A005248 -id:A005248 - cp's OEIS frontend

A003423 a(n) = a(n-1)^2 - 2, with a(0) = 6.

6, 34, 1154, 1331714, 1773462177794, 3145168096065837266706434, 9892082352510403757550172975146702122837936996354

Offset: 0

If x is either of the roots of x^2 - 6*x + 1 = 0 (i.e., x = 3 +- 2*sqrt(2)), then x^(2^n) + 1 = a(n)*x^(2^(n-1)). For example, x^8 + 1 = 1154*x^4. - James East, Oct 05 2018

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 376.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10
P. Liardet and P. Stambul, Series d'Engel et fractions continuees, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
E. Lucas, Théorie des Fonctions Numériques Simplement Périodiques, II, Amer. J. Math., 1 (1878), 289-321.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
J. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
J. Shallit & N. J. A. Sloane, Correspondence 1974-1975
Wikipedia, Engel Expansion
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (starting with 3), A003010 (starting with 4), A003487 (starting with 5).

Cf. A145505.

a:= n-> simplify(2*ChebyshevT(2^n, 3), 'ChebyshevT'):
seq(a(n), n=0..7);

a[1] := 6; a[n_] := a[n - 1]^2 - 2; Table[a[n], {n, 1, 8}] (* Stefan Steinerberger, Apr 11 2006 *)
Table[Round[(1 + Sqrt[2])^(2^n)], {n, 1, 7}] (* Artur Jasinski, Sep 25 2008 *)
NestList[#^2-2&,6,10] (* Harvey P. Dale, Nov 11 2011 *)

PARI
```
a(n)=if(n<1, 6*(n==0), a(n-1)^2-2)
```

a(n) = ceiling(c^(2^n)) where c = 3 + 2*sqrt(2) is the largest root of x^2 - 6x + 1 = 0. - Benoit Cloitre, Dec 03 2002
From Paul D. Hanna, Aug 11 2004: (Start)
a(n) = (3+sqrt(8))^(2^n) + (3-sqrt(8))^(2^n).
Sum_{n>=0} 1/(Product_{k=0..n} a(k) ) = 3 - sqrt(8). (End)
a(n) = 2*A001601(n+1).
a(n-1) = Round((1 + sqrt(2))^(2^n)). - Artur Jasinski, Sep 25 2008
a(n) = 2*T(2^n,3) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 3 - 2*sqrt(2). Thus 3 - 2*sqrt(2) = 1/6 + 1/(6*34) + 1/(6*34*1154) + .... See Liardet and Stambul. - Peter Bala, Oct 31 2012
From Peter Bala, Nov 11 2012: (Start)
4*sqrt(2)/7 = Product_{n >= 0} (1 - 1/a(n))
sqrt(2) = Product_{n >= 0} (1 + 2/a(n)).
a(n) - 1 = A145505(n+1). (End)
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2 + 4*Product_{k = 0 ..n-1} (a(k) + 2) for n >= 1.
Let b(n) = a(n) - 6. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A054486 Expansion of (1+2x)/(1-3x+x^2).

Original entry on oeis.org

1, 5, 14, 37, 97, 254, 665, 1741, 4558, 11933, 31241, 81790, 214129, 560597, 1467662, 3842389, 10059505, 26336126, 68948873, 180510493, 472582606, 1237237325, 3239129369, 8480150782, 22201322977, 58123818149, 152170131470, 398386576261, 1042989597313

Offset: 0

Barry E. Williams, May 06 2000

Binomial transform of A000285. - R. J. Mathar, Oct 26 2011

			G.f. = 1 + 5*x + 14*x^2 + 37*x^3 + 97*x^4 + 254*x^5 + 665*x^6 + 1741*x^7 + ...

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

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.
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 (3,-1).

Cf. A000032, A000045, A000285, A002878, A100545, A104449, A228208.

F:=Fibonacci;; List([0..30], n-> F(2*n+2) +2*F(2*n) ); # G. C. Greubel, Nov 08 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/(1-3*x+x^2)) ); // Marius A. Burtea, Nov 05 2019

a:=[1,5]; [n le 2 select a[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // Marius A. Burtea, Nov 05 2019

with(combinat); f:=fibonacci; seq(f(2*n+2)+2*f(2*n), n=0..30); # G. C. Greubel, Nov 08 2019

CoefficientList[Series[(2*z+1)/(z^2-3*z+1), {z, 0, 30}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 15 2011 *)
a[ n_]:= 3 Fibonacci[2n] + Fibonacci[2n+1]; (* Michael Somos, Mar 17 2015 *)
LinearRecurrence[{3,-1},{1,5},40] (* Harvey P. Dale, Apr 24 2019 *)

Vec((1+2*x)/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jul 15 2011

{a(n) = 3*fibonacci(2*n) + fibonacci(2*n+1)}; /* Michael Somos, Mar 17 2015 */

f=fibonacci; [f(2*n+2) +2*f(2*n) for n in (0..30)] # G. C. Greubel, Nov 08 2019

a(n) = 3*a(n-1) - a(n-2), a(0)=1, a(1)=5.
a(n) = (5*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - (((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)))/sqrt(5).
a(n) + 7*A001519(n) = A005248(n). - Creighton Dement, Oct 30 2004
a(n) = Lucas(2*n+1) + Fibonacci(2*n) = A002878(n) + A001906(n) = A025169(n-1) + A001906(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-6)^k. - Philippe Deléham, Mar 05 2014
0 = -11 + a(n)^2 - 3*a(n)*a(n+1) + a(n+1)^2 for all n in Z. - Michael Somos, Mar 17 2015
a(n) = -2*F(n)^2 + 6*F(n)*F(n+1) + F(n+1)^2 for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = 3*F(2*n) + F(2*n+1) for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = -A100545(-2-n) for all n in Z. - Michael Somos, Mar 17 2015
a(n) = A000285(2*n) = A228208(2*n+1) = A104449(2*n+1) for all n in Z. - Michael Somos, Mar 17 2015
From Klaus Purath, Nov 05 2019: (Start)
a(n) = (a(n-m) + a(n+m))/Lucas(2*m), m <= n.
a(n) = sum of 2*m+1 consecutive terms starting with a(n-m) divided by Lucas(2*m+1), m <= n.
a(n) = alternating sum of 2*m+1 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+1), m <= n.
a(n) + a(n+1) = sum of 2*m+2 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+2), m <= n.
a(n) + a(n+1) = (a(n-m) + a(n+m+1))/Fibonacci(2*m+1), m <= n.
The following formulas are extended to negative indexes:
a(n) = 3*Fibonacci(2*n+1) - Fibonacci(2*n-3).
a(n) = (Fibonacci(2*n+5) - 3* Fibonacci(2*n-1))/2.
a(n) = (4*Lucas(2*n+2) - Lucas(2*n-4))/5.
a(n) = Fibonacci(2*n+5) - 4*Fibonacci(2*n+1).
a(n) = (5*Fibonacci(2*n+5) - Fibonacci(2*n-7))/12. (End)
E.g.f.: exp(-(1/2)*(-3+sqrt(5))*x)*(-7 + sqrt(5) + (7 + sqrt(5))*exp(sqrt(5)*x))/(2*sqrt(5)). - Stefano Spezia, Nov 19 2019
a(n) = 3*n + 1 + Sum_{k=1..n} k*a(n-k). - Yu Xiao, Jun 20 2020

"a(1)=5", not "a(0)=5" from Dan Nielsen (nielsed(AT)uah.edu), Sep 10 2009

A065705 a(n) = Lucas(10*n).

Original entry on oeis.org

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..477
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem,Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A089775 Lucas numbers L(12n).

Original entry on oeis.org

2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

a(n+1)/a(n) converges to (322 + sqrt(103680))/2 = 321.996894379... a(0)/a(1) = 2/322; a(1)/a(2) = 322/103682; a(2)/a(3) = 103682/33385282; a(3)/a(4) = 33385282/10749957122; etc. Lim_{n -> inf} a(n)/a(n+1) = 0.00310562... = 2/(322 + sqrt(103680)) = (322 - sqrt(103680))/2.

			a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Nathaniel Johnston, Table of n, a(n) for n = 0..100
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 (322, -1).

Cf. A000032, A060964.
a(n) = A000032(12n).
Row 9 * 2 of array A188644
Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11).

[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011

Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *)

Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017

a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. (End)

a(11) - a(13) from Vincenzo Librandi, Apr 15 2011

A098149 a(0)=-1, a(1)=-1, a(n)=-3*a(n-1)-a(n-2) for n>1.

Original entry on oeis.org

-1, -1, 4, -11, 29, -76, 199, -521, 1364, -3571, 9349, -24476, 64079, -167761, 439204, -1149851, 3010349, -7881196, 20633239, -54018521, 141422324, -370248451, 969323029, -2537720636, 6643838879, -17393796001, 45537549124, -119218851371

Offset: 0

Creighton Dement, Aug 29 2004

Sequence relates bisections of Lucas and Fibonacci numbers.
2*a(n) + A098150(n) = 8*(-1)^(n+1)*A001519(n) - (-1)^(n+1)*A005248(n+1). Apparently, if (z(n)) is any sequence of integers (not all zero) satisfying the formula z(n) = 2(z(n-2) - z(n-1)) + z(n-3) then |z(n+1)/z(n)| -> golden ratio phi + 1 = (3+sqrt(5))/2.
Pisano period lengths: 1, 3, 4, 6, 1, 12, 8, 6, 12, 3, 10, 12, 7, 24, 4, 12, 9, 12, 18, 6, ... . - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Oct 12 2020: (Start)
[X(n) = (-1)^n*(S(n, 3) + S(n-1, 3)), Y(n) = X(n-1)] gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 3*X*Y = +5, for n = -oo..+oo, with Chebyshev S polynomials (A049310), with S(-1, x) = 0,  S(-|n|, x) = - S(|n|-2, x), for |n| >= 2, and S(n,-x) = (-1)^n*S(n, x). The present sequence is a(n) = -X(n-1), for n >= 0. See the formula section.
This binary indefinite quadratic form of discriminant 5, representing 5, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication) See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seong Ju Kim, R. Stees, and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
Tanya Khovanova, Recursive Sequences
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Index entries for linear recurrences with constant coefficients, signature (-3,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A098150, A001519, A005248, A000045, A001906, A049310, A027941.

a[0] = a[1] = -1; a[n_] := a[n] = -3a[n - 2] - a[n - 1]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Sep 01 2004 *)
LinearRecurrence[{-3,-1},{-1,-1},30] (* Harvey P. Dale, Apr 19 2014 *)
CoefficientList[Series[-(1 + 4 x)/(1 + 3 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 19 2014 *)

G.f.: -(1+4*x)/(1+3*x+x^2). - Philippe Deléham, Nov 19 2006
a(n) = (-1)^n*A002878(n-1). - R. J. Mathar, Jan 30 2011
-a(n+1) = Sum_{k, 0<=k<=n}(-5)^k*Binomial(n+k, n-k) = Sum_{k, 0<=k<=n}(-5)^k*A085478(n, k). - Philippe Deléham, Nov 28 2006
a(n) = (-1)^n*(S(n-1, 3) + S(n-2, 3)) = (-1)^n*S(2*(n-1), sqrt(5)), for n >= 0, with Chebyshev S polynomials (A049310), with S(-1, x) = 0 and S(-2, x) = -1. S(n, 3) = A001906(n+1) = F(2*(n+1)), with F = A000045. - Wolfdieter Lang, Oct 12 2020

Simpler definition from Philippe Deléham, Nov 19 2006

A110391 a(n) = L(3*n)/L(n), where L(n) = Lucas number.

Original entry on oeis.org

1, 4, 6, 19, 46, 124, 321, 844, 2206, 5779, 15126, 39604, 103681, 271444, 710646, 1860499, 4870846, 12752044, 33385281, 87403804, 228826126, 599074579, 1568397606, 4106118244, 10749957121, 28143753124, 73681302246, 192900153619, 505019158606, 1322157322204

Offset: 0

Amarnath Murthy, Jul 27 2005

Subsidiary sequences: a(n) = L((2k+1)*n)/L(n) for k = 2,3, etc. This is the sequence for k = 1.

			a(1) = L(3)/L(1) = 4/1 = 4.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000204, A000032, A002878, A114525, A153173, A153175, A153177, A153179, A153180.

[Lucas(3*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 17 2017

with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(3*n)/L(n),n=0..30); # Emeric Deutsch, Jul 31 2005

Table[LucasL[3 n]/LucasL[n], {n, 0, 27}] (* Michael De Vlieger, Mar 18 2015 *)
LinearRecurrence[{2,2,-1},{1,4,6},40] (* Harvey P. Dale, Aug 20 2020 *)

Vec((1+2*x-4*x^2)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Jun 03 2016

for(n=0,30, print1((fibonacci(3*n+1) + fibonacci(3*n-1))/( fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ G. C. Greubel, Dec 17 2017

From R. J. Mathar, Oct 18 2010: (Start)
a(n) = A005248(n) - (-1)^n.
a(n) = +2*a(n-1) +2*a(n-2) -a(n-3).
G.f.: ( 1+2*x-4*x^2 ) / ( (1+x)*(x^2-3*x+1) ). (End)
Exp( Sum_{n >= 1} a(n)*t^n/n ) = 1 + 4*t + 11*t^2 + 29*t^3 + ... is the o.g.f. for A002878. This is the case x = 1 of the general result exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n, where L(n,x) is the n-th Lucas polynomial of A114525. - Peter Bala, Mar 18 2015
a(n) = 2^(-n)*(-(-2)^n+(3-sqrt(5))^n+(3+sqrt(5))^n). - Colin Barker, Jun 03 2016

Corrected and extended by Emeric Deutsch and Erich Friedman, Jul 31 2005

A198632 Triangle version of the array of the number of closed paths of even length on the graph P_n (n vertices, n-1 edges).

Original entry on oeis.org

1, 0, 2, 0, 2, 3, 0, 2, 4, 4, 0, 2, 8, 6, 5, 0, 2, 16, 14, 8, 6, 0, 2, 32, 36, 20, 10, 7, 0, 2, 64, 94, 56, 26, 12, 8, 0, 2, 128, 246, 164, 76, 32, 14, 9, 0, 2, 256, 644, 488, 234, 96, 38, 16, 10, 0, 2, 512, 1686, 1460, 740, 304, 116, 44, 18, 11, 0, 2, 1024, 4414, 4376, 2372, 992, 374, 136, 50, 20, 12, 0, 2, 2048, 11556, 13124, 7654, 3296, 1244, 444, 156, 56, 22, 13

Offset: 0

Wolfdieter Lang, Nov 02 2011

This array is an example of counting walks on a graph whose adjacency matrix is given by a special symmetric tridiagonal matrix with nonnegative integer entries, appropriate for orthogonal polynomials. These are quadratic Jacobi matrices J_n with nonnegative entries. The corresponding graphs could be called Jacobi graphs. Here Chebyshev S-polynomials (coefficients A049310) are considered, which belong to the Jacobi class of the classical orthogonal polynomial systems. The corresponding graph has adjacency matrix [[0,1,0,...],[1,0,1,...],[0,1,0,1,...]...[0,...0,1,0]] (n rows and n columns), with characteristic polynomial S(n,x) (see also a comment by Michael Somos on A049310).
w(n,l;p_k->p_m) = ((J_n)^l)(k,m) is the number of walks of length l from vertex p_k to vertex p_m on such a Jacobi graph. w(n,0; p_k->p_m) = delta(k,m), with the Kronecker symbol delta. The total number of closed walks of length l is w(n,l):=Sum_{i=1..n} w(n,l; p_i->p_i) = trace(J_n^l), which is the l-th power sum of the eigenvalues of J_n, i.e., the zeros of the characteristic polynomial for J_n. There are theorems for the o.g.f. of the normalized power sums of these zeros. See, e.g., the given W. Lang reference, p. 244. This results for the S-polynomial in the o.g.f. G(n,x) = Sum_{l=0..infinity} w(n,l)*x^l = y*(d/dy)S(n,y)/S(y) with y=1/x. This can be rewritten in the form given in the formula section (this results from eq. (3.8b) of the W. Lang reference, and in eq. (3.8d) it should be coth, not tanh).
From Wolfdieter Lang, Oct 10 2012: (Start)
For an accompanying paper on path counting on Jacobi graphs see the W. Lang link under A201198.
The total number of round trips of length L on the graph P_n, taken per site, becomes for n -> infinity A126869(L). See the just mentioned link, p. 8. This limit is derived from the limit of G(n,x)/n with G(n,x) given in the formula section.
Thanks go to Clyde P. Kruskal for asking a question which led to this comment.
(End)

			The array w(n,2*k) is
n\k  0  1   2   3   4    5    6     7     8      9 ...
1:   1  0   0   0   0    0    0     0     0      0
2:   2  2   2   2   2    2    2     2     2      2
3:   3  4   8  16  32   64  128   256   512   1024
4:   4  6  14  36  94  246  644  1686  4414  11556
5:   5  8  20  56 164  488 1460  4376 13124  39368
6:   6 10  26  76 234  740 2372  7654 24778  80338
7:   7 12  32  96 304  992 3296 11072 37440 127104
8:   8 14  38 116 374 1244 4220 14504 50294 175454
9:   9 16  44 136 444 1496 5144 17936 63164 224056
...
The triangle is
k\n 1  2    3    4    5    6   7    8   9 10 11 12 ...
0:  1
1:  0  2
2:  0  2    3
3:  0  2    4    4
4:  0  2    8    6    5
5:  0  2   16   14    8    6
6:  0  2   32   36   20   10   7
7:  0  2   64   94   56   26  12    8
8:  0  2  128  246  164   76  32   14   9
9:  0  2  256  644  488  234  96   38  16 10
10: 0  2  512 1686 1460  740 304  116  44 18 11
11: 0  2 1024 4414 4376 2372 992  374 136 50 20 12
...
n=3, l=2*k = 4: graph P_3 as 1-2-3, with eight walks of length 4, namely 12121, 12321, 21212, 23232, 21232, 23212, 32323 and 32123.

S. Barbero, Dickson Polynomials, Chebyshev Polynomials, and Some Conjectures of Jeffery, Journal of Integer Sequences, 17 (2014), #14.3.8.
S. Barbero, U. Cerruti, N. Murru, Identities Involving Zeros of Ramanujan and Shanks Cubic Polynomials, J. Integer Seq., Vol. 16 (2013), Article 13.8.1.
Carlos M. da Fonseca, M. Lawrence Glasser, Victor Kowalenko, Basic trigonometric power sums with applications, arXiv:1601.07839 [math.NT], 2016. See Theorem 5.1.
W. Lang, On sums of powers of zeros of polynomials, J. Comp. Appl. Math., 89 (1998) 237-256.

Column sequences: A000007, 2*A000012, A198633, 2*A005248, A198635, ...

a(k,n)=w(n,2*(k-n+2)), the total number of closed walks (paths) of length 2*(k-n+2) on the graph P_n, which looks like o-o-o...-o, with n vertices (nodes) and n-1 edges (lines), k+1>=n>=1.
O.g.f. G(n,x) for w(n,l), which vanishes for odd l, is
((n+1)*coth((n+1)*log((2*x)/(1-sqrt(1-(2*x)^2)))) - 1/sqrt(1-(2*x)^2))/sqrt(1-(2*x)^2). See the comment above for a version with Chebyshev S-polynomials.
Conjecture: For the array w(n,2*k) in the example below, w(2*q,2*k)/2 = A185095(q,k), q >= 1, k >= 0. - L. Edson Jeffery, Nov 23 2013

A005592 a(n) = F(2n+1) + F(2n-1) - 1.

Original entry on oeis.org

1, 2, 6, 17, 46, 122, 321, 842, 2206, 5777, 15126, 39602, 103681, 271442, 710646, 1860497, 4870846, 12752042, 33385281, 87403802, 228826126, 599074577, 1568397606, 4106118242, 10749957121, 28143753122, 73681302246, 192900153617, 505019158606, 1322157322202

Offset: 0

N. J. A. Sloane

For any m, the maximum element in the continued fraction of F(2n+m)/F(m) is a(n). - Benoit Cloitre, Jan 10 2006
The continued fraction [a(n);1,a(n)-1,1,a(n)-1,...] = phi^(2n), where phi = 1.618... is the golden ratio, A001622. - Thomas Ordowski, Jun 07 2013
a(n) is the number of labeled subgraphs of the n-cycle C_n. For example, a(3)=17. There are 7 subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself); here 7+6+3+1 = 17. - John P. McSorley, Oct 31 2016
a(n) equals the sum of the n-th row of triangle A277919. - John P. McSorley, Nov 25 2016

			G.f. = 1 + 2*x + 6*x^2 + 17*x^3 + 46*x^4 + 122*x^5 + 321*x^6 + 842*x^7 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (terms n = 1..200 from Vincenzo Librandi)
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 41, 56.
M. D. McIlroy, The number of states of a dynamic storage system, Computer J., Vol. 25, No. 3 (1982), pp. 388-392.
M. D. McIlroy, The number of states of a dynamic storage system, Computer J., Vol. 25, No. 3 (1982), pp. 388-392. (Annotated scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Jesús Salas and Alan D. Sokal, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models. V. Further Results for the Square-Lattice Chromatic Polynomial, J. Stat. Phys., Vol. 135 (2009), pp. 279-373; arXiv preprint, arXiv:0711.1738 [cond-mat.stat-mech], 2007-2009. Mentions this sequence. - N. J. A. Sloane, Mar 14 2014
Robert S. Seamons, Problem B-89, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 4, No. 2 (1966), p. 190; A Close Approximation, Solution to Problem B-89 by Douglas Lind, ibid., Vol. 5, No. 1 (1967), pp. 108-109.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Equals A004146+1 and A005248+1.
Bisection of A014217; the other bisection is A002878, which also bisects A000032.
Cf. A000045, A065034.

a005592 n = a005592_list !! (n-1)
a005592_list = map (subtract 1) $
                   tail $ zipWith (+) a001519_list $ tail a001519_list
-- Reinhard Zumkeller, Aug 09 2013

[Fibonacci(2*n+1)+Fibonacci(2*n-1)-1: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

A005592:=-(2-2*z+z**2)/(z-1)/(z**2-3*z+1); # conjectured by Simon Plouffe in his 1992 dissertation
# second Maple program:
F:= n-> (<<0|1>, <1|1>>^n)[1,2]:
a:= n-> F(2*n+1)+F(2*n-1)-1:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 04 2016

Table[Fibonacci[2n+1]+Fibonacci[2n-1]-1,{n,30}] (* Harvey P. Dale, Aug 22 2011 *)
a[n_] := LucasL[2n]-1; Array[a, 30] (* Jean-François Alcover, Dec 09 2015 *)

a(n)=fibonacci(2*n+1)+fibonacci(2*n-1)-1 \\ Charles R Greathouse IV, Aug 23 2011

[lucas_number2(n,3,1)-1 for n in range(1,29)] # Zerinvary Lajos, Jul 06 2008

a(n) = Lucas(2*n)-1, with Lucas(n)=A000032(n).
a(n) = floor(r^(2*n)), where r = golden ratio = (1+sqrt(5))/2.
a(n) = floor(Fibonacci(5*n)/Fibonacci(3*n)). - Gary Detlefs, Mar 11 2011
a(n) = +4*a(n-1) -4*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 11 2011
a(n) = A001519(2*n-1) + A001519(2*n+1) - 1. - Reinhard Zumkeller, Aug 09 2013
a(n) = 3*a(n) - a(n-1) + 1; a(n) = A004146(n) + 1, n>0. - Richard R. Forberg, Sep 04 2013
a(n) = 2*cosh(2*n*arcsinh(1/2)) - 1. - Ilya Gutkovskiy, Oct 31 2016
a(n) = floor(sqrt(5)*Fibonacci(2*n)), for n > 0 (Seamons, 1966). - Amiram Eldar, Feb 05 2022

Formulae and comments by Clark Kimberling, Nov 24 2010

a(0)=1 prepended by Alois P. Heinz, Nov 04 2016

A056918 a(n) = 9*a(n-1)-a(n-2); a(0)=2, a(1)=9.

Original entry on oeis.org

2, 9, 79, 702, 6239, 55449, 492802, 4379769, 38925119, 345946302, 3074591599, 27325378089, 242853811202, 2158358922729, 19182376493359, 170483029517502, 1515164889164159, 13466000972959929, 119678843867475202

Offset: 0

Barry E. Williams, Aug 21 2000

All nonnegative integer solutions of Pell equation a(n)^2 - 77*b(n)^2 = +4 together with b(n)=A018913(n), n>=0. - Wolfdieter Lang, Aug 31 2004

Except for the first term, positive values of x (or y) satisfying x^2 - 9xy + y^2 + 77 = 0. - Colin Barker, Feb 13 2014

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A018913. a(n)=sqrt(77*A018913(n)^2 + 4). A005248.

a056918 n = a056918_list !! n
a056918_list = 2 : 9 :
   zipWith (-) (map (* 9) $ tail a056918_list) a056918_list
-- Reinhard Zumkeller, Jan 06 2013

a[0] = 2; a[1] = 9; a[n_] := 9a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)

[lucas_number2(n,9,1) for n in range(23)] # Zerinvary Lajos, Jun 25 2008

a(n) = 9*S(n-1, 9) - 2*S(n-2, 9) = S(n, 9) - S(n-2, 9) = 2*T(n, 9/2), with S(n, x) := U(n, x/2) (see A049310), S(-1, x) := 0, S(-2, x) := -1. S(n-1, 9)=A018913(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind.
a(n) = {9*[((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n] - 2*[((9+sqrt(77))/2)^(n-1) - ((9-sqrt(77))/2)^(n-1)]}/sqrt(77).
G.f.: (2-9*x)/(1-9*x+x^2).
a(n) = ap^n + am^n, with ap := (9+sqrt(77))/2 and am := (9-sqrt(77))/2.
G.f.: (2-9*x)/(1-9*x+x^2). - Philippe Deléham, Nov 03 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = product {n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(9 - sqrt(77)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.11095 50589 89701 91909 ... = 2 + 1/(9 + 1/(79 + 1/(702 + ...))).
Also F(-alpha) = 0.88873 23915 40314 47623 ... has the continued fraction representation 1 - 1/(9 - 1/(79 - 1/(702 - ...))) and the simple continued fraction expansion 1/(1 + 1/((9-2) + 1/(1 + 1/((79-2) + 1/(1 + 1/((702-2) + 1/(1 + ...))))))). F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((9^2-4) + 1/(1 + 1/((79^2-4) + 1/(1 + 1/((702^2-4) + 1/(1 + ...))))))). Cf. A005248.
(End)

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A081714 a(n) = F(n)*L(n+1) where F=Fibonacci and L=Lucas numbers.

Original entry on oeis.org

0, 3, 4, 14, 33, 90, 232, 611, 1596, 4182, 10945, 28658, 75024, 196419, 514228, 1346270, 3524577, 9227466, 24157816, 63245987, 165580140, 433494438, 1134903169, 2971215074, 7778742048, 20365011075, 53316291172, 139583862446, 365435296161, 956722026042

Offset: 0

Ralf Stephan, Apr 03 2003

Also convolution of Fibonacci and Lucas numbers.
For n>2, a(n) represents twice the area of the triangle created by the three points (L(n-3), L(n-2)), (L(n-1), L(n)) and (F(n+3), F(n+2)) where L(k)=A000032(k) and F(k)=A000045(k). - J. M. Bergot, May 20 2014
For n>1, a(n) is the remainder when F(n+3)*F(n+4) is divided by F(n+1)*F(n+2). - J. M. Bergot, May 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A000204.

List([0..30], n -> Fibonacci(n)*(Fibonacci(n+2)+Fibonacci(n))); # G. C. Greubel, Jan 07 2019

[Fibonacci(n)*Lucas(n+1): n in [0..30]]; // Vincenzo Librandi, Sep 08 2012

with(combinat): F:=n-> fibonacci(n): L:= n-> F(n+1)+F(n-1):
a:= n-> F(n)*L(n+1): seq(a(n), n=0..30);

Fibonacci[Range[0,50]]*LucasL[Range[0,50]+1] (* Vladimir Joseph Stephan Orlovsky, Mar 17 2011*)

my(x='x+O('x^51));for(n=0,50,print1(polcoeff(serconvol(Ser((1+2*x)/(1-x-x*x)),Ser(x/(1-x-x*x))),n)", "))

a(n)=fibonacci(n)*(fibonacci(n+2)+fibonacci(n))

a(n) = round((-(-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/sqrt(5))) \\ Colin Barker, Sep 28 2016

[fibonacci(n)*(fibonacci(n+2)+fibonacci(n)) for n in (0..30)] # G. C. Greubel, Jan 07 2019

G.f.: x*(3-2*x)/((1+x)*(1-3*x+x^2)).
a(n) = A122367(n) - (-1)^n. - R. J. Mathar, Jul 23 2010
a(n) = (L(n+1)^2 - F(2*n+2))/2 = ( A001254(n+1) - A001906(n+1) )/2. - Gary Detlefs, Nov 28 2010
a(n+1) = - A186679(2*n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A035513(1,n-1)*A035513(2,n-1). - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A005248(n+1). - R. J. Mathar, Sep 04 2016
a(n) = (-(-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5)). - Colin Barker, Sep 28 2016

Simpler definition from Michael Somos, Mar 16 2004

cp's OEIS Frontend

A003423 a(n) = a(n-1)^2 - 2, with a(0) = 6.

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A054486 Expansion of (1+2*x)/(1-3*x+x^2).

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A065705 a(n) = Lucas(10*n).

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A089775 Lucas numbers L(12n).

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A098149 a(0)=-1, a(1)=-1, a(n)=-3*a(n-1)-a(n-2) for n>1.

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A110391 a(n) = L(3*n)/L(n), where L(n) = Lucas number.

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A054486 Expansion of (1+2x)/(1-3x+x^2).