A011943 -id:A011943 - cp's OEIS frontend

A011922 a(n) = 15a(n-1) - 15a(n-2) + a(n-3) with a(0)=1, a(1)=3, and a(2)=33.

1, 3, 33, 451, 6273, 87363, 1216801, 16947843, 236052993, 3287794051, 45793063713, 637815097923, 8883618307201, 123732841202883, 1723376158533153, 24003533378261251, 334326091137124353, 4656561742541479683, 64857538304443591201, 903348974519668797123, 12582028104970919568513

Offset: 0

Mario Velucchi (mathchess(AT)velucchi.it)

Mario Velucchi, Seeing couples, in Recreational and Educational Computing, to appear 1997.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 31, 56.
Z. Franusic, On the Extension of the Diophantine Pair {1,3} in Z[surd d], J. Int. Seq. 13 (2010) # 10.9.6.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001353, A001075, A007655, A011916, A011918, A011920, A011943, A103974.

I:=[1,3,33]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..17]]; // Bruno Berselli, Jul 09 2011

a:= gfun:-rectoproc({a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), a(0)=1,a(1)=3,a(2)=33},a(n),remember):
map(a,[$0..100]); # Robert Israel, Jul 02 2015

RecurrenceTable[{a[n] == 15 a[n - 1] - 15 a[n - 2] + a[n - 3], a[0] == 1, a[1] == 3, a[2] == 33}, a, {n, 0, 15}] (* Michael De Vlieger, Jul 02 2015 *)
LinearRecurrence[{15,-15,1},{1,3,33},30] (* Harvey P. Dale, Dec 04 2018 *)

a[0]:1$ a[1]:3$ a[2]:33$ a[n]:=15*a[n-1]-15*a[n-2]+a[n-3]$ makelist(a[n], n, 0, 16); /* Bruno Berselli, Jul 09 2011 */

a(n)=([0,1,0; 0,0,1; 1,-15,15]^n*[1;3;33])[1,1] \\ Charles R Greathouse IV, Jul 02 2015

a(n) = (2+sqrt(1+((((2+sqrt(3))^(2*n)-(2-sqrt(3))^(2*n))^2)/4)))/3. [corrected by Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 30 2001]
a(n) = ((7+4*sqrt(3))^n+(7-4*sqrt(3))^n+4)/6. - Bruno Berselli, Jul 09 2011
G.f.: (1-12*x+3*x^2)/ ((1-x) * (x^2-14*x+1)). - R. J. Mathar, Apr 15 2010
Sqrt(3) = 1 + Sum_{n>=1} 2/a(n) = 1 + 2/3 + 2/33 + ... - Gary W. Adamson, Jun 12 2003
a(n)^2 = A103974(n+1)^2 - (4*A007655(n+1))^2. - Paul D. Hanna, Mar 06 2005
a(n) = (A011943(n+1) + 2)/3. - Ralf Stephan, Aug 13 2013
a(n) = A001075(n)^2 - A001353(n)^2. - Richard R. Forberg, Aug 24 2013
E.g.f.: exp(x)*(2 + exp(6*x)*cosh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 11 2022

Recurrence in definition by R. J. Mathar, Apr 15 2010

A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.

Original entry on oeis.org

2, 14, 194, 2702, 37634, 524174, 7300802, 101687054, 1416317954, 19726764302, 274758382274, 3826890587534, 53301709843202, 742397047217294, 10340256951198914, 144021200269567502, 2005956546822746114, 27939370455248878094, 389145229826661547202, 5420093847118012782734

Offset: 0

Lekraj Beedassy, May 13 2003

Solves for x in x^2 - 3*y^2 = 4. [Complete nonnegative solutions are in A003500 and A052530. - Wolfdieter Lang, Sep 05 2021]
For n>0, a(n)+2 is the number of dimer tilings of a 4 X 2n Klein bottle (cf. A103999).
This is the Lucas sequence V(14,1). In addition to the comment above: If x = a(n) then y(n) = (a(n+1) - a(n-1))/24, n >= 1. - Klaus Purath, Aug 17 2021

			G.f. = 2 + 14*x + 194*x^2 + 2702*x^3 + 37634*x^4 + 524174*x^5 + ...

G. C. Greubel, Table of n, a(n) for n = 0..850
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (14,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A067900, A302332.
Row 2 * 2 of array A188644.
Cf. A001075, A001353, A003500, A052530.

m:=7;; a:=[2,14];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[Floor((2+Sqrt(3))^(2*n)+(1+Sqrt(3))^(-n)): n in [0..19]]; // Vincenzo Librandi, Mar 31 2011

a := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(14) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od:
seq( simplify(2*ChebyshevT(n, 7)), n=0..20); # G. C. Greubel, Dec 23 2019

a[0]=2; a[1]=14; a[n_]:= 14a[n-1] -a[n-2]; Table[a[n], {n,0,20}] (* Robert G. Wilson v, Jan 30 2004 *)
2*ChebyshevT[Range[21] -1, 7] (* G. C. Greubel, Dec 23 2019 *)

vector( 21, n, 2*polchebyshev(n-1, 1, 7) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number2(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 26 2008

[2*chebyshev_T(n,7) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: 2*(1-7*x)/(1-14*x+x^2). - N. J. A. Sloane, Nov 22 2006
a(n) = p^n + q^n, where p = 7 + 4*sqrt(3) and q = 7 - 4*sqrt(3). - Tanya Khovanova, Feb 06 2007
a(n) = 2*A011943(n+1). - R. J. Mathar, Sep 27 2014
From Peter Bala, Oct 16 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 7 - 4*sqrt(3). This sequence gives the partial denominators in the simple continued fraction expansion of 1 + F(alpha) = 2.07140228197873197080... = 2 + 1/(14 + 1/(194 + 1/(2702 + ...))). Cf. A005248.
12*Sum_{n >= 1} 1/(a(n) - 16/a(n)) = 1.
16*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 12/a(n)) = 1.
Series acceleration formula for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/12 - 16*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(7-4*sqrt(3)))^2 - 1 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
(End)
From Klaus Purath, Aug 17 2021: (Start)
a(n) = (a(n-1)*a(n-2) + 2688)/a(n-3), n >= 3.
a(n) = (a(n-1)^2 + 192)/a(n-2), n >= 2.
a(2*n) = A302332(n-1) + A302332(n), n >= 1.
a(2*n+1) = 14*A302332(n). (End)
a(n) = A003500(2*n) = S(2*n,4) - S(2*n-2, 4) = 2*T(2*n,2), for n >= 0, with Chebyshev S and T. S(n, 4) = A001353(n+1) and T(n, 2) = A001075(n). - Wolfdieter Lang, Sep 06 2021

A101124 Number triangle associated to Chebyshev polynomials of first kind.

Original entry on oeis.org

1, 0, 1, -1, 1, 1, 0, 1, 2, 1, 1, 1, 7, 3, 1, 0, 1, 26, 17, 4, 1, -1, 1, 97, 99, 31, 5, 1, 0, 1, 362, 577, 244, 49, 6, 1, 1, 1, 1351, 3363, 1921, 485, 71, 7, 1, 0, 1, 5042, 19601, 15124, 4801, 846, 97, 8, 1, -1, 1, 18817, 114243, 119071, 47525, 10081, 1351, 127, 9, 1, 0, 1, 70226, 665857, 937444, 470449, 120126, 18817, 2024, 161

Offset: 0

Paul Barry, Dec 02 2004

			As a number triangle, rows begin:
  {1},
  {0,1},
  {-1,1,1},
  {0,1,2,1},
  ...
As a square array, rows begin
   1, 1,  1,   1,    1, ...
   0, 1,  2,   3,    4, ...
  -1, 1,  7,  17,   31, ...
   0, 1, 26,  99,  244, ...
   1, 1, 97, 577, 1921, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Index entries for sequences related to Chebyshev polynomials.

Columns include A001075, A001541, A001091, A001079, A023038, A011943.
Row sums are A101125.
Diagonal sums are A101126.
Main diagonal gives A115066.
Mirror of A322836.
Cf. A053120.

T[n_, k_] := SeriesCoefficient[x^k (1 - k x)/(1 - 2 k x + x^2), {x, 0, n}];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 12 2017 *)

Number triangle S(n, k)=T(n-k, k), k

Columns have g.f. x^k(1-kx)/(1-2kx+x^2).

Also, square array if(n=0, 1, T(n, k)) read by antidiagonals.

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1

Offset: 0

Seiichi Manyama, Dec 26 2018

			Square array begins:
   1,  1,   1,    1,      1,       1,         1, ...
   1,  3,  17,   99,    577,    3363,     19601, ...
   1,  5,  49,  485,   4801,   47525,    470449, ...
   1,  7,  97, 1351,  18817,  262087,   3650401, ...
   1,  9, 161, 2889,  51841,  930249,  16692641, ...
   1, 11, 241, 5291, 116161, 2550251,  55989361, ...
   1, 13, 337, 8749, 227137, 5896813, 153090001, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-3 give A000012, A005408, A069129(n+1), A322830.
Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986.
Main diagonal gives A173174.
A(n-1,n) gives A173148(n).
Cf. A322699, A322747.

A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *)

a(n) = 2 * A322699(n) + 1.
A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k).
A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k).
A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1.
A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = 2*n+1.

A094347 a(n) = 14*a(n-1) - a(n-2); a(0) = a(1) = 2.

Original entry on oeis.org

2, 2, 26, 362, 5042, 70226, 978122, 13623482, 189750626, 2642885282, 36810643322, 512706121226, 7141075053842, 99462344632562, 1385331749802026, 19295182152595802, 268747218386539202, 3743165875258953026

Offset: 0

Lekraj Beedassy, Jun 03 2004

Even x satisfying the Pellian x^2 - 3*y^2 = 1. For corresponding y see A028230.

Michael De Vlieger, Table of n, a(n) for n = 0..874
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = 2*A001570(n).
Bisection of A001075.
Cf. A028230.

LinearRecurrence[{14,-1},{2,2},40] (* or *) CoefficientList[ Series[2(1-13x)/(1-14x+x^2),{x,0,39}],x] (* Harvey P. Dale, Apr 23 2011 *)

(a[0]:2, a[1]:2, a[n] := 14*a[n - 1] - a[n-2], makelist(a[n], n, 0, 50)); /* Franck Maminirina Ramaharo, Nov 12 2018 */

G.f.: 2*(1 - 13*x)/(1 - 14*x + x^2). [Philippe Deléham, Nov 17 2008]
a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1))/2. - Gerry Martens, Jun 03 2015
a(n) = (1/2)*sqrt(4 + (-2*sqrt(-2 + (7 - 4*sqrt(3))^(2*n) + (7 + 4*sqrt(3))^(2*n)) + sqrt(3)*sqrt(2 + (7 - 4*sqrt(3))^(2*n) + (7 + 4*sqrt(3))^(2*n)))^2). - Gerry Martens, Jun 03 2015
E.g.f.: exp(7*x)*(2*cosh(4*sqrt(3)*x) - sqrt(3)*sinh(4*sqrt(3)*x)). - Franck Maminirina Ramaharo, Nov 12 2018

Corrected by Lekraj Beedassy, Jun 11 2004

A141041 a(n) = ((3 + 2sqrt(3))^n + (3 - 2sqrt(3))^n)/2.

Original entry on oeis.org

1, 3, 21, 135, 873, 5643, 36477, 235791, 1524177, 9852435, 63687141, 411680151, 2661142329, 17201894427, 111194793549, 718774444575, 4646231048097, 30033709622307, 194140950878133, 1254946834135719, 8112103857448713

Offset: 0

Roger L. Bagula, Aug 18 2008

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,3).

Cf. A011943, A034478, A081336, A099842, A201701.

[n le 2 select 3^(n-1) else 6*Self(n-1) +3*Self(n-2): n in [1..31]]; // G. C. Greubel, Oct 10 2022

a[n_]= ((3+2*Sqrt[3])^n + (3-2*Sqrt[3])^n)/2; Table[FullSimplify[a[n]], {n,0,30}]
LinearRecurrence[{6,3},{1,3},30] (* Harvey P. Dale, Aug 25 2014 *)

A141041 = BinaryRecurrenceSequence(6,3,1,3)
[A141041(n) for n in range(31)] # G. C. Greubel, Oct 10 2022

a(n) = 3*abs(A099842(n-1)), for n > 0.
G.f.: (1-3*x)/(1-6*x-3*x^2). - Philippe Deléham, Mar 03 2012
a(n) = 6*a(n-1) + 3*a(n-2), a(0) = 1, a(1) = 3. - Philippe Deléham, Mar 03 2012
a(n) = Sum_{k=0..n} A201701(n,k)*3^(n-k). - Philippe Deléham, Mar 03 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(4*k-3)/(x*(4*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) = (-i*sqrt(3))^n * ChebyshevT(n, i*sqrt(3)). - G. C. Greubel, Oct 10 2022

Edited by N. J. A. Sloane, Aug 24 2008

A011944 a(n) = 14*a(n-1) - a(n-2) with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 28, 390, 5432, 75658, 1053780, 14677262, 204427888, 2847313170, 39657956492, 552364077718, 7693439131560, 107155783764122, 1492487533566148, 20787669686161950, 289534888072701152

Offset: 0

E. K. Lloyd

Standard deviation of A011943.
Product x*y, where the pair (x, y) solves for x^2 - 3y^2 = 1, i.e., a(n)=A001075(n)*A001353(n). - Lekraj Beedassy, Jul 13 2006
Solutions m to the Diophantine equation where square m^2 = k*(k+1)/3, corresponding solutions k are in A007654. - Bernard Schott, Apr 10 2021
All solutions for y in Pell equation x^2 - 12*y^2 = 1. Corresponding values for x are in A011943. - Herbert Kociemba, Jun 05 2022

Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = 2 * A007655 = {A001353(2n)}/2. Cf. A011943.

Cf. A007654.

LinearRecurrence[{14,-1},{0,2},20] (* Harvey P. Dale, Oct 17 2019 *)
Table[2 ChebyshevU[-1 + n, 7], {n, 0, 18}] (* Herbert Kociemba, Jun 05 2022 *)

For all members x of the sequence, 12*x^2 +1 is a square. Lim_{n->infinity} a(n)/a(n-1) = 7 + sqrt(12). - Gregory V. Richardson, Oct 13 2002
a(n) = ((7+2*sqrt(12))^(n-1) - (7-2*sqrt(12))^(n-1)) / (2*sqrt(12)). - Gregory V. Richardson, Oct 13 2002
a(n) = 13*(a(n-1) + a(n-2)) - a(n-3). a(n) = 15*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, Sep 20 2006
a(n) = sinh(2n*arcsinh(sqrt(3)))/sqrt(12). - Herbert Kociemba, Apr 24 2008
G.f.: 2x/(1-14*x+x^2). - Philippe Deléham, Nov 17 2008

A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1).

Original entry on oeis.org

1, -3, 13, -48, 181, -675, 2521, -9408, 35113, -131043, 489061, -1825200, 6811741, -25421763, 94875313, -354079488, 1321442641, -4931691075, 18405321661, -68689595568, 256353060613, -956722646883, 3570537526921, -13325427460800, 49731172316281

Offset: 0

Creighton Dement, Jul 21 2005

In reference to program code, 2baseiseq[X](n) = ((-1)^n)*A001353(n) (a(n)^2 + 1 is a perfect square.) 1tesseq[X](n) = (-1^(n+1))*A097948(n).

Floretion Algebra Multiplication Program, FAMP Code: 1ibaseiseq[X] with X = .5'i + .5i' + 'ii' - .5'jj' + 1.5'kk' - 1 (* Corrected by Creighton Dement, Dec 11 2009 *)

Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (-4,0,4,1).

Cf. A007654, A001570, A076139. See also A117808, A122571 (same except for signs).

Cf. A001353, A097948.

/* By definition: */
m:=15; R:=PowerSeriesRing(Integers(), m);
A001570:=Coefficients(R!((1-x)/(1-14*x+x^2)));
A007654:=Coefficients(R!(-3*x^2*(1+x)/(-1+x)/(1-14*x+x^2)));
&cat[[A001570[i],-A007654[i]]: i in [1..m-2]]; // Bruno Berselli, Feb 05 2013

seriestolist(series((x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)), x=0,25));

LinearRecurrence[{-4,0,4,1},{1,-3,13,-48},30] (* Harvey P. Dale, Jun 15 2018 *)

G.f.: (x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)).
Floor(((2 + sqrt(3))^n + (2 - sqrt(3))^n)/4) produces this sequence with a different offset and without signs. - James R. Buddenhagen, May 20 2010
Define c(n) = a(n) - 4*a(n+1) - a(n+2) and d(n) = -a(n) - 4*a(n+1) - a(n+2); Conjectures: I: c(2n) = 24*A076139(n); (Triangular numbers that are one-third of another triangular number) II: c(2n+1) = -A011943(n+1); (Numbers n such that any group of n consecutive integers has integral standard deviation) III: d(2n) = -2; IV: d(2n+1) = -1

A144536 Denominators of continued fraction convergents to sqrt(3)/2.

Original entry on oeis.org

1, 1, 7, 15, 97, 209, 1351, 2911, 18817, 40545, 262087, 564719, 3650401, 7865521, 50843527, 109552575, 708158977, 1525870529, 9863382151, 21252634831, 137379191137, 296011017105, 1913445293767, 4122901604639, 26650854921601, 57424611447841, 371198523608647

Offset: 0

N. J. A. Sloane, Dec 29 2008

			0, 1, 6/7, 13/15, 84/97, 181/209, 1170/1351, 2521/2911, 16296/18817, 35113/40545, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
John Elias, Double-Snowflake Fraction
Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A126664, A144535, A002531/A002530.

Bisections give A011943, A028230.

with(numtheory); Digits:=200: cf:=convert(evalf(sqrt(3)/2,confrac); [seq(nthconver(cf,i), i=0..100)];

Denominator[Convergents [Sqrt[3]/2, 30]] (* Vincenzo Librandi, Feb 01 2014 *)
LinearRecurrence[{0,14,0,-1},{1,1,7,15},30] (* Harvey P. Dale, Sep 15 2017 *)

G.f.: (1 + x - 7*x^2 + x^3)/(1 - 14*x^2 + x^4). - Colin Barker, Jan 01 2012
a(n) = 14*a(n-2) - a(n-4). - Sergei N. Gladkovskii, Jun 07 2015
a(n) = ((3+sqrt(3))*((-2+sqrt(3))^n + (2+sqrt(3))^n) - (-3+sqrt(3))*((-2-sqrt(3))^n + (2-sqrt(3))^n))/12. - Vaclav Kotesovec, Jun 08 2015
From John Elias, Dec 02 2021: (Start)
a(2*n) = 6*A001353(n)^2 + 1. See illustration in links.
a(2*n+1) = 2*a(2*n) + a(2*n-1). (End)

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

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cp's OEIS Frontend

A011922 a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) with a(0)=1, a(1)=3, and a(2)=33.

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A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.

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A101124 Number triangle associated to Chebyshev polynomials of first kind.

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A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j.

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A094347 a(n) = 14*a(n-1) - a(n-2); a(0) = a(1) = 2.

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A141041 a(n) = ((3 + 2*sqrt(3))^n + (3 - 2*sqrt(3))^n)/2.

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A011944 a(n) = 14*a(n-1) - a(n-2) with a(0) = 0, a(1) = 2.

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A011922 a(n) = 15a(n-1) - 15a(n-2) + a(n-3) with a(0)=1, a(1)=3, and a(2)=33.

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

A141041 a(n) = ((3 + 2sqrt(3))^n + (3 - 2sqrt(3))^n)/2.