

A283269


Number of ways to write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is a square


4



1, 1, 2, 2, 0, 3, 5, 0, 3, 4, 4, 1, 0, 3, 7, 0, 1, 5, 3, 3, 0, 5, 3, 0, 6, 2, 8, 2, 0, 8, 3, 0, 2, 3, 6, 7, 0, 2, 6, 0, 6, 8, 4, 1, 0, 4, 3, 0, 2, 3, 7, 5, 0, 4, 13, 0, 8, 5, 2, 3, 0, 6, 6, 0, 0, 7, 13, 2, 0, 7, 3, 0, 5, 4, 9, 1, 0, 5, 3, 0, 3
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OFFSET

0,3


COMMENTS

Conjecture: (i) a(2^k*m) > 0 for any positive odd integers k and m. Also, a(4*n+1) = 0 only for n = 63.
(ii) For any positive odd integers k and m, we can write 2^k*m as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is twice a square.
(iii) For any positive odd integer n not congruent to 7 modulo 8, we can write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 4*z is a square.
(iv) Let n be any nonnegative integer. Then we can write 8*n + 1 as x^2 + y^2 + z^2 with x + 3*y a square, where x and y are integers, and z is a positive integer. Also, except for n = 2255, 4100 we can write 4*n + 2 as x^2 + y^2 + z^2 with x + 3*y a square, where x,y,z are integers.
(v) For each n = 0,1,2,..., we can write 8*n + 6 as x^2 + y^2 + z^2 with x + 2*y a square, where x,y,z are integers with y nonnegative and z odd.
The GaussLegendre theorem asserts that a nonnegative integer can be written as the sum of three squares if and only if it is not of the form 4^k*(8m+7) with k and m nonnegative integers. Thus a(4^k*(8m+7)) = 0 for all k,m = 0,1,2,....
See also A283273 for a similar conjecture.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, Refining Lagrange's foursquare theorem, J. Number Theory 175(2017), 167190.
ZhiWei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.


EXAMPLE

a(11) = 1 since 11 = 3^2 + 1^2 + (1)^2 with 3 + 3*1 + 5*(1) = 1^2.
a(43) = 1 since 43 = (5)^2 + (3)^2 + 3^2 with (5) + 3*(3) + 5*3 = 1^2.
a(75) = 1 since 75 = (1)^2 + 5^2 + 7^2 with (1) + 3*5 + 5*7 = 7^2.
a(262) = 1 since 262 = 1^2 + 15^2 + (6)^2 with 1 + 3*15 + 5*(6) = 4^2.
a(277) = 1 since 277 = (6)^2 + 4^2 + 15^2 with (6) + 3*4 + 5*15 = 9^2.
a(617) = 1 since 617 = 17^2 + 18^2 + 2^2 with 17 + 3*18 + 5*2 = 9^2.
a(1430) = 1 since 1430 = (13)^2 + (6)^2 + 35^2 with (13) + 3*(6) + 5*35 = 12^2.
a(5272) = 1 since 5272 = (66)^2 + 30^2 + (4)^2 with (66) + 3*30 + 5*(4) = 2^2.
a(7630) = 1 since 7630 = (78)^2 + 39^2 + 5^2 with (78) + 3*39 + 5*5 = 8^2.
a(7933) = 1 since 7933 = (56)^2 + 69^2 + (6)^2 with (56) + 3*69 + 5*(6) = 11^2.
a(14193) = 1 since 14193 = (7)^2 + 112^2 + 40^2 with (7) + 3*112 + 5*40 = 23^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[nx^2y^2]&&SQ[(1)^i*x+(1)^j*3y+(1)^k*5*Sqrt[nx^2y^2]], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[nx^2]}, {i, 0, Min[x, 1]}, {j, 0, Min[y, 1]}, {k, 0, Min[Sqrt[nx^2y^2], 1]}]; Print[n, " ", r]; Continue, {n, 0, 80}]


CROSSREFS

Cf. A000290, A005875, A271518, A283170, A283196, A283204, A283205, A283239, A283273.
Sequence in context: A294598 A077264 A188333 * A201947 A098816 A214639
Adjacent sequences: A283266 A283267 A283268 * A283270 A283271 A283272


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 04 2017


STATUS

approved



