A158490 -id:A158490 - cp's OEIS frontend

A158491 a(n) = 20*n^2 - 1.

19, 79, 179, 319, 499, 719, 979, 1279, 1619, 1999, 2419, 2879, 3379, 3919, 4499, 5119, 5779, 6479, 7219, 7999, 8819, 9679, 10579, 11519, 12499, 13519, 14579, 15679, 16819, 17999, 19219, 20479, 21779, 23119, 24499, 25919, 27379, 28879, 30419, 31999, 33619, 35279

Offset: 1

Vincenzo Librandi, Mar 20 2009

The identity (20*n^2 - 1)^2 - (100*n^2 - 10)*(2*n)^2 = 1 can be written as a(n)^2 - A158490(n)*A005843(n)^2 = 1.

Sequence found by reading the line from 19, in the direction 19, 79, ... in the square spiral whose vertices are the generalized dodecagonal numbers A195162. - Omar E. Pol, Nov 05 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A134538, A158490, A195162.

I:=[19, 79, 179]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

LinearRecurrence[{3,-3,1},{19,79,179},50]
20*Range[40]^2-1 (* Harvey P. Dale, Aug 24 2021 *)

a(n)=20*n^2-1 \\ Charles R Greathouse IV, Dec 23 2011

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f: x*(-19-22*x+x^2)/(x-1)^3.
From Amiram Eldar, Mar 06 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) - 1)/2. (End)
From Elmo R. Oliveira, Jan 25 2025: (Start)
E.g.f.: exp(x)*(20*x^2 + 20*x - 1) + 1.
a(n) = A134538(2*n). (End)

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

cp's OEIS Frontend

A158491 a(n) = 20*n^2 - 1.

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