A367191 a(n) is the largest positive number k such that k^2 + k*n + n^2 is a perfect square, or 0 if no such k exists.

0, 0, 0, 5, 0, 16, 10, 33, 7, 56, 32, 85, 20, 120, 66, 161, 39, 208, 112, 261, 64, 320, 170, 385, 95, 456, 240, 533, 132, 616, 322, 705, 175, 800, 416, 901, 224, 1008, 522, 1121, 279, 1240, 640, 1365, 340, 1496, 770, 1633, 407, 1776, 912, 1925, 480, 2080, 1066, 2241, 559, 2408, 1232, 2581, 644, 2760, 1410, 2945

Offset

0,4

Comments

A companion to A300728 where "smallest" is replaced with "largest".

Links

Table of n, a(n) for n=0..63.

Formula

Conjectured formulas according to n mod 4 in first column.

0, (n/4 - 1)(3n/4 + 1),

1 or 3, 4((n + 3)/4 - 1)(3(n + 3)/4 - 2),

2, 2((n + 2)/4 - 1)(3(n + 2)/4 - 1).

Example

With n = 7, the solutions to k^2 + k*n + n^2 = j^2 are k = 8 and k = 33, therefore A300728(7) = 8 and a(7) = 33.

Mathematica

s[n_] := Solve[j > 0 && k > 0 && k^2 + k*n + n^2 == j^2, {j, k}, Integers];
a[n_] := If[n == 0, 0, With[{sn = s[n]}, Which[sn == {}, 0, IntegerQ[k /. sn[[1]]], Max[k /. sn], True, 0]]];
Table[a[n], {n, 0, 100}]

Prog

(Python)
from sympy.abc import x, y
from sympy.solvers.diophantine.diophantine import diop_quadratic
def A367191(n): return max(diop_quadratic(x*(x+n)+n**2-y**2))[0] if n else 0 # Chai Wah Wu, Nov 11 2023

Crossrefs

Cf. A003136, A300728.

Sequence in context: A291218 A321416 A226372 * A093782 A085105 A368061

Adjacent sequences: A367188 A367189 A367190 * A367192 A367193 A367194

Keyword

nonn

Author

Jean-François Alcover, Nov 09 2023

Status

approved