A269400 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 6*w^2*x^2 + 12*x^2*y^2 + 52*y^2*z^2 + 27*z^2*w^2 a square, where w,x,y are nonnegative integers and z is a positive integer.

1, 1, 1, 1, 2, 2, 2, 1, 3, 3, 2, 3, 4, 3, 1, 1, 4, 5, 2, 3, 3, 4, 4, 2, 5, 5, 2, 5, 5, 2, 1, 1, 3, 6, 2, 3, 4, 8, 1, 3, 8, 7, 3, 3, 4, 5, 2, 3, 6, 9, 4, 6, 10, 4, 3, 3, 3, 8, 5, 4, 5, 5, 5, 1, 7, 4, 2, 7, 4, 5, 1, 5, 7, 5, 2, 4, 8, 1, 1, 3

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 = 3, 15, 31, 39, 71, 79, 195, 311, 319, 403, 559, 591, 683, 719, 1031, 1439, 1643, 2519, 6879, 2^k, 2^(2k+1)*39 (k = 0,1,2,...). Also, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x a positive integer and w,y,z nonnegative integer such that 6*w^2*x^2 + 12*x^2*y^2 + 52*y^2*z^2 + 27*z^2*w^2 is a square.

(ii) For each triple (a,b,c) = (1,3,2), (1,11,9), (1,14,4),(1,20,25), (1,27,18), (1,36,9), (1,56,4), (4,32,25), (9,15,25), (9,35,25), (25,8,64), (25,15,54), (25,32,28), (25,35,49), (28,32,49), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that a*w^2*x^2 + b*x^2*y^2 + c*y^2*z^2 is a square.

See also A268507, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 1 > 0 and 6*0^2*0^2 + 12*0^2*0^2 + 52*0^2*1^2 + 27*1^2*0^2 = 0^2.
a(2) = 1 since 2 = 0^2 + 1^2 + 0^2 + 1^2 with 1 > 0 and 6*0^2*1^2 + 12*1^2*0^2 + 52*0^2*1^2 + 27*1^2*0^2 = 0^2.
a(3) = 1 since 3 = 0^2 + 1^2 + 1^2 + 1^2 with 1 > 0 and 6*0^2*1^2 + 12*1^2*1^2 + 52*1^2*1^2 + 27*1^2*0^2 = 8^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 1 > 0 and 6*2^2*3^2 + 12*3^2*1^2 + 52*1^2*1^2 = 22^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 5 > 0 and 6*1^2*1^2 + 12*1^2*2^2 + 52*2^2*5^2 + 27*5^2*1^2 = 77^2.
a(39) = 1 since 39 = 2^2 + 1^2 + 5^2 + 3^2 with 3 > 0 and 6*2^2*1^2 + 12*1^2*5^2 + 52*5^2*3^2 + 27*3^2*2^2 = 114^2.
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 5 > 0 and 6*3^2*1^2 + 12*1^2*6^2 + 52*6^2*5^2 + 27*5^2*3^2 = 231^2.
a(78) = 1 since 78 = 2^2 + 7^2 + 4^2 + 3^2 with 3 > 0 and 6*2^2*7^2 + 12*7^2*4^2 + 52*7^2*4^2 + 27*3^2*2^2 = 138^2.
a(79) = 1 since 79 = 2^2 + 5^2 + 7^2 + 1^2 with 1 > 0 and 6*2^2*5^2 + 12*5^2*7^2 + 52*7^2*1^2 + 27*1^2*2^2 = 134^2.
a(195) = 1 since 195 = 3^2 + 7^2 + 4^2 + 11^2 with 11 > 0 and 6*3^2*7^2 + 12*7^2*4^2 + 52*4^2*11^2 + 27*11^2*3^2 = 377^2.
a(311) = 1 since 311 = 14^2 + 9^2 + 3^2 + 5^2 with 5 > 0 and 6*14^2*9^2 + 12*9^2*3^2 + 52*3^2*5^2 + 27*5^2*14^2 = 498^2.
a(319) = 1 since 319 = 6^2 + 3^2 + 7^2 + 15^2 with 15 > 0 and 6*6^2*3^2 + 12*3^2*7^2 + 52*7^2*15^2 + 27*15^2*6^2 = 894^2.
a(403) = 1 since 403 = 3^2 + 13^2 + 12^2 + 9^2 with 9 > 0 and 6*3^2*13^2 + 12*13^2*12^2 + 52*12^2*9^2 + 27*9^2*3^2 = 963^2.
a(559) = 1 since 559 = 5^2 + 23^2 + 2^2 + 1^2 with 1 > 0 and 6*5^2*23^2 + 12*23^2*2^2 + 52*2^2*1^2 + 27*1^2*5^2 = 325^2.
a(591) = 1 since 591 = 21^2 + 11^2 + 2^2 + 5^2 with 5 > 0 and 6*21^2*11^2 + 12*11^2*2^2 + 52*2^2*5^2 + 27*5^2*21^2 = 793^2.
a(683) = 1 since 683 = 0^2 + 11^2 + 21^2 + 11^2 with 11 > 0 and 6*0^2*11^2 + 12*11^2*21^2 + 52*21^2*11^2 + 27*11^2*0^2 = 1848^2.
a(719) = 1 since 719 = 10^2 + 3^2 + 21^2 + 13^2 with 13 > 0 and 6*10^2*3^2 + 12*3^2*21^2 + 52*21^2*13^2 + 27*13^2*10^2 = 2094^2.
a(1031) = 1 since 1031 = 26^2 + 15^2 + 9^2 + 7^2 with 7 > 0 and 6*26^2*15^2 + 12*15^2*9^2 + 52*9^2*7^2 + 27*7^2*26^2 = 1494^2.
a(1439) = 1 since 1439 = 13^2 + 27^2 + 10^2 + 21^2 with 21 > 0 and 6*13^2*27^2 + 12*27^2*10^2 + 52*10^2*21^2 + 27*21^2*13^2 = 2433^2.
a(1643) = 1 since 1643 = 36^2 + 17^2 + 3^2 + 7^2 with 7 > 0 and 6*36^2*17^2 + 12*17^2*3^2 + 52*3^2*7^2 + 27*7^2*36^2 = 2004^2.
a(2519) = 1 since 2519 = 27^2 + 7^2 + 30^2 + 29^2 with 29 > 0 and 6*27^2*7^2 + 12*7^2*30^2 + 52*30^2*29^2 + 27*29^2*27^2 = 7527^2.
a(6879) = 1 since 6879 = 38^2 + 53^2 + 49^2 + 15^2 with 15 > 0 and 6*38^2*53^2 + 12*53^2*49^2 + 52*49^2*15^2 + 27*15^2*38^2 = 11922^2.
		

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, 2016.

Crossrefs

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

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[6*(n-x^2-y^2-z^2)*x^2+12*x^2*y^2+52*y^2*z^2+27*z^2*(n-x^2-y^2-z^2)],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}]