

A283299


Number of ways to write 2*n + 1 as x^2 + 2*y^2 + 3*z^2 with x,y,z integers such that x + y + z is a square or twice a square.


2



1, 4, 3, 1, 6, 3, 1, 7, 1, 2, 8, 4, 4, 4, 3, 5, 4, 1, 4, 9, 3, 3, 9, 1, 4, 10, 3, 3, 11, 7, 4, 8, 5, 6, 7, 6, 2, 10, 3, 3, 14, 1, 2, 5, 3, 6, 12, 2, 4, 11, 3, 2, 5, 5, 7, 14, 6, 4, 6, 7, 4, 5, 4, 3, 13, 3, 3, 12, 3, 2, 15, 2, 2, 12, 3, 7, 4, 5, 4, 11, 8
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OFFSET

0,2


COMMENTS

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 6, 8, 17, 23, 41, 128, 197, 372, 764, 1143, 1893, 3761, 4307, 6408, 6918.
(ii) Any positive odd integer can be written as x^2 + y^2 + 2*z^2 with x,y,z integers such that x + 2*y + 3*z is a square or twice a square.
By Dickson's book in the reference, any positive odd integer can be written as x^2 + 2*y^2 + 3*z^2 (or x^2 + y^2 + 2*z^2) with x,y,z integers.
We have verified a(n) > 0 for all n = 0..10^6.
See also A283366 for a similar conjecture.


REFERENCES

L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939. (See pages 112113.)


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(0) = 1 since 2*0 + 1 = 1^2 + 2*0^2 + 3*0^2 with 1 + 0 + 0 = 1^2.
a(3) = 1 since 2*3 + 1 = 2^2 + 2*0^2 + 3*(1)^2 with 2 + 0 + (1) = 1^2.
a(8) = 1 since 2*8 + 1 = 3^2 + 2*(2)^2 + 3*0^2 with 3 + (2) + 0 = 1^2.
a(17) = 1 since 2*17 + 1 = 0^2 + 2*(2)^2 + 3^2 with 0 + (2) + 3 = 1^2.
a(41) = 1 since 2*41 + 1 = 9^2 + 2*(1)^2 + 3*0^2 with 9 + (1) + 0 = 2*2^2.
a(128) = 1 since 2*128 + 1 = 3^2 + 2*10^2 + 3*(4)^2 with 3 + 10 + (4) = 3^2.
a(197) = 1 since 2*197 + 1 = 12^2 + 2*(2)^2 + 3*(9)^2 with 12 + (2) + (9) = 1^2.
a(372) = 1 since 2*372 + 1 = 22^2 + 2*3^2 + 3*(9)^2 with 22 + 3 + (9) = 4^2.
a(764) = 1 since 2*764 + 1 = 18^2 + 2*(23)^2 + 3*7^2 with 18 + (23) + 7 = 2*1^2.
a(3761) = 1 since 2*3761 + 1 = (57)^2 + 2*31^2 + 3*28^2 with (57) + 31 + 28 = 2*1^2.
a(6408) = 1 since 2*6408 + 1 = (22)^2 + 2*75^2 + 3*19^2 with (22) + 75 + 19 = 2*6^2.
a(6918) = 1 since 2*6918 + 1 = 100^2 + 2*9^2 + 3*35^2 with 100 + 9 + 35 = 12^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
TQ[n_]:=TQ[n]=SQ[n]SQ[2n];
Do[r=0; Do[If[SQ[2n+13x^22y^2]&&TQ[(1)^i*x+(1)^j*y+(1)^k*Sqrt[2n+13x^22y^2]], r=r+1], {x, 0, Sqrt[(2n+1)/3]}, {y, 0, Sqrt[(2n+13x^2)/2]}, {i, 0, Min[x, 1]}, {j, 0, Min[y, 1]},
{k, 0, Min[Sqrt[2n+13x^22y^2], 1]}]; Print[n, " ", r]; Continue, {n, 0, 80}]


CROSSREFS

Cf. A000290, A271518, A283239, A283269, A283273, A283366.
Sequence in context: A275981 A087274 A253182 * A190157 A103552 A127673
Adjacent sequences: A283296 A283297 A283298 * A283300 A283301 A283302


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 05 2017


STATUS

approved



