

A272977


Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 3*x^2*y + z^2*w a square, where w is a nonzero integer and x,y,z are nonnegative integers with x >= z.


17



2, 4, 1, 3, 8, 1, 1, 4, 3, 11, 3, 1, 9, 5, 3, 3, 10, 7, 6, 9, 3, 6, 1, 1, 11, 15, 4, 2, 13, 2, 2, 4, 4, 16, 5, 4, 13, 5, 2, 10, 12, 6, 5, 1, 12, 6, 1, 3, 7, 19, 2, 10, 10, 6, 3, 1, 2, 12, 7, 3, 15, 7, 4, 3, 16, 8, 6, 9, 5, 6, 1, 7, 12, 19, 3, 3, 7, 2, 4, 9
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OFFSET

1,1


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 23, 47, 71, 147, 199, 263, 439, 16^k*m (k = 0,1,2,... and m = 6, 12, 24, 44, 56, 140, 156, 174, 204, 284, 4652).
(ii) For each ordered pair (a,b) = (7,1), (8,1), (9,2), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with a*x^2*y + b*z^2*w a square, where x,y,z are nonnegative integers and w is a nonzero integer.
Compare this conjecture with the one in A270073.
See arXiv:1604.06723 for more refinements of Lagrange's foursquare theorem.


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.
ZhiWei Sun, Refine Lagrange's foursquare theorem, a message to Number Theory List, April 26, 2016.


EXAMPLE

a(1) = 2 since 1 = 1^2 + 0^2 + 0^2 + 1^2 with 1 > 0 and 3*1^2*0 + 0^2*1 = 0^2, and also 1 = 1^2 + 0^2 + 0^2 + (1)^2 with 1 > 0 and 3*1^2*0 + 0^2*(1) = 0^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1 = 1 and 3*1^2*0 + 1^2*1 = 1^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with 2 > 1 and 3*2^2*0 + 1^2*1 = 1^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + (2)^2 with 1 = 1 and 3*1^2*1 + 1^2*(2) = 1^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + (3)^2 with 1 = 1 and 3*1^2*1 + 1*(3) = 0^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 3^2 + (2)^2 with 3 = 3 and 3*3^2*1 + 3^2*(2) = 3^2.
a(24) = 1 since 24 = 2^2 + 0^2 + 2^2 + 4^2 with 2 = 2 and 3*2^2*0 + 2^2*4 = 4^2.
a(44) = 1 since 44 = 3^2 + 5^2 + 3^2 + 1^2 with 3 = 3 and 3*3^2*5 + 3^2*1 = 12^2.
a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + (5)^2 with 3 = 3 and 3*3^2*2 + 3^2*(5) = 3^2.
a(56) = 1 since 56 = 6^2 + 0^2 + 2^2 + 4^2 with 6 > 2 and 3*6^2*0 + 2^2*4 = 4^2.
a(71) = 1 since 71 = 5^2 + 6^2 + 3^2 + (1)^2 with 5 > 3 and 3*5^2*6 + 3^2*(1) = 21^2.
a(140) = 1 since 140 = 5^2 + 3^2 + 5^2 + (9)^2 with 5 = 5 and 3*5^2*3 + 5^2*(9) = 0^2.
a(147) = 1 since 147 = 11^2 + 0^2 + 5^2 + 1^2 with 11 > 5 and 3*11^2*0 + 5^2*1 = 5^2.
a(156) = 1 since 156 = 7^2 + 3^2 + 7^2 + 7^2 with 7 = 7 and 3*7^2*3 + 7^2*7 = 28^2.
a(174) = 1 since 174 = 13^2 + 0^2 + 2^2 + 1^2 with 13 > 2 and 3*13^2*0 + 2^2*1 = 2^2.
a(199) = 1 since 199 = 9^2 + 1^2 + 9^2 + 6^2 with 9 = 9 and 3*9^2*1 + 9^2*6 = 27^2.
a(204) = 1 since 204 = 1^2 + 9^2 + 1^2 + (11)^2 with 1 = 1 and 3*1^2*9 + 1^2*(11) = 4^2.
a(263) = 1 since 263 = 3^2 + 14^2 + 3^2 + 7^2 with 3 = 3 and
3*3^2*14 + 3^2*7 = 21^2.
a(284) = 1 since 284 = 13^2 + 3^2 + 5^2 + (9)^2 with 13 > 5 and 3*13^2*3 + 5^2*(9) = 36^2.
a(439) = 1 since 439 = 13^2 + 5^2 + 7^2 + (14)^2 with 13 > 7 and 3*13^2*5 + 7^2*(14) = 43^2.
a(4652) = 1 since 4652 = 11^2 + 21^2 + 11^2 + (63)^2 with 11 = 11 and 3*11^2*21 + 11^2*(63) = 0^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[nx^2y^2z^2]&&SQ[3x^2*y+z^2*(1)^k*Sqrt[nx^2y^2z^2]], r=r+1], {z, 0, Sqrt[(n1)/2]}, {x, z, Sqrt[n1z^2]}, {y, 0, Sqrt[n1x^2z^2]}, {k, 0, 1}]; 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.
Sequence in context: A338530 A248212 A290824 * A230505 A208526 A275895
Adjacent sequences: A272974 A272975 A272976 * A272978 A272979 A272980


KEYWORD

nonn


AUTHOR

ZhiWei Sun, May 11 2016


STATUS

approved



