Emilio Martín - cp's OEIS frontend

User: Emilio Martín

Emilio Martín has authored 2 sequences.

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

Emilio Martín, May 07 2025

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

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

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

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).

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

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))

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

A382209 Numbers k such that 10+k and 10*k are perfect squares.

Original entry on oeis.org

90, 136890, 197402490, 284654260890, 410471246808090, 591899253243012090, 853518312705176632890, 1230772815021611461622490, 1774773545742851022483004890, 2559222222188376152809031436090, 3690396669622092669499600847844090, 5321549438372835441042271613559748890

Offset: 1

Emilio Martín, Mar 18 2025

The limit of a(n+1)/a(n) is 1441.99930651839... = 721+228*sqrt(10) = (19+6*sqrt(10))^2.

If 10*A158490(n) is a perfect square, then A158490(n) is a term.

			90 is a term because 10+90=100 is a square and 10*90=900 is a square.
(3,1) is a solution to x^2 - 10*y^2 = -1, from which a(n) = 100*y^2-10 = 10*x^2 = 90.

Emilio Martín, Table of n, a(n) for n = 1..100
Wikipedia, Negative Pell equation (in German)
Wikipedia, Pell's equation
Index entries for linear recurrences with constant coefficients, signature (1443,-1443,1).

Subsequence of A158490.
Cf. A097315, A005667, A173127, A081071.
Cf. A383734 = 2*A008843 (2+k and 2*k are squares).
Cf. 5*A075796^2 (5+k and 5*k are squares).
Cf. 5*A081071 (20+k and 20*k are squares).
Cf. A245226 (m such that k+m and k*m are squares).

CoefficientList[Series[ 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)),{x,0,11}],x] (* or *) LinearRecurrence[{1443,-1443,1},{90,136890,197402490},12] (* James C. McMahon, May 08 2025 *)

from itertools import islice
def A382209_gen(): # generator of terms
    x, y = 30, 10
    while True:
        yield x**2//10
        x, y = x*19+y*60, x*6+y*19
A382209_list = list(islice(A382209_gen(),30)) # Chai Wah Wu, Apr 24 2025

a(n) = 10 * ((1/2) * (3+sqrt(10))^(2*n-1) + (1/2) * (3-sqrt(10))^(2*n-1))^2.
a(n) = 10 * (sinh((2n-1) * arcsinh(3)))^2.
a(n) = 10 * A173127(n)^2 = 100 * A097315(n)^2 - 10 (negative Pell's equation solutions).
a(n+2) = 1442 * a(n+1) - a(n) + 7200.
G.f.: 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)). - Stefano Spezia, Apr 24 2025

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User: Emilio Martín

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

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