

A300728


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


2



1, 0, 0, 5, 0, 3, 10, 8, 7, 15, 6, 24, 20, 35, 16, 9, 5, 63, 30, 80, 12, 24, 48, 120, 11, 15, 70, 45, 32, 195, 18, 224, 10, 7, 126, 13, 60, 323, 160, 16, 24, 399, 48, 440, 96, 27, 240, 528, 15, 56, 30, 40, 140, 675, 90, 33, 9, 55, 390, 840, 36, 899, 448, 17, 20, 39, 14, 1088, 252, 91, 26, 1224, 33, 1295, 646, 45
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OFFSET

0,4


COMMENTS

A positive a(n) cannot be 1 or a multiple of n for n > 0 since there is no square in A002061 except 1. Also it is easy to show that a(n) cannot be 2 or 4 since a(2) = a(4) = 0.
If n >= 5 is odd, a(n) <= n^2/4  n/2  3/4, with a(n) = n^2/4  n/2  3/4 if n is a prime >= 5.
If n >= 10 and n == 2 (mod 4), a(n) <= n^2/8  n/2  3/2, with equality if n/2 is a prime >= 5.
If n >= 16 and n == 0 (mod 4), 1 < a(n) <= n^2/16  n/2  3, with equality if n/4 is 4 or a prime >= 5. (End)


LINKS



EXAMPLE

a(2) = 0 because k^2 + 2*k + 4 = (k + 1)^2 + 3 cannot be a square for k > 0.
a(4) = 0 because k^2 + 4*k + 16 = (k + 2)^2 + 12 cannot be a square for k > 0.
a(5) = 3 because 3^2 + 3*5 + 5^2 = 7^2 and 3 is the least positive number with this property.


MAPLE

f:= proc(n) local k;
for k from 1 do if issqr(k^2 + k*n + n^2) then return k fi od
end proc:
f(1):= 0: f(2):= 0: f(4):= 0:


MATHEMATICA

f[n_] := Module[{k},
For[k = 1, True, k++, If[IntegerQ[Sqrt[k^2 + k*n + n^2]], Return[k]]]];
f[1] = 0; f[2] = 0; f[4] = 0;


PROG

(Python)
from sympy.abc import x, y
from sympy.solvers.diophantine.diophantine import diop_quadratic
def A300728(n): return min((d[0] for d in diop_quadratic(x*(x+n)+n**2y**2) if d[0]>0), default=0) if n else 1 # Chai Wah Wu, Nov 11 2023


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



