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A271824
Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (x+2*y)^2 + 8*z^2 + 40*w^2 a square, where x is a positive integer and y,z,w are nonnegative integers.
32
1, 2, 2, 1, 2, 2, 2, 2, 1, 4, 1, 3, 3, 2, 1, 1, 3, 6, 3, 3, 4, 1, 1, 2, 3, 4, 3, 3, 2, 5, 4, 2, 1, 3, 3, 3, 5, 1, 5, 4, 2, 6, 3, 2, 5, 3, 3, 3, 2, 8, 3, 6, 6, 4, 4, 2, 4, 6, 3, 3, 5, 3, 4, 1, 5, 5, 4, 4, 2, 6, 1, 6, 2, 4, 7, 4, 3, 5, 7, 3
OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 9, 11, 15, 23, 33, 71, 129, 167, 187, 473, 4^k*m (k = 0,1,2,... and m = 1, 22, 38, 278). Also, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with 9*(x+2*y)^2 + 16*z^2 + 24*w^2 a square, where x is a positive integer and y,z,w are nonnegative integers.
(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (a*x-b*y)^2 + c*z^2 + d*w^2 a square, provided that (a,b,c,d) is among the quadruples (4,8,1,8), (12,24,1,24), (2,4,5,40), (3,6,7,9), (3,9,7,9), (3,6,7,63), (1,2,8,16), (1,2,8,40), (3,6,8,40), (2,6,9,12), (3,5,9,15), (4,8,9,16), (12,24,9,16), (3,6,15,25), (3,6,16,24), (3,12,16,24), (6,9,16,24), (9,12,16,24), (4,8,16,41), (8,12,16,41), (3,6,16,48), (6,9,16,48), (2,3,16,56), (3,6,28,63), (2,4,36,45), (6,12,40,45), (7,14,56,64) and (2,6,57,60).
(iii) Let a and be positive integers with a <= b and gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (a*x+b*y)*z a square, if and only if (a,b) is among the ordered pairs (1,1), (1,2), (1,3), (2,5), (3,3), (3,6), (3,15), (5,6), (5,11), (5,13), (6,15), (8,46) and (9,23).
(iv) Let a and b be positive integers with a <= b and gcd(a,b) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (a*x^2+b*y^2)*z a square, if and only if (a,b) is among the ordered pairs (3,13), (5,11), (15,57), (15,165) and (138,150).
There are many ordered pairs (a,b) of integers with gcd(a,b) squarefree such that any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and a*x^2 + b*y^2 a square. For example, we have shown that (1,-1), (2,-2), (3,-3) and (1,2) are indeed such ordered pairs.
See also A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775 and A271778 for other conjectures refining Lagrange's four-square theorem.
LINKS
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.
EXAMPLE
a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (3+2*0)^2 + 8*0^2 + 40*02 = 3^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 3^2 + 0^2 with (1+2*1)^2 + 8*3^2 + 40*0^2 = 9^2.
a(15) = 1 since 15 = 1^2 + 3^2 + 2^2 + 1^2 with (1+2*3)^2 + 8*2^2 + 40*1^2 = 11^2.
a(22) = 1 since 22 = 3^2 + 2^2 + 3^2 + 0^2 with (3+2*2)^2 + 8*3^2 + 40*0^2 = 11^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with (1+2*3)^2 + 8*2^2 + 40*3^2 = 21^2.
a(33) = 1 since 33 = 4^2 + 1^2 + 0^2 + 4^2 with (4+2*1)^2 + 8*0^2 + 40*4^2 = 26^2.
a(38) = 1 since 38 = 5^2 + 2^2 + 0^2 + 3^2 with (5+2*2)^2 + 8*0^2 + 40*3^2 = 21^2.
a(71) = 1 since 71 = 1^2 + 6^2 + 5^2 + 3^2 since (1+2*6)^2 + 8*5^2 + 40*3^2 = 27^2.
a(129) = 1 since 129 = 5^2 + 6^2 + 8^2 + 2^2 with (5+2*6)^2 + 8*8^2 + 40*2^2 = 31^2.
a(167) = 1 since 167 = 11^2 + 1^2 + 3^2 + 6^2 with (11+2*1)^2 + 8*3^2 + 40*6^2 = 41^2.
a(187) = 1 since 187 = 3^2 + 5^2 + 12^2 + 3^2 with (3+2*5)^2 + 8*12^2 + 40*3^2 = 41^2.
a(278) = 1 since 278 = 3^2 + 0^2 + 10^2 + 13^2 with (3+2*0)^2 + 8*10^2 + 40*13^2 = 87^2.
a(473) = 1 since 473 = 7^2 + 10^2 + 0^2 + 18^2 with (7+2*10)^2 + 8*0^2 + 40*18^2 = 117^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(x+2y)^2+8z^2+40(n-x^2-y^2-z^2)], r=r+1], {x, 1, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 14 2016
STATUS
approved