A114047 -id:A114047 - cp's OEIS frontend

A075871 Numbers k such that 13*k^2 + 1 is a square.

0, 180, 233640, 303264540, 393637139280, 510940703520900, 663200639532988920, 860833919173116097260, 1117361763886065161254560, 1450334708690193406192321620, 1882533334518107155172472208200, 2443526817869794397220462733921980

Offset: 1

Gregory V. Richardson, Oct 16 2002

Limit_{n->infinity} a(n)/a(n-1) = 649 + 180*sqrt(13).

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 13*y^2 = 1. The corresponding x values are in A114047. - Vincenzo Librandi, Aug 08 2010, edited by Jon E. Schoenfield, May 04 2014

Colin Barker, Table of n, a(n) for n = 1..322
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1298,-1).

Cf. A114047.

Cf. A202156.

I:=[0,180]; [n le 2 select I[n] else 1298*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jun 14 2015

LinearRecurrence[{1298, -1}, {0, 180}, 50] (* Vincenzo Librandi, Jun 14 2015 *)

concat(0, Vec(180*x^2/(1-1298*x+x^2) + O(x^20))) \\ Colin Barker, Jun 13 2015

a(n) = 1298*a(n-1) - a(n-2), n>1. - Michael Somos, Oct 30 2002
a(n) = ((649 + 180*sqrt(13))^n - (649 - 180*sqrt(13))^n) / (2*sqrt(13)).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 1297*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 1299*(a(n-1) - a(n-2)) + a(n-3). (End)
G.f.: 180*x^2/(1-1298*x+x^2). - Philippe Deléham, Nov 18 2008

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

Original entry on oeis.org

0, 2, 36, 646, 11592, 208010, 3732588, 66978574, 1201881744, 21566892818, 387002188980, 6944472508822, 124613502969816, 2236098580947866, 40125160954091772, 720016798592704030

Offset: 0

Gary Detlefs, Feb 20 2012

Denote as {a,b,c,d} the second-order linear recurrence a(n) = c*a(n-1) + d*a(n-2) with initial terms a, b. The following sequences and recurrence formulas are related to integer solutions of k*x^2 + 1 = y^2.
.
   k             x                        y
   -  -----------------------  -----------------------
A001542 {0,2,6,-1}       A001541 {1,3,6,-1}
A001353 {0,1,4,-1}       A001075 {1,2,4,-1}
A060645 {0,4,18,-1}      A023039 {1,9,18,-1}
A001078 {0,2,10,-1}      A001079 {1,5,10,-1}
A001080 {0,3,16,-1}      A001081 {1,8,16,-1}
A001109 {0,1,6,-1}       A001541 {1,3,6,-1}
A084070 {0,1,38,-1}      A078986 {1,19,38,-1}
A001084 {0,3,20,-1}      A001085 {1,10,20,-1}
A011944 {0,2,14,-1}      A011943 {1,7,14,-1}
A075871 {0,180,1298,-1}  A114047 {1,649,1298,-1}
A068204 {0,4,30,-1}      A069203 {1,15,30,-1}
A001090 {0,1,8,-1}       A001091 {1,4,8,-1}
A121740 {0,8,66,-1}      A099370 {1,33,66,-1}
A202299 {0,4,34,-1}      A056771 {1,17,34,-1}
A174765 {0,39,340,-1}    A114048 {1,179,340,-1}
a(n)    {0,2,18,-1}      A023039 {1,9,18,-1}
A174745 {0,12,110,-1}    A114049 {1,55,110,-1}
A174766 {0,42,394,-1}    A114050 {1,197,394,-1}
A174767 {0,5,48,-1}      A114051 {1,24,48,-1}
A004189 {0,1,10,-1}      A001079 {1,5,10,-1}
A174768 {0,10,102,-1}    A099397 {1,51,102,-1}
The sequence of the c parameter is listed in A180495.

Bruno Berselli, Table of n, a(n) for n = 0..500
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039, A049660, A079962.

m:=16; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2*x/(1-18*x+x^2))); // Bruno Berselli, Jun 19 2019

readlib(issqr):for x from 1 to 720016798592704030 do if issqr(20*x^2+1) then print(x) fi od;

LinearRecurrence[{18, -1}, {0, 2}, 16] (* Bruno Berselli, Feb 21 2012 *)
Table[2 ChebyshevU[-1 + n, 9], {n, 0, 16}]  (* Herbert Kociemba, Jun 05 2022 *)

makelist(expand(((2+sqrt(5))^(2*n)-(2-sqrt(5))^(2*n))/(4*sqrt(5))), n, 0, 15); /* Bruno Berselli, Jun 19 2019 */

a(n) = 18*a(n-1) - a(n-2).
From Bruno Berselli, Feb 21 2012: (Start)
G.f.: 2*x/(1-18*x+x^2).
a(n) = -a(-n) = 2*A049660(n) = ((2 + sqrt(5))^(2*n)-(2 - sqrt(5))^(2*n))/(4*sqrt(5)). (End)
a(n) = Fibonacci(6*n)/4. - Bruno Berselli, Jun 19 2019
For n>=1, a(n) = A079962(6n-3). - Christopher Hohl, Aug 22 2021

A202155 x-values in the solution to x^2 - 13*y^2 = -1.

Original entry on oeis.org

18, 23382, 30349818, 39394040382, 51133434066018, 66371158023650982, 86149711981264908618, 111822259780523827735182, 145145207045407947135357618, 188398366922679734857866452982, 244540935120431250437563520613018, 317413945387952840388222591889244382

Offset: 1

Bruno Berselli, Dec 15 2011

The corresponding values of y of this Pell equation are in A202156.

A. H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, Dover (New York), 1966, p. 264.

Bruno Berselli, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
A. M. S. Ramasamy, Polynomial solutions for the Pell's equation, Indian Journal of Pure and Applied Mathematics 25 (1994), p. 579 (Theorem 4, case t=1).
J. P. Robertson, Solving the generalized Pell equation x^2-D*y^2=N, pp. 9, 24.
Index entries for linear recurrences with constant coefficients, signature (1298,-1).

Cf. A002313, A003654, A031396, A114047, A202156.

m:=13; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(18*x*(1+x)/(1-1298*x+x^2)));

LinearRecurrence[{1298, -1}, {18, 23382}, 12]

makelist(expand(((18+5*sqrt(13))^(2*n-1)+(18-5*sqrt(13))^(2*n-1))/2), n, 1, 12);

G.f.: 18*x*(1+x)/(1-1298*x+x^2).

a(n) = -a(-n+1) = (r^(2n-1)-1/r^(2n-1))/2, where r=18+5*sqrt(13).

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A075871 Numbers k such that 13*k^2 + 1 is a square.

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A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

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A202155 x-values in the solution to x^2 - 13*y^2 = -1.

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