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A271724
Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w*(x+2*y+3*z) a square, where w,x,y,z are nonnegative integers with x > 0.
38
1, 3, 2, 1, 4, 4, 1, 3, 4, 6, 4, 2, 4, 7, 1, 1, 10, 8, 5, 6, 8, 5, 1, 4, 7, 10, 7, 2, 11, 13, 2, 3, 8, 9, 8, 6, 7, 13, 3, 6, 15, 8, 4, 4, 13, 8, 1, 2, 8, 15, 11, 4, 14, 18, 5, 7, 6, 6, 12, 5, 12, 17, 5, 1, 16, 21, 3, 11, 16, 12, 1, 8, 8, 18, 16, 5, 16, 12, 4, 6
OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 15, 47, 151, 4^k*q (k = 0,1,2,... and q = 1, 23, 71).
(ii) For positive integers a,b,c with gcd(a,b,c) squarefree, any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and w*(a*x+b*y+c*z) a square, if and only if {a,b,c} is among {1,2,3}, {1,3,6}, {1,6,9}, {5,6,9}, {18,30,114}.
(iii) For each quadruple (a,b,c,d) = (1,1,2,12), (1,2,7,60), (1,3,9,48), (1,4,11,48), (1,5,8,24), (1,8,11,24), (2,6,8,15), (3,5,6,24), (3,6,15,40), (3,6,18,40), (3,12,15,20), (4,4,8,15), (4,8,12,21), (4,8,12,45), (4,8,20,15), (4,8,36,45), (5,10,15,24), (6,9,15,20), (7,14,28,60), (7,21,28,60), (7,21,42,60), (12,36,48,55), (14,21,28,60), (3,9,18,112), (3,21,33,80), (4,5,9,120), (4,12,16,105), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (a*x+b*y+c*z)^2 + (d*w)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271714 and A271721 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(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 1 > 0 and 0*(1+2*0+3*0) = 0^2.
a(3) = 2 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1*(1+2*0+3*1) = 2^2, and 3 = 0^2 + 1^2 + 1^2 + 1^2 with 0*(1+2*1+3*1) = 0^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1*(1+2*1+3*2) = 3^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 2*(3+2*1+3*1) = 4^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1*(3+2*2+3*3) = 4^2.
a(31) = 2 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with 2*(1+2*1+3*5) = 6^2, and also 31 = 2^2 + 3^2 + 3^2 + 3^2 with 2*(3+2*3+3*3) = 6^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1*(1+2*3+3*6) = 5^2.
a(71) = 1 since 71 = 1^2 + 6^2 + 5^2 + 3^2 with 1*(6+2*5+3*3) = 5^2.
a(151) = 1 since 151 = 9^2 + 6^2 + 5^2 + 3^2 with 9*(6+2*5+3*3) = 15^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[Sqrt[n-x^2-y^2-z^2](x+2y+3z)], r=r+1], {x, 1, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Label[aa]; Continue, {n, 1, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 13 2016
STATUS
approved