

A273107


Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (8*x+12*y)^2 + (15*z)^2 a square, where x,y,z,w are nonnegative integers with x+y > 0 and z > 0.


15



0, 1, 2, 1, 0, 1, 1, 1, 1, 1, 3, 4, 2, 3, 2, 1, 2, 3, 3, 1, 2, 3, 1, 1, 1, 1, 4, 3, 3, 4, 1, 1, 5, 3, 2, 3, 3, 5, 2, 1, 2, 1, 3, 3, 3, 3, 2, 4, 5, 5, 2, 4, 5, 6, 1, 3, 7, 3, 5, 4, 2, 6, 4, 1, 5, 4, 5, 4, 7, 7, 4, 3, 5, 4, 5, 6, 2, 10, 3, 1
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OFFSET

1,3


COMMENTS

Conjecture: a(n) > 0 for all n > 5, and a(n) = 1 only for n = 7, 9, 23, 25, 31, 55, 2^k*m (k = 1,2,... and m = 1, 5), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 13, 21).
This conjecture implies that any integer n > 5 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 8*x+12*y and 15*z are the two legs of a right triangle with positive integer sides.
See also A271714, A273108, A273110 and A273134 for similar conjectures related to Pythagorean triples. For more conjectural refinements of Lagrange's foursquare theorem, one may consult arXiv:1604.06723.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Refining Lagrange's foursquare theorem, arXiv:1604.06723 [math.GM], 2016.


EXAMPLE

a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 + 0 > 0 < 1 and (8*1+12*0)^2 + (15*1)^2 = 17^2.
a(4) = 1 since 4 = 1^2 + 1^2 + 1^2 + 1^2 with 1 + 1 > 0 < 1 and (8*1+12*1)^2 + (15*1)^2 = 25^2.
a(6) = 1 since 6 = 1^2 + 0^2 + 1^2 + 2^2 with 1 + 0 > 0 < 1 and (8*1+12*0)^2 + (15*1)^2 = 17^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 + 1 > 0 < 1 and (8*1+12*1)^2 + (15*1)^2 = 25^2.
a(9) = 1 since 9 = 2^2 + 0^2 + 2^2 + 1^2 with 2 + 0 > 0 < 2 and (8*2+12*0)^2 + (15*2)^2 = 34^2.
a(10) = 1 since 10 = 0^2 + 3^2 + 1^2 + 0^2 with 0 + 3 > 0 < 1 and (8*0+12*3)^2 + (15*1)^2 = 39^2.
a(20) = 1 since 20 = 3^2 + 1^2 + 1^2 + 3^2 with 3 + 1 > 0 < 1 and (8*3+12*1)^2 + (15*1)^2 = 39^2.
a(23) = 1 since 23 = 2^2 + 1^2 + 3^2 + 3^2 with 2 + 1 > 0 < 3 and (8*2+12*1)^2 + (15*3)^2 = 53^2.
a(25) = 1 since 25 = 1^2 + 2^2 + 4^2 + 2^2 with 1 + 2 > 0 < 4 and (8*1+12*2)^2 + (15*4)^2 = 68^2.
a(26) = 1 since 26 = 0^2 + 3^2 + 1^2 + 4^2 with 0 + 3 > 0 < 1 and (8*0+12*3)^2 + (15*1)^2 = 39^2.
a(31) = 1 since 31 = 3^2 + 3^2 + 3^2 + 2^2 with 3 + 3 > 0 < 3 and (8*3+12*3)^2 + (15*3)^2 = 75^2.
a(42) = 1 since 42 = 2^2 + 2^2 + 5^2 + 3^2 with 2 + 2 > 0 < 5 and (8*2+12*2)^2 + (15*5)^2 = 85^2.
a(55) = 1 since 55 = 6^2 + 1^2 + 3^2 + 3^2 with 6 + 1 > 0 < 3 and (8*6+12*1)^2 + (15*3)^2 = 75^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[nx^2y^2z^2]&&SQ[(8x+12y)^2+(15z)^2], r=r+1], {x, 0, Sqrt[n1]}, {y, Max[0, 1x], Sqrt[n1x^2]}, {z, 1, Sqrt[nx^2y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273108, A273110, A273134.
Sequence in context: A031214 A130654 A053259 * A194329 A321749 A143842
Adjacent sequences: A273104 A273105 A273106 * A273108 A273109 A273110


KEYWORD

nonn


AUTHOR

ZhiWei Sun, May 15 2016


STATUS

approved



