A283366 Number of ways to write 2*n + 1 as x^2 + y^2 + 2*z^2 with x,y,z integers such that 2*x + y + z is a square or a power of two.

2, 4, 3, 5, 4, 2, 7, 5, 5, 6, 3, 5, 7, 8, 3, 9, 7, 3, 11, 1, 2, 8, 9, 7, 6, 2, 3, 11, 7, 7, 7, 7, 1, 12, 7, 4, 12, 6, 7, 4, 8, 4, 8, 7, 7, 9, 3, 1, 15, 8, 2, 12, 4, 4, 4, 8, 5, 12, 11, 5, 7, 6, 5, 11, 2, 3, 12, 12, 9, 9, 9, 4, 12, 8, 5, 5, 7, 3, 18, 8, 6

Offset: 0

Author

Zhi-Wei Sun, Mar 06 2017

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 19, 32, 47, 115, 200, 974, 1271, 2240, 2549, 3185, 4865, 9254, 15881.

By the Gauss-Legendre theorem, for any nonnegative integer n, we can write 4*n + 2 as u^2 + v^2 + (2*z)^2 with u,v,z integers and u == v (mod 2), and hence 2*n + 1 = x^2 + y^2 + 2*z^2 with x = (u+v)/2 and y = (u-v)/2.

The conjecture implies that any positive integer with even 2-adic order can be written as x^2 + y^2 + 2*z^2 with x,y,z integers such that 2*x + y + z is a square or twice a square.

See also A283299 for a similar conjecture.

Examples

			a(0) = 2 since 2*0 + 1 = 1^2 + 0^2 + 2*0^2 with 2*1 + 0 + 0 = 2^1, and 2*0 + 1 = 0^2 + 1^2 + 2*0^2 with 2*0 + 1 + 0 = 1^2.
a(19) = 1 since 2*19 + 1 = 1^2 + 6^2 + 2*1^2 with 2*1 + 6 + 1 = 3^2.
a(32) = 1 since 2*32 + 1 = 4^2 + (-7)^2 + 2*0^2 with 2*4 + (-7) + 0 = 1^2.
a(47) = 1 since 2*47 + 1 = 6^2 + (-3)^2 + 2*(-5)^2 with 2*6 + (-3) + (-5) = 2^2.
a(115) = 1 since 2*115 + 1 = 10^2 + (-9)^2 + 2*5^2 with 2*10 + (-9) + 5 = 4^2.
a(200) = 1 since 2*200 + 1 = (-3)^2 + 0^2 + 2*14^2 with 2*(-3) + 0 + 14 = 2^3.
a(974) = 1 since 2*974 + 1 = 26^2 + (-25)^2 + 2*(-18)^2 with 2*26 + (-25) + (-18) = 3^2.
a(1271) = 1 since 2*1271 + 1 = 14^2 + 13^2 + 2*(-33)^2 with 2*14 + 13 + (-33) = 2^3.
a(2240) = 1 since 2*2240 + 1 = 28^2 + (-13)^2 + 2*(-42)^2 with 2*28 + (-13) + (-42) = 1^2.
a(2549) = 1 since 2*2549 + 1 = 59^2 + (-40)^2 + 2*3^2 with 2*59 + (-40) + 3 = 9^2.
a(3185) = 1 since 2*3185 + 1 = 33^2 + (-72)^2 + 2*7^2 with 2*33 + (-72) + 7 = 1^2.
a(4865) = 1 since 2*4865 + 1 = 72^2 + (-63)^2 + 2*(-17)^2 with 2*72 + (-63) + (-17) = 8^2.
a(9254) = 1 since 2*9254 + 1 = 61^2 + 26^2 + 2*(-84)^2 with 2*61 + 26 + (-84) = 8^2.
a(15881) = 1 since 2*15881 + 1 = (-48)^2 + 153^2 + 2*(-55)^2 with 2*(-48) + 153 + (-55) = 2^1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
• Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Crossrefs

Cf. A000079, A000290, A283239, A283269, A283273, A283299.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=n>0&&IntegerQ[Log[2,n]];
TQ[n_]:=TQ[n]=SQ[n]||Pow[n];
Do[r=0;Do[If[SQ[2n+1-2x^2-y^2]&&TQ[(-1)^i*x+(-1)^j*y+(-1)^k*2*Sqrt[2n+1-2x^2-y^2]],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[2n+1-2x^2]},{i,0,Min[x,1]},{j,0,Min[y,1]},{k,0,Min[Sqrt[2n+1-2x^2-y^2],1]}];Print[n," ",r],{n,0,80}]