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

0,2

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.

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.

Example

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.

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}]

Crossrefs

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

Sequence in context: A271830 A193815 A104509 * A306801 A117212 A208153

Adjacent sequences: A271510 A271511 A271512 * A271514 A271515 A271516

Keyword

nonn

Author

Zhi-Wei Sun, Apr 09 2016

Status

approved