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

0,1

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.

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.

Example

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.

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}]

Crossrefs

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

Sequence in context: A134017 A292285 A163984 * A048186 A352196 A060762

Adjacent sequences: A283363 A283364 A283365 * A283367 A283368 A283369

Keyword

nonn

Author

Zhi-Wei Sun, Mar 06 2017

Status

approved