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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.
53

%I #44 Jul 05 2024 08:14:14

%S 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,

%T 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,

%U 1,5,8,7

%N 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.

%C 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]

%C (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]

%C (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.

%C 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.

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

%C See also A271510 and A271518 for related conjectures.

%H Zhi-Wei Sun, <a href="/A271513/b271513.txt">Table of n, a(n) for n = 0..10000</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190. Also available from <a href="http://arxiv.org/abs/1604.06723"> arXiv:1604.06723 [math.NT]</a>, 2016-2017.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]

%t 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}]

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

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Apr 09 2016