A344058 Number of ways to write n as x + y + z with x*y + 2*y*z + 3*z*x a square, where x,y,z are positive integers with x or y a power of two (including 2^0 = 1).

0, 0, 0, 1, 2, 1, 1, 1, 3, 2, 2, 5, 3, 2, 5, 1, 5, 5, 2, 8, 5, 3, 9, 5, 3, 8, 4, 7, 7, 6, 11, 1, 8, 5, 4, 14, 6, 2, 5, 8, 9, 6, 8, 11, 8, 10, 5, 5, 13, 5, 7, 18, 17, 6, 9, 7, 5, 7, 6, 14, 11, 12, 7, 1, 12, 10, 14, 9, 13, 6, 10, 14, 14, 11, 10, 9, 7, 6, 10, 8, 8, 12, 7, 12, 12, 10, 11, 11, 8, 10, 10, 25, 15, 7, 18, 5, 11, 13, 13, 12

Offset

1,5

Comments

Conjecture: a(n) > 0 for all n > 3.

We have verified a(n) > 0 for all n = 4..10^5. Clearly, a(2*n) > 0 if a(n) > 0.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

Chao Huang and Zhi-Wei Sun, On partitions of integers with restrictions involving squares, arXiv:2105.03416 [math.NT], 2021.

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].

Example

a(6) = 1 with 6 = 3 + 2^0 + 2 and 3*2^0 + 2*2^0*2 + 3*2*3 = 5^2.
a(7) = 1 with 7 = 3 + 2^0 + 3 and 3*2^0 + 2*2^0*3 + 3*3*3 = 6^2.
For each k > 1, we have a(2^k) = 1 with 2^k = 2^(k-2) + 2^(k-1) + 2^(k-2) and 2^(k-2)*2^(k-1) + 2*2^(k-1)*2^(k-2) + 3*2^(k-2)*2^(k-2) = (3*2^(k-2))^2.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]];
Pow[x_]:=x>0&&IntegerQ[Log[2, x]];
tab={}; Do[r=0; Do[If[(Pow[x]||Pow[y])&&SQ[x*y+(2y+3x)*(n-x-y)], r=r+1], {x, 1, n-2}, {y, 1, n-1-x}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

Crossrefs

Cf. A000290, A271644, A340274, A343862, A343897, A343950.

Sequence in context: A161092 A029332 A358172 * A134431 A211098 A070879

Adjacent sequences: A344055 A344056 A344057 * A344059 A344060 A344061

Keyword

nonn

Author

Zhi-Wei Sun, May 08 2021

Status

approved