A383734 Numbers k such that 2+k and 2*k are squares.

2, 98, 3362, 114242, 3880898, 131836322, 4478554082, 152139002498, 5168247530882, 175568277047522, 5964153172084898, 202605639573839042, 6882627592338442562, 233806732499933208098, 7942546277405390632802, 269812766699283348307202, 9165691521498228451812098

Offset: 1

Author

Emilio Martín, May 07 2025

Keywords

• nonn
• easy

Comments

The limit of a(n+1)/a(n) is 33.97056... = 17+12*sqrt(2) = (3+2*sqrt(2))^2 (see A156164).

Examples

			98 is a term becouse 98+2=100 is a square and 98*2=196 is a square.
		

Links

• Eric Weisstein's World of Mathematics, NSW numbers.
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A002315, A088165, A008843, A008844, A075870, A031396, A156164.

Cf. A382209 (10+k and 10*k are squares).

Cf. A245226 (m such that k+m and k*m are squares).

Programs

Mathematica

LinearRecurrence[{35, -35, 1}, {2, 98, 3362}, 20] (* Amiram Eldar, May 07 2025 *)

Python

from itertools import islice
def A383734_gen(): # generator of terms
    x, y = 1, 7
    while True:
        yield 2*x**2
        x, y = y, 6*y - x
A383734_list = list(islice(A383734_gen(), 100))

Formula

a(n) = (1/2) * ((3+2*sqrt(2))^(2*n-1) + (3-2*sqrt(2))^(1-2*n)) - 1.

a(n) = -2*sqrt(2)*sinh(n*log(17+12*sqrt(2))) + 3*cosh(n*log(17+12*sqrt(2))) - 1.

a(n) = 2*A002315(n-1)^2.

a(n) = A075870(n)^2 - 2.

a(n) = 34*a(n-1) - a(n-2) + 32.

G.f.: 2 * (1 + 14*x + x^2) / ((1 - x)*(1 - 34*x + x^2)). - Stefano Spezia, May 08 2025