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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 16 2016
STATUS
approved