cp's OEIS Frontend

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.

%I A283299 #23 Mar 06 2017 06:03:17
%S A283299 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,
%T A283299 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,
%U A283299 13,3,3,12,3,2,15,2,2,12,3,7,4,5,4,11,8
%N 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.
%C A283299 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.
%C A283299 (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.
%C A283299 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.
%C A283299 We have verified a(n) > 0 for all n = 0..10^6.
%C A283299 See also A283366 for a similar conjecture.
%D A283299 L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939. (See pages 112-113.)
%H A283299 Zhi-Wei Sun, <a href="/A283299/b283299.txt">Table of n, a(n) for n = 0..10000</a>
%H A283299 Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.
%H A283299 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.
%e A283299 a(0) = 1 since 2*0 + 1 = 1^2 + 2*0^2 + 3*0^2 with 1 + 0 + 0 = 1^2.
%e A283299 a(3) = 1 since 2*3 + 1 = 2^2 + 2*0^2 + 3*(-1)^2 with 2 + 0 + (-1) = 1^2.
%e A283299 a(8) = 1 since 2*8 + 1 = 3^2 + 2*(-2)^2 + 3*0^2 with 3 + (-2) + 0 = 1^2.
%e A283299 a(17) = 1 since 2*17 + 1 = 0^2 + 2*(-2)^2 + 3^2 with 0 + (-2) + 3 = 1^2.
%e A283299 a(41) = 1 since 2*41 + 1 = 9^2 + 2*(-1)^2 + 3*0^2 with 9 + (-1) + 0 = 2*2^2.
%e A283299 a(128) = 1 since 2*128 + 1 = 3^2 + 2*10^2 + 3*(-4)^2 with 3 + 10 + (-4) = 3^2.
%e A283299 a(197) = 1 since 2*197 + 1 = 12^2 + 2*(-2)^2 + 3*(-9)^2 with 12 + (-2) + (-9) = 1^2.
%e A283299 a(372) = 1 since 2*372 + 1 = 22^2 + 2*3^2 + 3*(-9)^2 with 22 + 3 + (-9) = 4^2.
%e A283299 a(764) = 1 since 2*764 + 1 = 18^2 + 2*(-23)^2 + 3*7^2 with 18 + (-23) + 7 = 2*1^2.
%e A283299 a(3761) = 1 since 2*3761 + 1 = (-57)^2 + 2*31^2 + 3*28^2 with (-57) + 31 + 28 = 2*1^2.
%e A283299 a(6408) = 1 since 2*6408 + 1 = (-22)^2 + 2*75^2 + 3*19^2 with (-22) + 75 + 19 = 2*6^2.
%e A283299 a(6918) = 1 since 2*6918 + 1 = 100^2 + 2*9^2 + 3*35^2 with 100 + 9 + 35 = 12^2.
%t A283299 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t A283299 TQ[n_]:=TQ[n]=SQ[n]||SQ[2n];
%t A283299 Do[r=0;Do[If[SQ[2n+1-3x^2-2y^2]&&TQ[(-1)^i*x+(-1)^j*y+(-1)^k*Sqrt[2n+1-3x^2-2y^2]],r=r+1],{x,0,Sqrt[(2n+1)/3]},{y,0,Sqrt[(2n+1-3x^2)/2]},{i,0,Min[x,1]},{j,0,Min[y,1]},
%t A283299 {k,0,Min[Sqrt[2n+1-3x^2-2y^2],1]}];Print[n," ",r];Continue,{n,0,80}]
%Y A283299 Cf. A000290, A271518, A283239, A283269, A283273, A283366.
%K A283299 nonn
%O A283299 0,2
%A A283299 _Zhi-Wei Sun_, Mar 05 2017