A086902 -id:A086902 - cp's OEIS frontend

A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593

Offset: 0

Henry Bottomley, May 10 2000

In general, sequences with recurrence a(n) = k*a(n-1) + a(n-2) and a(0)=1 (and a(-1)=0) have the generating function 1/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]
a(p-1) == 53^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
From Johannes W. Meijer, Jun 12 2010: (Start)
For the sequence given above k=7 which implies that it is associated with A041091.
For a similar statement about sequences with recurrence a(n) = k*a(n-1) + a(n-2) but with a(0) = 2, and a(-1) = 0, see A086902; a sequence that is associated with A041090.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,7} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 7-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 7 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for linear recurrences with constant coefficients, signature (7,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A000129, A001076, A005668, A006190, A052918, A243399.
Row n=7 of A073133, A172236 and A352361.
Cf. A099367 (squares).

I:=[1, 7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

LinearRecurrence[{7, 1}, {1, 7}, 30] (* Vincenzo Librandi, Feb 23 2013 *)

a(n)=([0,1; 1,7]^n*[1;7])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,7,-1) for n in range(1, 19)] # Zerinvary Lajos, Apr 24 2009

a(3n) = A041091(5n), a(3n+1) = A041091(5n+3), a(3n+2) = 2*A041091(5n+4).
G.f.: 1/(1 - 7x - x^2).
a(n) = U(n, 7*i/2)*(-i)^n with i^2=-1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See A049310.
a(n) = F(n, 7), the n-th Fibonacci polynomial evaluated at x=7. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = (sigma^n - (-sigma)^(-n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*7^(n-1-2i). (End)
a(n) = ((7 + sqrt(53))^n - (7 - sqrt(53))^n)/(2^n*sqrt(53)). Offset 1. a(3)=50. - Al Hakanson (hawkuu(AT)gmail.com), Jan 17 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097836(n), a(2n) = A097838(n).
Lim_{k->oo} a(n+k)/a(k) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Lim_{n->oo} A086902(n)/A054413(n-1) = sqrt(53).
(End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(53)-7)/2. - Vladimir Shevelev, Feb 23 2013
From Kai Wang, Feb 24 2020: (Start)
Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.
Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)-7)/14.
In general, for sequences with recurrence f(n)= k*f(n-1)+f(n-2) and f(0)=1,
Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).
Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4) - k)/(2*k). (End)
E.g.f.: (1/53)*exp(7*x/2)*(53*cosh(sqrt(53)*x/2) + 7*sqrt(53)*sinh(sqrt(53)*x/2)). - Stefano Spezia, Feb 26 2020
G.f.: x/(1 - 7*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 7 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010

Extended by T. D. Noe, May 23 2011

A049666 a(n) = Fibonacci(5*n)/5.

Original entry on oeis.org

0, 1, 11, 122, 1353, 15005, 166408, 1845493, 20466831, 226980634, 2517253805, 27916772489, 309601751184, 3433536035513, 38078498141827, 422297015595610, 4683345669693537, 51939099382224517, 576013438874163224

Offset: 0

Clark Kimberling

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 11's along the main diagonal and 1's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,11} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
For n >= 1, a(n) equals the denominator of the continued fraction [11, 11, ..., 11] (with n copies of 11). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
From Michael A. Allen, Mar 30 2023: (Start)
Also called the 11-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 11 kinds of squares available. (End)

			G.f. = x + 11*x^2 + 122*x^3 + 1353*x^4 + 15005*x^5 + 166408*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..950
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (11,1).

A column of array A028412.
Cf. A000045, A102312, A243399.
Row n=11 of A073133, A172236 and A352361, and column k=11 of A157103.

[Fibonacci(5*n)/5: n in [0..30]]; // G. C. Greubel, Dec 02 2017

A049666 := proc(n)
    combinat[fibonacci](5*n)/5 ;
end proc: # R. J. Mathar, May 07 2024

Table[Fibonacci[5*n]/5, {n, 0, 100}] (* T. D. Noe, Oct 29 2009 *)
a[ n_] := Fibonacci[n, 11]; (* Michael Somos, May 28 2014 *)

numlib::fibonacci(5*n)/5 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(5*n)/5 \\ Charles R Greathouse IV, Feb 03 2014

from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(0,1,11,11,1,0)
[next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,11,-1) for n in range(0, 19)] # Zerinvary Lajos, Apr 27 2009

[fibonacci(5*n)/5 for n in range(0, 19)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 11*x - x^2).
a(n) = A102312(n)/5.
a(n) = 11*a(n-1) + a(n-2) for n > 1, a(0)=0, a(1)=1. With a=golden ratio and b=1-a, a(n) = (a^(5n)-b^(5n))/(5*sqrt(5)). - Mario Catalani (mario.catalani(AT)unito.it), Jul 24 2003
a(n) = F(n, 11), the n-th Fibonacci polynomial evaluated at x=11. - T. D. Noe, Jan 19 2006
a(n) = ((11+sqrt(125))^n-(11-sqrt(125))^n)/(2^n*sqrt(125)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n) = 11*A049670(n), a(2n+1) = A097843(n).
a(3n+1) = A041227(5n), a(3n+2) = A041227(5n+3), a(3n+3) = 2*A041227(5n+4).
Limit_{k->oo} a(n+k)/a(k) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit_{n->oo} A001946(n)/A049666(n) = sqrt(125).
(End)
a(n) = F(n) + (-1)^n*5*F(n)^3 + 5*F(n)^5, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012
a(-n) = -(-1)^n * a(n). - Michael Somos, May 28 2014
E.g.f.: (exp((1/2)*(11-5*sqrt(5))*x)*(-1 + exp(5*sqrt(5)*x)))/(5*sqrt(5)). - Stefano Spezia, Aug 02 2019

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

Original entry on oeis.org

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127

Offset: 0

Paul Barry, Aug 16 2003

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{5,n}.
Tanya Khovanova, Recursive Sequences
Wikipedia, Metallic mean
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 (5,1).

Cf. A006497, A014448, A085447.

Cf. A086902, A000032, A052918.

I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)

{a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */

[lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A099371 Expansion of g.f.: x/(1 - 9*x - x^2).

Original entry on oeis.org

0, 1, 9, 82, 747, 6805, 61992, 564733, 5144589, 46866034, 426938895, 3889316089, 35430783696, 322766369353, 2940328107873, 26785719340210, 244011802169763, 2222891938868077, 20250039251982456, 184473245206710181, 1680509246112374085, 15309056460218076946

Offset: 0

Wolfdieter Lang, Oct 18 2004

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 9's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,9} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Mar 10 2023: (Start)
Also called the 9-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 9 kinds of squares available. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
J. H. Han and M. D. Hirschhorn, Another Look at an Amazing Identity of Ramanujan, Mathematics Magazine, Vol. 79 (2006), pp. 302-304. See equation 6 on page 303.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,1).

Row n=9 of A073133, A172236 and A352361.
Cf. A099372 (squares).
Cf. A041151, A054413, A086902, A087798, A099371, A097839, A178765, A243399.

I:=[0,1]; [n le 2 select I[n] else 9*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

F:= gfun:-rectoproc({a(n)=9*a(n-1)+a(n-2),a(0)=0,a(1)=1},a(n),remember):
seq(F(n),n=0..30); # Robert Israel, Feb 01 2015

CoefficientList[Series[x/(1-9*x-x^2), {x,0,30}], x] (* G. C. Greubel, Apr 16 2017 *)
LinearRecurrence[{9,1}, {0,1}, 30] (* G. C. Greubel, Jan 24 2018 *)

my(x='x+O('x^30)); concat([0], Vec(1/(1-9*x-x^2)) ) \\ Charles R Greathouse IV, Feb 03 2014

from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(0,1,9,9,1,0)
[next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,9,-1) for n in range(0, 20)] # Zerinvary Lajos, Apr 26 2009

G.f.: x/(1 - 9*x - x^2).
a(n) = 9*a(n-1) + a(n-2), n >= 2, a(0)=0, a(1)=1.
a(n) = (-i)^(n-1)*S(n-1, 9*i) with S(n, x) Chebyshev's polynomials of the second kind (see A049310) and i^2=-1.
a(n) = (ap^n - am^p)/(ap-am) with ap:= (9+sqrt(85))/2 and am:= (9-sqrt(85))/2 = -1/ap (Binet form).
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1-k, k)*9^(n-1-2*k) n >= 1.
a(n) = F(n, 9), the n-th Fibonacci polynomial evaluated at x=9. - T. D. Noe, Jan 19 2006
a(n) = ((9+sqrt(85))^n - (9-sqrt(85))^n)/(2^n*sqrt(85)). Offset 1. a(3)=82. - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 85^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+2) = 9*A097839(n), a(2n+1) = A097841(n).
a(3n+1) = A041151(5n), a(3n+2) = A041151(5n+3), a(3n+3) = 2*A041151(5n+4).
Limit_{k -> infinity} (a(n+k)/a(k)) = (A087798(n) + A099371(n)*sqrt(85))/2.
Lim_{n->infinity} A087798(n)/A099371(n) = sqrt(85). (End)
a(n) ~ 1/sqrt(85)*((9+sqrt(85))/2)^n. - Jean-François Alcover, Dec 04 2013
a(n) = [1,0] (M^n) [0,1]^T where M is the matrix [9,1; 1,0]. - Robert Israel, Feb 01 2015
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(85)*x/2)/sqrt(85). - Stefano Spezia, Apr 06 2023

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

Original entry on oeis.org

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1

Offset: 0

Eric W. Weisstein, Dec 06 2005

Unsigned version of A108045.

The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
B. Johnson, Fibonacci numbers and matrices, 2009.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Eric Weisstein's World of Mathematics, Lucas Polynomial

Cf. A108045 (signed version).
Cf. A034807, A162514.
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Lucas := proc(n,x)
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
end proc:
A114525 := proc(n,k)
    coeftayl(Lucas(n,x),x=0,k) ;
end proc:
seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019

row[n_] := CoefficientList[LucasL[n, x], x];
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A001946 a(n) = 11*a(n-1) + a(n-2).

Original entry on oeis.org

2, 11, 123, 1364, 15127, 167761, 1860498, 20633239, 228826127, 2537720636, 28143753123, 312119004989, 3461452808002, 38388099893011, 425730551631123, 4721424167835364, 52361396397820127, 580696784543856761, 6440026026380244498, 71420983074726546239

Offset: 0

N. J. A. Sloane, Simon Plouffe

For odd n there is the Aurifeuillian factorization a(n) = Lucas[5n] = Lucas[n]*A[n]*B[n] = A000032[n]*A124296[n]*A124297[n], where A[n] = A124296[n] = 5*F(n)^2 - 5*F(n) + 1 and B[n] = A124297[n] = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - Alexander Adamchuk, Oct 25 2006

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 139.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
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
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (11, 1).

Cf. A000032, A000045, A121171, A124296, A124297.

[ Lucas(5*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

A001946:=(-2+11*z)/(-1+11*z+z**2); # Conjectured by Simon Plouffe in his 1992 dissertation

Table[Fibonacci[5n-1]+Fibonacci[5n+1],{n,0,30}] (* Alexander Adamchuk, Oct 25 2006 *)
LinearRecurrence[{11,1},{2,11},20] (* Harvey P. Dale, Jan 25 2024 *)

a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - Alexander Adamchuk, Oct 25 2006
a(n) = ((11 + 5*sqrt(5))/2)^n + ((11 - 5*sqrt(5))/2)^n. - Tanya Khovanova, Feb 06 2007
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 11*A097842(n), a(2n) = A065705(n).
a(3n+1) = A041226(5n), a(3n+2) = A041226(5n+3), a(3n+3) = 2* A041226(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit(A001946(n)/A049666(n), n=infinity) = sqrt(125). (End)
From Peter Bala, Mar 22 2015: (Start)
a(n) = Fibonacci(10*n)/Fibonacci(5*n) for n >= 1.
a(n) = ( Fibonacci(5*n + 2*k) - F(5*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(5*n + 2*k + 1) + F(5*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k.
a(n) = Sum_{k = 0..2*n} binomial(2*n,k)*Lucas(n + k). (End)
a(n) = [x^n] ( (1 + 11*x + sqrt(1 + 22*x + 125*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 26 2015
E.g.f.: 2*cosh(5*sqrt(5)*x/2)*(cosh(11*x/2) + sinh(11*x/2)). - Stefano Spezia, Jan 18 2025

A087798 a(n) = 9*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 9.

Original entry on oeis.org

2, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, 432083484, 3936182123, 35857722591, 326655685442, 2975758891569, 27108485709563, 246952130277636, 2249677658208287, 20494051054152219, 186696137145578258

Offset: 0

Nikolay V. Kosinov, Dmitry V. Poljakov (kosinov(AT)unitron.com.ua), Oct 10 2003

a(n+1)/a(n) converges to (9 + sqrt(85))/2.

For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010

			a(4) = 9*a(3) + a(2) = 9*756 + 83 = 6887.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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 (9,1).

Cf. A014511.

I:=[2,9]; [n le 2 select I[n] else 9*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 9, a[n] == 9 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{9,1}, {2,9}, 30] (* G. C. Greubel, Nov 07 2018 *)

x='x+O('x^30); Vec((2-9*x)/(1-9*x-x^2)) \\ G. C. Greubel, Nov 07 2018

a(n) = ((9 + sqrt(85))/2)^n + ((9 - sqrt(85))/2)^n.
G.f.: (2 - 9*x)/(1 - 9*x - x^2). - Philippe Deléham, Nov 02 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 9*A097840(n), a(2n) = A099373(n).
a(3n+1) = A041150(5n), a(3n+2) = A041150(5n+3), a(3n+3) = 2*A041150(5n+4).
Lim_{k->infinity} a(n+k)/a(k) = (A087798(n) + A099371(n)*sqrt(85))/2.
Lim_{n->infinity} A087798(n)/A099371(n) = sqrt(85). (End)

More terms from Ray Chandler, Nov 06 2003

A088316 a(n) = 13*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 13.

Original entry on oeis.org

2, 13, 171, 2236, 29239, 382343, 4999698, 65378417, 854919119, 11179326964, 146186169651, 1911599532427, 24996980091202, 326872340718053, 4274337409425891, 55893258663254636, 730886700031736159, 9557420359075824703, 124977351368017457298, 1634262988143302769577

Offset: 0

Nikolay V. Kosinov, Dmitry V. Polyakov (kosinov(AT)unitron.com.ua), Nov 06 2003

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

G. C. Greubel, Table of n, a(n) for n = 0..890
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 (13,1).

Cf. A006905, A041318, A054413, A086902, A088316, A097845, A140455.

I:=[2,13]; [n le 2 select I[n] else 13*Self(n-1) +Self(n-2): n in [1..31]]; // G. C. Greubel, Dec 13 2022

LinearRecurrence[{13,1}, {2,13}, 31] (* Stefano Spezia, Sep 20 2022 *)

A088316=BinaryRecurrenceSequence(13,1,2,13)
[A088316(n) for n in range(31)] # G. C. Greubel, Dec 13 2022

a(n) = ((13+sqrt(173))/2)^n + ((13-sqrt(173))/2)^n.
Lim_{n -> oo} a(n+1)/a(n) = (13 + sqrt(173))/2.
Lim_{n -> oo} a(n)/a(n+1) = 2/(13+sqrt(173)).
G.f.: (2-13*x)/(1-13*x-x^2). - Philippe Deléham, Nov 02 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2*n+1) = 13*A097845(n).
a(3*n+1) = A041318(5n), a(3n+2) = A041318(5n+3), a(3n+3) = 2*A041318(5n+4).
Limit_{k->oo} a(n+k)/a(k) = (A088316(n) + A140455(n)*sqrt(173))/2.
Limit_{n->oo} A088316(n)/A140455(n) = sqrt(173). (End)

A090301 a(n) = 15*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 15.

Original entry on oeis.org

2, 15, 227, 3420, 51527, 776325, 11696402, 176222355, 2655031727, 40001698260, 602680505627, 9080209282665, 136805819745602, 2061167505466695, 31054318401746027, 467875943531657100, 7049193471376602527

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.066372... = 2/(15+sqrt(229)) = (sqrt(229)-15)/2.
Lim_{n-> infinity} a(n+1)/a(n) = 15.066372... = (15+sqrt(229))/2 = 2/(sqrt(229)-15).
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010

			a(4) = 15*a(3) + a(2) = 15*3420 + 227 = ((15+sqrt(229))/2)^4 + ((15-sqrt(229))/2)^4 = 51526.9999805 + 0.0000194 = 51527.

G. C. Greubel, Table of n, a(n) for n = 0..500
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 (15,1).

Cf. A058087, A071416.
Lucas polynomials: A114525.
Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), this sequence (m=15), A090305 (m=16), A090306 (m=17), A090307 (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25), A087281 (m=29), A087287 (m=76), A089772 (m=199).

m:=15;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 31 2019

m:=15; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 31 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 15*I/2)), n = 0..20); # G. C. Greubel, Dec 31 2019

LucasL[Range[20]-1, 15] (* G. C. Greubel, Dec 31 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 15*I/2) ) \\ G. C. Greubel, Dec 31 2019

[2*(-I)^n*chebyshev_T(n, 15*I/2) for n in (0..20)] # G. C. Greubel, Dec 31 2019

a(n) = 15*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 15.
a(n) = ((15+sqrt(229))/2)^n + ((15-sqrt(229))/2)^n.
(a(n))^2 = a(2n) - 2 if n=1, 3, 5...
(a(n))^2 = a(2n) + 2 if n=2, 4, 6...
G.f.: (2-15*x)/(1-15*x-x^2). - Philippe Deléham, Nov 02 2008
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
Lim_{k-> infinity} a(n+k)/a(k) = (A090301(n) + A154597(n)*sqrt(229))/2.
Lim_{n-> infinity} A090301(n)/ A154597(n) = sqrt(229).
a(2n+1) = 15*A098246(n).
a(3n+1) = A041426(5n), a(3n+2) = A041426(5n+3), a(3n+3) = 2*A041426(5n+4).
(End)
a(n) = Lucas(n, 15) = 2*(-i)^n * ChebyshevT(n, 15*i/2). - G. C. Greubel, Dec 31 2019
E.g.f.: 2*exp(15*x/2)*cosh(sqrt(229)*x/2). - Stefano Spezia, Jan 01 2020

More terms from Ray Chandler, Feb 14 2004

A090306 a(n) = 17*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 17.

Original entry on oeis.org

2, 17, 291, 4964, 84679, 1444507, 24641298, 420346573, 7170533039, 122319408236, 2086600473051, 35594527450103, 607193567124802, 10357885168571737, 176691241432844331, 3014108989526925364, 51416544063390575519

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.058621... = 2/(17+sqrt(293)) = (sqrt(293)-17)/2.
Lim_{n-> infinity} a(n+1)/a(n) = 17.058621... = (17+sqrt(293))/2 = 2/(sqrt(293)-17).
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010

			a(4) = 17*a(3) + a(2) = 17*4964 + 291=((17+sqrt(293))/2)^4 + ((17-sqrt(293))/2)^4 = 84678.999988190 + 0.000011809 = 84679.

G. C. Greubel, Table of n, a(n) for n = 0..500
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 (17,1).

Cf. A005074.

Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), A090301 (m=15), A090305 (m=16), this sequence (m=17), A090307 (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25).

m:=17;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 30 2019

m:=17; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 30 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 17*I/2)), n = 0..20); # G. C. Greubel, Dec 30 2019

LinearRecurrence[{17,1},{2,17},30] (* Harvey P. Dale, Jan 24 2018 *)
LucasL[Range[20]-1, 17] (* G. C. Greubel, Dec 30 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 17*I/2) ) \\ G. C. Greubel, Dec 30 2019

[2*(-I)^n*chebyshev_T(n, 17*I/2) for n in (0..20)] # G. C. Greubel, Dec 30 2019

a(n) = 17*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 17.
a(n) = ((17+sqrt(293))/2)^n + ((17-sqrt(293))/2)^n.
(a(n))^2 = a(2n) - 2 if n=1, 3, 5, ...
(a(n))^2 = a(2n) + 2 if n=2, 4, 6, ...
G.f.: (2-17*x)/(1-17*x-x^2). - Philippe Deléham, Nov 02 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 17*A098249(n).
a(3n+1) = A041550(5n), a(3n+2) = A041550(5n+3), a(3n+3) = 2*A041550(5n+4).
Lim_{k-> infinity} a(n+k)/a(k) = (A090306(n) + A178765(n)*sqrt(293))/2.
Lim_{n-> infinity} A090306(n)/A178765(n) = sqrt(293). (End)
a(n) = Lucas(n, 17) = 2*(-i)^n * ChebyshevT(n, 17*i/2). - G. C. Greubel, Dec 30 2019
E.g.f.: 2*exp(17*x/2)*cosh(sqrt(293)*x/2). - Stefano Spezia, Dec 31 2019

More terms from Ray Chandler, Feb 14 2004

cp's OEIS Frontend

A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

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A049666 a(n) = Fibonacci(5*n)/5.

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A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

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A099371 Expansion of g.f.: x/(1 - 9*x - x^2).

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A114525 Triangle of coefficients of the Lucas (w-)polynomials.

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A001946 a(n) = 11*a(n-1) + a(n-2).

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