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A271665 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that w^2 + 4*x*y + 8*y*z + 32*z*x is a square, where w is a positive integer and x,y,z are nonnegative integers. 38
1, 3, 1, 1, 6, 3, 1, 3, 1, 6, 2, 1, 7, 10, 1, 1, 9, 3, 2, 6, 2, 2, 3, 3, 8, 10, 1, 1, 10, 2, 2, 3, 5, 8, 11, 1, 7, 13, 2, 6, 16, 6, 1, 2, 6, 2, 3, 1, 3, 16, 4, 7, 9, 3, 2, 10, 4, 9, 4, 1, 8, 15, 1, 1, 15, 5, 2, 9, 6, 8, 2, 3, 10, 13, 4, 2, 17, 7, 1, 6 (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 = 4^k*3^m, 4^k*3^m*43, 4^k*9^m*q (k,m = 0, 1, 2, ... and q = 7, 15, 79, 95, 141, 159, 183).

(ii) Any positive integer n can be written as w^2 + x^2 + y^2 + z^2 with w*x + x*y + 2*y*z + 3*z*x (or w*x + 3*x*y + 8*y*z + 5*z*x) twice a square, where w is a positive integer and x,y,z are nonnegative integers.

(iii) For each k = 1, 2, 8, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with w^2 + k*(x*y+y*z) a square, where w is a positive integer and x,y,z are nonnegative integers.

(iv) For each ordered pair (b,c) = (16,4), (24,4), (32,16), 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^2 + b*y^2 + c*x*z + c*y*z + c*z*w is a square.

We also guess that for each triple (b,c,d) = (1,3,4), (1,6,8), (1,7,24), (1,8,15), (1,10,24), (1,12,35), (1,14,48), (1,20,48), (2,1,2), (2,4,2), (2,4,7), (2,6,7), (2,8,4), (2,8,14), (2,8,31), (2,10,23), (2,12,14), (2,12,34), (2,14,47), (3,1,1), (3,1,4), (3,1,16), (3,2,14), (3,2,17), (3,2,38), (3,3,3), (3,3,13), (3,4,1), (3,4,4), (3,5,11), (3,6,3), (3,6,6), (3,6,26), (3,8,2), (3,8,13), (3,8,22), (3,9,39), (3,12,12), (3,12,33), (3,15,1), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and x^2 + b*y^2 + c*x*z + d*y*z a square.

See also A271510, A271513, A271518 and A271644 for other conjectures refining Lagrange's four-square theorem.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.

EXAMPLE

a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1^2 + 4*0*1 + 8*1*1 + 32*1*0 = 3^2.

a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 4*2*0 + 8*0*0 + 32*0*0 = 2^2.

a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1^2 + 4*2*1 + 8*1*1 + 32*1*2 = 9^2.

a(15) = 1 since 15 = 1^2 + 2^2 + 1^2 + 3^2 with 1^2 + 4*2*1 + 8*1*3 + 32*3*2 = 15^2.

a(43) = 1 since 43 = 3^2 + 3^2 + 4^2 + 3^2 with 3^2 + 4*3*4 + 8*4*3 + 32*3*3 = 21^2.

a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5^2 + 4*3*6 + 8*6*3 + 32*3*3 = 23^2.

a(95) = 1 since 95 = 5^2 + 6^2 + 5^2 + 3^2 with 5^2 + 4*6*5 + 8*5*3 + 32*3*6 = 29^2.

a(129) = 1 since 129 = 5^2 + 6^2 + 8^2 + 2^2 with 5^2 + 4*6*8 + 8*8*2 + 32*2*6 = 27^2.

a(141) = 1 since 141 = 8^2 + 5^2 + 4^2 + 6^2 with 8^2 + 4*5*4 + 8*4*6 + 32*6*5 = 36^2.

a(159) = 1 since 159 = 11^2 + 1^2 + 6^2 + 1^2 with 11^2 + 4*1*6 + 8*6*1 + 32*1*1 = 15^2.

a(183) = 1 since 183 = 1^2 + 9^2 + 10^2 + 1^2 with 1^2 + 4*9*10 + 8*10*1 + 32*1*9 = 27^2.

MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]

Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x*y+8*y*z+32*z*x+(n-x^2-y^2-z^2)], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 0, Sqrt[n-1-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]

CROSSREFS

Cf. A000118, A000290, A271510, A271513, A271518, A271608, A271644.

Sequence in context: A067433 A256697 A133567 * A184049 A125230 A208334

Adjacent sequences:  A271662 A271663 A271664 * A271666 A271667 A271668

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Apr 12 2016

STATUS

approved

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Last modified February 24 11:12 EST 2018. Contains 299603 sequences. (Running on oeis4.)