A255236 -id:A255236 - cp's OEIS frontend

A054490 Expansion of (1+5x)/(1-6x+x^2).

1, 11, 65, 379, 2209, 12875, 75041, 437371, 2549185, 14857739, 86597249, 504725755, 2941757281, 17145817931, 99933150305, 582453083899, 3394785353089, 19786259034635, 115322768854721, 672150354093691, 3917579355707425, 22833325780150859

Offset: 0

Barry E. Williams, May 04 2000

A Pellian-related second-order recursive sequence.
Third binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
Binomial transform of A164607. - R. J. Mathar, Oct 26 2011
Pisano period lengths: 1, 1, 4, 2, 6, 4, 3, 2, 12, 6, 12, 4, 14, 3, 12, 2, 8, 12, 20, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = x1 = a(n) of the first class of the (generalized) Pell equation x^2 - 2*y^2 = -7. For the corresponding y1 terms see 2*A038723(n). All positive solutions of the second class are given by (x2(n), y2(n)) = (A255236(n), A038725(n+1)), n >= 0. See (A254938(1), 2*A255232(1)) for the fundamental solution (1, 2) of the first class. See the Nagell reference, Theorem 111, p. 210, Theorem 110, p. 208, Theorem 108a, pp. 206-207.
This sequence also gives all positive solutions y = y1 of the first class of the Pell equation x^2 - 2*y^2 = 14. The corresponding solutions x1 are given in 4*A038723. This follows from the preceding comment. (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(0) = -(2*A038761(0) - A038762(0)), a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1.
This follows from the general trivial fact that if X^2 - D*Y^2 = N (X, Y positive integers, D > 1, not a square, and N a non-vanishing integer) then x:= D*Y +/- X and y:= Y +/- X (correlated signs) satisfy x^2 - D*y^2 = -(D-1)*N. with integers x and y. Here D = 2 and N = 7. (End)

			n = 2: sqrt(8*23^2-7) = 65.
2*19 + 27  = 65. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See y(n) in Lemma 7.9 p. 21.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000129, A001109, A038723, A038725, A054488, A054489, A101386, A253811, A255236.

a:=[1,11];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 20 2020

I:=[1,11]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

a[0]:=1: a[1]:=11: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006

CoefficientList[Series[(1+5x)/(1-6x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)
LinearRecurrence[{6, -1}, {1, 11}, 30] (* G. C. Greubel, Jul 26 2018 *)

my(x='x+O('x^30)); Vec((1+5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

[lucas_number1(2*n+1,2,-1) + 3*lucas_number1(2*n,2,-1) for n in (0..30)] # G. C. Greubel, Jan 20 2020

a(n) = 6*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=11.
a(n) = sqrt(8*A038723(n)^2 - 7).
a(n) = (11*((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) - ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = 11*S(n, 6) + 5*S(n-1, 6), n >= 0, with Chebyshev's polynomials S(n, x) (A049310) evaluated at x=6: S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comments above. - Wolfdieter Lang, Feb 26 2015
a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1. See the Mar 19 2015 comment above. - Wolfdieter Lang, Mar 19 2015
From G. C. Greubel, Jan 20 2020: (Start)
a(n) = Pell(2*n+1) + 3*Pell(2*n).
a(n) = ChebyshevU(n,3) + 5*ChebyshevU(n-1,3).
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 2*sqrt(2)*sinh(2*sqrt(2)*x) ). (End)

More terms from James Sellers, May 05 2000

More terms from Vincenzo Librandi, Mar 20 2015

A038725 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 64, 373, 2174, 12671, 73852, 430441, 2508794, 14622323, 85225144, 496728541, 2895146102, 16874148071, 98349742324, 573224305873, 3340996092914, 19472752251611, 113495517416752, 661500352248901, 3855506596076654, 22471539224211023, 130973728749189484

Offset: 0

Barry E. Williams, May 02 2000

From Wolfdieter Lang, Feb 26 2015: (Start)
The sequence {2*a(n+1)}_{n >= 0}, gives all positive solutions y = y2(n) = 2*a(n+1) of the second class of the Pell equation x^2 - 2*y^2  = -7. For the corresponding terms x = x2(n) see A255236(n).
See A255236 for comments on the first class solutions and the relation to the Pell equation x^2 - 2*y^2 = 14. (End)

			n = 2: a(3) = sqrt((181^2 + 7)/2)/2 = 64.
a(3) = (53 + 75)/2 = 64. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See x'(n) in Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001653 and A001541. Cf. A001109.

A038723(n) = a(-n).

a[0]:=1: a[1]:=2: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

Union[Flatten[NestList[{#[[2]],#[[3]],6#[[3]]-#[[2]]}&,{1,2,11},25]]]  (* Harvey P. Dale, Mar 04 2011 *)
LinearRecurrence[{6,-1},{1,2},30] (* Harvey P. Dale, Jun 12 2017 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 - quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) - polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = ((4-sqrt(2))/8)*(3+2*sqrt(2))^(n-1)+((4+sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
From Michael Somos, Sep 28 2008: (Start)
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v.
G.f.: (1 - 4*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). (End)
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n) = S(n, 6) - 4*S(n-1, 6), n>=0, with the Chebyshev polynomials S(n, x) (A049310), with S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comment above.
a(n) = 6*a(n-1) - a(n-2), n >= 1, a(-1) = 4, a(0) = 1. (See the name.) (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(n+1) = sqrt((A255236(n)^2 + 7)/2)/2, n >= 0.
a(n+1) = (A038761(n) + A038762(n))/2, n >= 0. See the Mar 19 2015 comment on A054490. (End)
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) - sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, May 01 2020

A077446 Numbers k such that 2*k^2 + 14 is a square.

Original entry on oeis.org

1, 5, 11, 31, 65, 181, 379, 1055, 2209, 6149, 12875, 35839, 75041, 208885, 437371, 1217471, 2549185, 7095941, 14857739, 41358175, 86597249, 241053109, 504725755, 1404960479, 2941757281, 8188709765, 17145817931, 47727298111

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "2*n^2 + 14 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = (7+n^2)/2. - Ctibor O. Zizka, Nov 09 2009
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = a(n+1), n >= 0, of the Pell equation x^2 - 2*y^2 = -7. For the corresponding y-solutions see y(n) = 2*A006452(n+2) = A077447(n+1)/2. This implies that X^2 - 2*Y^2 = 14 has the general solutions (X(n),Y(n)) = (2*y(n), x(n)). See the first comment above.
For the positive first class solutions see (A054490(n), 2*A038723(n)) and for the second class solutions (A255236(n), 2*A038725(n+1)). (End)
For n > 0, a(n) is the n-th almost Lucas-balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: (A077447(3))^2 - 2*a(3)^2 = 16^2 - 2*11^2  = 14;
a(3)^2 - 2*(2*A006452(3+1))^2 = 11^2 - 2*(2*4)^2 = -7. - _Wolfdieter Lang_, Feb 26 2015

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Michael De Vlieger, Table of n, a(n) for n = 1..2612
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
J. J. O'Connor and E. F. Robertson, Pell's Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001109, A006452, A038723, A038725, A049310, A054490, A077447, A155765, A156035, A156649, A255236.

LinearRecurrence[{0,6,0,-1},{1,5,11,31},50] (* Sture Sjöstedt, Oct 08 2012 *)

2*(a(n))^2 + 14 = (A077447(n))^2.
Lim. n-> Inf. a(n)/a(n-2) = 5.8284271247461... = 3 + 2*sqrt(2) = A156035 = RG (Great Ratio).
Lim. k-> Inf. a(2*k+1)/a(2*k) = 2.09383632135605... = (9 + 4*sqrt(2))/7 = A156649 = R1 (Ratio 1).
Lim. k -> Inf. a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2 (Ratio 2); RG = R1*R2.
a(2*k-1) = [ 2*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] - [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] + [(3+2*Sqrt(2))^(n-2) - (3-2*Sqrt(2))^(n-2)] ] / (4*Sqrt(2)) a(2*k) = [ 5*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] + [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] ] / (4*Sqrt(2)).
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1+x)*(x^2+4*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=5, a(3)=11, a(4)=31. - Sture Sjöstedt, Oct 08 2012
Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015
E.g.f.: 1 - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x). - Stefano Spezia, Nov 26 2022
a(n) = a(n-1) + 2*A217975(n-1) + A123335(n-2) - A123335(n-3) for n > 1 and with A123335(-1) = 1. - Vladimir Pletser, Aug 30 2025

A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

Original entry on oeis.org

2, 2, 1, 3, 4, 8, 9, 19, 22, 46, 53, 111, 128, 268, 309, 647, 746, 1562, 1801, 3771, 4348, 9104, 10497, 21979, 25342, 53062, 61181, 128103, 147704, 309268, 356589, 746639, 860882, 1802546, 2078353, 4351731, 5017588, 10506008, 12113529, 25363747, 29244646, 61233502

Offset: 0

Raphie Frank, Dec 30 2015

This sequence gives all x in N | 2*x^2 - 7(-1)^x = y^2. The companion sequence to this sequence, giving y values, is A266505.
A266505(n)/a(n) converges to sqrt(2).
Alternatively, 1/4*(3*A002203(floor[n/2]) - A002203(n-(-1)^n)), where A002203 gives the Companion Pell numbers, or, in Lucas sequence notation, V_n(2, -1).
Alternatively, bisection of A266506.
Alternatively, A048654(n -1) and A078343(n + 1) interlaced.
Alternatively, A100525(n-1), A266507(n), A038761(n) and A253811(n) interlaced.
Let b(n) = (a(n) - a(n)(mod 2))/2, that is b(n) = {1, 1, 0, 1, 2, 4, 4, 9, 11, 23, 26, 55, 64, ...}. Then:
A006452(n) = {b(4n+0) U b(4n+1)} gives n in N such that n^2 - 1 is triangular;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that n^2 + n + 1 is triangular (indices of Sophie Germain triangular numbers);
A216162(n) = {b(4n+0) U b(4n+2) U b(4n+1) U b(4n+3)}, sequences A006452 and A216134 interlaced.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[2,2,1,3]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

LinearRecurrence[{0, 2, 0, 1}, {2, 2, 1, 3}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(1 - x) (2 + 4 x + x^2)/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 41}] (* Michael De Vlieger, Dec 31 2015 *)

Vec((1-x)*(2+4*x+x^2)/(1-2*x^2-x^4) + O(x^50)) \\ Colin Barker, Dec 31 2015

a(n) = 1/sqrt(8)*(+sqrt(2)*(1+sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n - 3*(1-sqrt(2))^(floor(n/2)-(-1)^n) + sqrt(2)*(1-sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n + 3*(1+sqrt(2))^(floor(n/2)-(-1)^n)).
a(n) = 1/4*((3*((1+sqrt(2))^floor(n/2)+(1-sqrt(2))^floor(n/2))) - (-1)^n*((1+sqrt(2))^(floor(n/2)-(-1)^n)+(1-sqrt(2))^(floor(n/2)-(-1)^n))).
a(2n) = (+sqrt(2)*(1+sqrt(2))^(n-1) - 3 *(1-sqrt(2))^(n-1) + sqrt(2)*(1-sqrt(2))^(n-1) + 3*(1 + sqrt(2))^(n-1))/sqrt(8) = A048654(n -1).
a(2n) = 1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) - ((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))) = A048654(n -1).
a(2n + 1) = (-sqrt(2)*(1+sqrt(2))^(n+1) - 3 *(1-sqrt(2))^(n+1) - sqrt(2)*(1-sqrt(2))^(n+1) + 3*(1+sqrt(2))^(n+1))/sqrt(8) = A078343(n + 1).
a(2n + 1) =1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) + ((1+sqrt(2))^(n+1)+(1-sqrt(2))^(n+1))) =  A078343(n + 1).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A100525(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A266507(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038761(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A253811(n).
sqrt(2*a(n)^2 - 7(-1)^a(n))*sgn(2*n - 1) = A266505(n).
(a(2n + 1) + a(2n))/2 = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 1) - a(2n))/2 = A000129(n), where A000129 gives the Pell numbers.
(a(2n+2) + a(2n+1))*2 = A002203(n+2)
(a(2n+2) - a(2n+1))*2 = A002203(n-1).
G.f.: (1-x)*(2+4*x+x^2) / (1-2*x^2-x^4). - Colin Barker, Dec 31 2015

A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

Original entry on oeis.org

-1, 1, 3, 5, 5, 11, 13, 27, 31, 65, 75, 157, 181, 379, 437, 915, 1055, 2209, 2547, 5333, 6149, 12875, 14845, 31083, 35839, 75041, 86523, 181165, 208885, 437371, 504293, 1055907, 1217471, 2549185, 2939235, 6154277, 7095941, 14857739, 17131117, 35869755, 41358175, 86597249, 99847467

Offset: 0

Raphie Frank, Dec 30 2015

a(n)/A266504(n) converges to sqrt(2).
Alternatively, bisection of A266506.
Alternatively, A135532(n) and A048655(n) interlaced.
Alternatively, A255236(n-1), A054490(n), A038762(n) and A101386(n) interlaced.
Let b(n) = (a(n) - (a(n) mod 2))/2, that is b(n) = {-1, 0, 1, 2, 2, 5, 6, 13, 15, 32, 37, 78, 90, ...}. Then:
A006451(n) = {b(4n+0) U b(4n+1)} gives n in N such that triangular(n) + 1 is square;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that triangular(n) follows form n^2 + n + 1 (twice a triangular number + 1).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

I:=[-1,1,3,5]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

a:=proc(n) option remember; if n=0 then -1 elif n=1 then 1 elif n=2 then 3 elif n=3 then 5 else 2*a(n-2)+a(n-4); fi; end:  seq(a(n), n=0..50); # Wesley Ivan Hurt, Jan 01 2016

LinearRecurrence[{0, 2, 0, 1}, {-1, 1, 3, 5}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(-1 + 3 x) (1 + x)^2/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 42}] (* Michael De Vlieger, Dec 31 2015 *)

my(x='x+O('x^40)); Vec((-1+3*x)*(1+x)^2/(1-2*x^2-x^4)) \\ G. C. Greubel, Jul 26 2018

G.f.: (-1 + 3*x)*(1 + x)^2/(1 - 2*x^2 - x^4).
a(n) = (-(1+sqrt(2))^floor(n/2)*(-1)^n - sqrt(8)*(1-sqrt(2))^floor(n/2) - (1-sqrt(2))^floor(n/2)*(-1)^n + sqrt(8)*(1+sqrt(2))^floor(n/2))/2.
a(n) = 3*(((1+sqrt(2))^floor(n/2)-(1-sqrt(2))^floor(n/2))/sqrt(8)) - (-1)^n*(((1+sqrt(2))^(floor(n/2)-(-1)^n)-(1-sqrt(2))^(floor(n/2)-(-1)^n))/sqrt(8)).
a(n) = (3*A000129(floor(n/2)) - A000129(n-(-1)^n)), where A000129 gives the Pell numbers.
a(n) = sqrt(2*A266504(n)^2 - 7*(-1)^A266504(n))*sgn(2*n-1), where A266504 gives all x in N such that 2*x^2 - 7*(-1)^x = y^2. This sequence gives associated y values.
a(2n) = (-(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n - (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n) = A135532(n).
a(2n) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) - (((1+sqrt(2))^(n-1)-(1-sqrt(2))^(n-1))/sqrt(8)) = A135532(n).
a(2n+1) = (+(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n + (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n + 1) = A048655(n).
a(2n+1) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) + (((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/sqrt(8)) = A048655(n).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A255236(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A054490(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038762(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A101386(n).
(sqrt(2*(a(2n + 1) )^2 + 14*(-1)^floor(n/2)))/2 = A266504(n).
(a(2n + 1) + a(2n))/8 = A000129(n), where A000129 gives the Pell numbers.
a(2n + 1) - a(2n) = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 2) + a(2n + 1))/2 = A000129(n+2).
(a(2n + 2) - a(2n + 1))/2 = A000129(n-1).

cp's OEIS Frontend

A054490 Expansion of (1+5*x)/(1-6*x+x^2).

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A038725 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=2.

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A077446 Numbers k such that 2*k^2 + 14 is a square.

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A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

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A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

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A054490 Expansion of (1+5x)/(1-6x+x^2).