A338096 Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + w^2 with x + 2*y + 3*z a positive power of two, where x, y, z, w are nonnegative integers.

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

Offset

0,3

Comments

Conjecture 1 (1-2-3 Conjecture): a(n) > 0 for all n >= 0. In other words, any positive odd integer m can be written as x^2 + y^2 + z^2 + w^2 with x, y, z, w nonnegative integers such that x + 2*y + 3*z = 2^k for some positive integer k.

Conjecture 2 (Strong Version of the 1-2-3 Conjecture): For any integer m > 4627 not congruent to 0 or 2 modulo 8, we can write m as x^2 + y^2 + z^2 + w^2 with x, y, z, w nonnegative integers such that x + 2*y + 3*z = 4^k for some positive integer k.

We have verified Conjectures 1 and 2 for m up to 5*10^6. Conjecture 2 implies that A299924(n) > 0 for all n > 0.

By Theorem 1.2(v) of the author's 2017 JNT paper, any positive integer n can be written as x^2 + y^2 + z^2 + 4^k with k, x, y, z nonnegative integers.

See also A338094 and A338095 for similar conjectures.

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. See also arXiv:1604.06723 [math.NT].

Zhi-Wei Sun, Restricted sums of four squares, Int. J. Number Theory 15(2019), 1863-1893. See also arXiv:1701.05868 [math.NT].

Zhi-Wei Sun, Sums of four squares with certain restrictions, arXiv:2010.05775 [math.NT], 2020.

Example

a(1) = 1, and 2*1 + 1 = 1^2 + 0^2 + 1^2 + 1^2 with 1 + 2*0 + 3*1 = 2^2.
a(3) = 1, and 2*3 + 1 = 1^2 + 2^2 + 1^2 + 1^2 with 1 + 2*2 + 3*1 = 2^3.
a(9) = 1, and 2*9 + 1 = 1^2 + 6^2 + 1^2 + 1^2 with 1 + 2*6 + 3*1 = 2^4.
a(21) = 1, and 2*21 + 1 = 5^2 + 4^2 + 1^2 + 1^2 with 5 + 2*4 + 3*1 = 2^4.
a(39) = 1, and 2*39 + 1 = 1^2 + 5^2 + 7^2 + 2^2 with 1 + 2*5 + 3*7 = 2^5.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[n_]:=PQ[n]=n>1&&IntegerQ[Log[2, n]];
tab={}; Do[r=0; Do[If[SQ[2n+1-x^2-y^2-z^2]&&PQ[x+2y+3z], r=r+1], {x, 0, Sqrt[2n+1]}, {y, Boole[x==0], Sqrt[2n+1-x^2]}, {z, 0, Sqrt[2n+1-x^2-y^2]}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]

Crossrefs

Cf. A000079, A000118, A000290, A000302, A279612, A299924, A338094, A338095, A338103, A338119, A338121.

Sequence in context: A342375 A055515 A363437 * A215010 A136744 A068237

Adjacent sequences: A338093 A338094 A338095 * A338097 A338098 A338099

Keyword

nonn

Author

Zhi-Wei Sun, Oct 09 2020

Status

approved