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A271644 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that w*x + 2*x*y + 2*y*z is a square, where w is a positive integer and x,y,z are nonnegative integers. 8
1, 3, 1, 1, 4, 4, 1, 3, 4, 4, 2, 2, 5, 2, 1, 1, 8, 8, 2, 5, 7, 3, 2, 4, 8, 7, 3, 2, 6, 4, 4, 3, 7, 6, 2, 4, 6, 4, 3, 4, 9, 4, 3, 4, 8, 4, 1, 2, 5, 7, 4, 7, 10, 11, 3, 2, 5, 5, 2, 2, 7, 4, 2, 1, 8, 9, 2, 8, 14, 9, 1, 8, 8, 6, 5, 4, 8, 2, 3, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 15, 47, 71, 379, 4^k (k = 0,1,2,...).
(ii) If a, b and c are positive integers with a <= b <= c, gcd(a,b,c) squarefree, and the triple (a,b,c) not equal to (1,2,2), then not all natural numbers can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and a*w*x + b*x*y + c*y*z a square.
(iii) Let a,b,c be positive integers with gcd(a,b,c) squarefree. Then every n = 0,1,2,... can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that a*x*y + b*y*z + c*z*x is a square, if and only if {a,b,c} is among {1,2,3}, {1,3,8}, {1,8,13}, {2,4,45}, {4,5,7}, {4,7,23}, {5,8,9}, {11,16,31}.
Clearly, part (i) of this conjecture is stronger than Lagrange's four-square theorem.
See also A271510, A271513, A271518 and A271608 for related conjectures.
LINKS
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
EXAMPLE
a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1*0 + 2*0*0 + 2*0*0 = 0^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1*1 + 2*1*0 + 2*0*1 = 1^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1*1 + 2*1*2 + 2*2*1 = 3^2.
a(11) = 2 since 11 = 1^2 + 1^2 + 0^2 + 3^2 with 1*1 + 2*1*0 + 2*0*3 = 1^2, and 11 = 1^2 + 3^2 + 1^2 + 0^2 with 1*3 + 2*3*1 + 2*1*0 = 3^2.
a(12) = 2 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1*1 + 2*1*1 + 2*1*3 = 3^2, and 12 = 2^2 + 2^2 + 0^2 + 2^2 with 2*2 + 2*2*0 + 2*0*2 = 2^2.
a(15) = 1 since 15 = 3^2 + 1^2 + 1^2 + 2^2 with 3*1 + 2*1*1 + 2*1*2 = 3^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 6^2 + 3^2 with 1*1 + 2*1*6 + 2*6*3 = 7^2.
a(71) = 1 since 71 = 3^2 + 3^2 + 2^2 + 7^2 with 3*3 + 2*3*2 + 2*2*7 = 7^2.
a(379) = 1 since 379 = 3^2 + 3^2 + 0^2 + 19^2 with 3*3 + 2*3*0 + 2*0*19 = 3^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-w^2-x^2-y^2]&&SQ[w*x+2*x*y+2*y*Sqrt[n-w^2-x^2-y^2]], r=r+1], {w, 1, Sqrt[n]}, {x, 0, Sqrt[n-w^2]}, {y, 0, Sqrt[n-w^2-x^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
Sequence in context: A134510 A335322 A171145 * A262191 A334844 A296457
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 11 2016
STATUS
approved

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