OFFSET
0,4
COMMENTS
Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 15, 29, 33, 47, 53, 65, 89, 129, 689, 1553, 2^(2k+1)*m (k = 0,1,2,... and m = 1, 29).
(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x^2 + b*y^2 + c*z^2 - d*w^2 a square, if (a,b,c,d) is among the following quadruples: (1,3,6,3), (1,3,9,3), (1,3,30,3), (1,4,12,4), (1,4,20,4), (1,5,20,5), (1,5,35,20), (3,4,9,3), (3,9,40,3), (4,5,16,4), (4,11,33,11), (4,12,16,7), (5,16,20,20), (5,25,36,5), (6,10,25,10), (9,12,28,12), (9,21,28,21), (15,21,25,15), (15,24,25,15), (1,5,60,5), (1,20,60,20), (9, 28,63,63), (9,28,84,84), (12,33,64,12), (16,21,105,21), (16,33,64,16), (21,25,45,45), (24,25,75,75), (24,25,96,96), (25,40,96,40), (25,48,96,48), (25,60,84,60), (25,60,96,60), (25,75,126,75), (32,64,105,32).
(iii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x^2 + b*y^2 - c*z^2 -d*w^2 a square, whenever (a,b,c,d) is among the quadruples (3,9,3,20), (5,9,5,20), (5,25,4,5), (9,81,9,20),(12,16,3,12), (16,64,15,16), (20,25,4,20), (27,81,20,27), (30,64,15,30), (32,64,15,32), (48,64,15,48), (60,64,15,60), (60,81,20,60), (64,80,15,80).
(iv) For each triple (a,b,c) = (21,5,15), (36,3,8), (48,8,39), (64,7,8), (40,15,144), (45,20,144), (69,20,60), any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x^2 - b*y^2 - c*z^2 a square.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
EXAMPLE
a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with 1^2 + 3*1^2 + 5*0^2 - 8*0^2 = 2^2.
a(15) = 1 since 15 = 1^2 + 3^2 + 1^2 + 2^2 with 1^2 + 3*3^2 + 5*1^2 - 8*2^2 = 1^2.
a(29) = 1 since 29 = 3^2 + 4^2 + 0^2 + 2^2 with 3^2 + 3*4^2 + 5*0^2 - 8*2^2 = 5^2.
a(33) = 1 since 33 = 2^2 + 4^2 + 2^2 + 3^2 with 2^2 + 3*4^2 + 5*2^2 - 8*3^2 = 0.
a(47) = 1 since 47 = 5^2 + 3^2 + 2^2 + 3^2 with 5^2 + 3*3^2 + 5*2^2 - 8*3^2 = 0^2.
a(53) = 1 since 53 = 3^2 + 2^2 + 6^2 + 2^2 with 3^2 + 3*2^2 + 5*6^2 - 8*2^2 = 13^2.
a(58) = 1 since 58 = 4^2 + 1^2 + 5^2 + 4^2 with 4^2 + 3*1^2 + 5*5^2 - 8*4^2 = 4^2.
a(65) = 1 since 65 = 3^2 + 6^2 + 2^2 + 4^2 with 3^2 + 3*6^2 + 5*2^2 - 8*4^2 = 3^2.
a(89) = 1 since 89 = 6^2 + 4^2 + 6^2 + 1^2 with 6^2 + 3*4^2 + 5*6^2 - 8*1^2 = 16^2.
a(129) = 1 since 129 = 9^2 + 4^2 + 4^2 + 4^2 with 9^2 + 3*4^2 + 5*4^2 - 8*4^2 = 9^2.
a(689) = 1 since 689 = 11^2 + 18^2 + 10^2 + 12^2 with 11^2 + 3*18^2 + 5*10^2 - 8*12^2 = 21^2.
a(1553) = 1 since 1553 = 21^2 + 6^2 + 26^2 + 20^2 with 21^2 + 3*6^2 + 5*26^2 - 8*20^2 = 27^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x^2+3*y^2+5*z^2-8*(n-x^2-y^2-z^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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 14 2016
STATUS
approved