A056771 -id:A056771 - cp's OEIS frontend

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

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

A280181 Indices of centered 9-gonal numbers (A060544) that are also squares (A000290).

Original entry on oeis.org

1, 17, 561, 19041, 646817, 21972721, 746425681, 25356500417, 861374588481, 29261379507921, 994025528680817, 33767606595639841, 1147104598723073761, 38967788749988868017, 1323757712900898438801, 44968794449880558051201, 1527615253583038075302017

Offset: 1

Colin Barker, Dec 28 2016

Also positive integers y in the solutions to 2*x^2 - 9*y^2 + 9*y - 2 = 0, the corresponding values of x being A046176.

Consider all ordered triples of consecutive integers (k, k+1, k+2) such that k is a square and k+1 is twice a square; then the values of k are the squares of the NSW numbers (A002315), the values of k+1 are twice the squares of the odd Pell numbers (A001653), and the values of k+2 are thrice the terms of this sequence. (See the Example section.) - Jon E. Schoenfield, Sep 06 2019

			17 is in the sequence because the 17th centered 9-gonal number is 1225, which is also the 35th square.
From _Jon E. Schoenfield_, Sep 06 2019: (Start)
The following table illustrates the relationship between the NSW numbers (A002315), the odd Pell numbers (A001653), and the terms of this sequence:
.
  |  A002315(n-1)^2  |   2*A001653(n)^2  |
n |   = 3*a(n) - 2   |    = 3*a(n) - 1   |       3*a(n)
--+------------------+-------------------+-------------------
1 |    1^2 =       1 |   1^2*2 =       2 |      1*3 =       3
2 |    7^2 =      49 |   5^2*2 =      50 |     17*3 =      51
3 |   41^2 =    1681 |  29^2*2 =    1682 |    561*3 =    1683
4 |  239^2 =   57121 | 169^2*2 =   57122 |  19041*3 =   57123
5 | 1393^2 = 1940449 | 985^2*2 = 1940450 | 646817*3 = 1940451
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001653, A002315, A046176, A046177, A056771, A060544, A077420.

Cf. A156164.

LinearRecurrence[{35, -35, 1}, {1, 17, 561}, 50] (* G. C. Greubel, Dec 28 2016 *)

Vec(x*(1 - 18*x + x^2) / ((1 - x)*(1 - 34*x + x^2)) + O(x^20))

a(n) = (6 + (3-2*sqrt(2))*(17+12*sqrt(2))^(-n) + (3+2*sqrt(2))*(17+12*sqrt(2))^n) / 12.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
G.f.: x*(1 - 18*x + x^2) / ((1 - x)*(1 - 34*x + x^2)).
a(n) = (A002315(n-1)^2 + 2)/3 = (2*A001653(n)^2 + 1)/3. - Jon E. Schoenfield, Sep 06 2019
a(n) = A077420(floor((n-1)/2)) * A056771(floor(n/2)). - Jon E. Schoenfield, Sep 08 2019
E.g.f.: -1+(1/12)*(6*exp(x)+(3-2*sqrt(2))*exp((17-12*sqrt(2))*x)+(3+2*sqrt(2))*exp((17+12*sqrt(2))*x)). - Stefano Spezia, Sep 08 2019
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2) = A156164. - Andrea Pinos, Oct 07 2022

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

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