|
| |
|
|
A271721
|
|
Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= z >= 0, x > 0 and w >= z such that (x-y)*(w-z) is a square.
|
|
37
|
|
|
|
1, 2, 1, 2, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 2, 4, 5, 3, 3, 3, 2, 1, 2, 3, 5, 4, 5, 2, 2, 4, 2, 3, 5, 1, 4, 4, 5, 3, 3, 4, 5, 4, 3, 4, 2, 2, 3, 3, 5, 3, 8, 4, 6, 3, 2, 4, 6, 3, 3, 4, 4, 5, 2, 3, 7, 6, 7, 2, 3, 2, 5, 6, 8, 3, 7, 3, 2, 2, 3, 6, 11, 5, 8, 5, 8, 4, 2, 3, 8, 4, 5, 5, 3, 1, 2, 9, 10, 5, 8
(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 = 1, 3, 5, 11, 15, 23, 35, 95, 4^k*190 (k = 0,1,2,...).
(ii) For each k = 4, 5, 6, 7, 8, 11, 12, 13, 15, 17, 18, 20, 22, 25, 27, 29, 33, 37, 38, 41, 50, 61, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that (x-y)*(w-k*z) is a square.
(iii) For each triple (a,b,c) = (3,1,1), (1,2,1), (2,2,1), (3,2,1), (2,2,2), (6,2,1), (1,3,1), (3,3,1), (15,3,1), (1,4,1), (2,4,1), (1,5,1), (3,5,1), (5,5,1), (1,5,2), (1,6,1), (2,6,1), (3,6,1), (15,6,1), (1,7,1), (5,7,1), (1,8,1), (1,8,5), (3,9,1), (1,10,1), (1,12,1), (1,13,1), (3,13,1), (1,14,1), (1,15,1), (1,15,2), (6,16,1), (2,18,1), (3,18,1), (1,20,2), (1,21,1), (3,21,1), (1,23,1), (1,24,1), (1,27,1), (3,27,1), (1,34,1), (1,45,1), (3,45,1), (3,48,1), (1,55,1), (1,60,1), (5,60,1), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*(x+b*y)*(w-c*z) is a square.
This is stronger than Lagrange's four-square theorem. Note that for k = 2 or 3, any natural number n can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and (x-y)*(w-k*z) = 0, for, if n cannot be represented by x^2 + y^2 + 2*z^2 then it has the form 4^k*(16*m+14) (k,m = 0,1,2,...) and hence it can be represented by x^2 + y^2 + (k^2+1)*z^2. It is known that natural numbers not represented by x^2 + y^2 + 5*z^2 have the form 4^k*(8*m+3), and that positive even numbers not represented by x^2 + y^2 + 10*z^2 have the form 4^k*(16*m+6) (as conjectured by S. Ramanujan and proved by L. E. Dickson).
|
|
|
REFERENCES
|
L. E. Dickson, Integers represented by positive ternary quadratic forms, Bull. Amer. Math. Soc. 33(1927), 63-70.
L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939, pp. 112-113.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 0^2 with 1 = 1 > 0 = 0 and (1-1)*(0-0) = 0^2.
a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 0^2 with 2 > 1 > 0 = 0 and (2-1)*(0-0) = 0^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 0^2 + 3^2 with 1 = 1 > 0 < 3 and (1-1)*(3-0) = 0^2.
a(14) = 2 since 14 = 3^2 + 1^2 + 0^2 + 2^2 with 3 > 1 > 0 < 2 and (3-1)*(2-0) = 2^2, and also 14 = 3^2 + 2^2 + 0^2 + 1^2 with 3 > 2 > 0 < 1 and (3-2)*(1-0) = 1^2.
a(15) = 1 since 15 = 3^2 + 2^2 + 1^2 + 1^2 with 3 > 2 > 1 = 1 and (3-2)*(1-1) = 0^2.
a(23) = 1 since 23 = 3^2 + 3^2 + 1^2 + 2^2 with 3 = 3 > 1 < 2 and (3-3)*(2-1) = 0^2.
a(35) = 1 since 35 = 3^2 + 3^2 + 1^2 + 4^2 with 3 = 3 > 1 < 4 and (3-3)*(4-1) = 0^2.
a(95) = 1 since 95 = 5^2 + 5^2 + 3^2 + 6^2 with 5 = 5 > 3 < 6 and (5-5)*(6-3) = 0^2.
a(190) = 1 since 190 = 13^2 + 4^2 + 1^2 + 2^2 with 13 > 4 > 1 < 2 and (13-4)*(2-1) = 3^2.
|
|
|
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]-z)*(x-y)], r=r+1], {z, 0, Sqrt[n/4]}, {y, z, Sqrt[(n-z^2)/2]}, {x, Max[1, y], Sqrt[(n-y^2-2z^2)]}]; Print[n, " ", r]; Continue, {n, 1, 100}]
|
|
|
CROSSREFS
|
Cf. A000118, A000290, A259789, A271510, A271513, A271518, A271608, A271644, A271665, A271714, A271719, A271724.
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
