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

1,5

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

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.

Example

3, 7, 23, 31, 39, 47, 55, 71, 79, 151, 191, 551).
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.

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}]

Crossrefs

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

Sequence in context: A095133 A334572 A126081 * A272351 A243612 A351902

Adjacent sequences: A268504 A268505 A268506 * A268508 A268509 A268510

Keyword

nonn

Author

Zhi-Wei Sun, Apr 16 2016

Status

approved