2 + 1^2 + 1^2 with 3*9^2 + 4*1^2 + 9*1^2 = 16^2.
a(107) = 1 since 107 = 9^2 + 3^2 + 4^2 + 1^2 with 3*3^2 + 4*4^2 + 9*1^2 = 10^2.
a(155) = 1 since 155 = 0^2 + 9^2 + 5^2 + 7^2 with 3*9^2 + 4*5^2 + 9*7^2 = 28^2.
a(323) = 1 since 323 = 3^2 + 15^2 + 8^2 + 5^2 with 3*15^2 + 4*8^2 + 9*5^2 = 34^2.
a(683) = 1 since 683 = 15^2 + 11^2 + 16^2 + 9^2 with 3*11^2 + 4*16^2 + 9*9^2 = 46^2.
a(803) = 1 since 803 = 24^2 + 13^2 + 7^2 + 3^2 with 3*13^2 + 4*7^2 + 9*3^2 = 28^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. Also available from arXiv:1604.06723 [math.NT], 2016-2017.

Crossrefs

Cf. A000118, A000290, A270969, A271510, A271518.

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+4y^2+9z^2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,0,70}]