A018913 -id:A018913 - cp's OEIS frontend

A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

1, 0, 1, -1, 0, 1, 0, -2, 0, 1, 1, 0, -3, 0, 1, 0, 3, 0, -4, 0, 1, -1, 0, 6, 0, -5, 0, 1, 0, -4, 0, 10, 0, -6, 0, 1, 1, 0, -10, 0, 15, 0, -7, 0, 1, 0, 5, 0, -20, 0, 21, 0, -8, 0, 1, -1, 0, 15, 0, -35, 0, 28, 0, -9, 0, 1, 0, -6, 0, 35, 0, -56, 0, 36, 0, -10, 0, 1, 1, 0, -21, 0, 70, 0, -84, 0

Offset: 0

Wolfdieter Lang

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).
Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.
Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011
S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002
S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017
|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005
S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010
In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010
From Wolfdieter Lang, Jul 12 2011: (Start)
In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.
The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):
  x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.
Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:
  S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).
(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]
  (End)
The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011
This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011
The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013
These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016
For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016
From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)
Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.
Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.
See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.
Then we have with the present T triangle
  A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.
  A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),
  A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),
  A181546(n) = Sum_{i=0..n+1} T(n,i)^4,
  A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).
S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.
(End)
For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016
The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016
LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)
For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016
For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017
The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017
From Wolfdieter Lang, May 08 2017: (Start)
The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.
In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)
The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017
The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017
From Wolfdieter Lang, Apr 12 2018: (Start)
Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:
S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).
S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).
Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.
The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).
For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)
The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019
Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022
From Wolfdieter Lang, Apr 26 2023: (Start)
Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.
O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).
See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)
S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023
Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

			The triangle T(n, k) begins:
  n\k  0  1   2   3   4   5   6    7   8   9  10  11
  0:   1
  1:   0  1
  2:  -1  0   1
  3:   0 -2   0   1
  4:   1  0  -3   0   1
  5:   0  3   0  -4   0   1
  6:  -1  0   6   0  -5   0   1
  7:   0 -4   0  10   0  -6   0    1
  8:   1  0 -10   0  15   0  -7    0   1
  9:   0  5   0 -20   0  21   0   -8   0   1
  10: -1  0  15   0 -35   0  28    0  -9   0   1
  11:  0 -6   0  35   0 -56   0   36   0 -10   0   1
  ... Reformatted and extended by _Wolfdieter Lang_, Oct 24 2012
For more rows see the link.
E.g., fourth row {0,-2,0,1} corresponds to polynomial S(3,x)= -2*x + x^3.
From _Wolfdieter Lang_, Jul 12 2011: (Start)
Zeros of S(3,x) with rho(4)= 2*cos(Pi/4) = sqrt(2):
  +- t(1,sqrt(2)) = +- sqrt(2) and
  +- t(2,sqrt(2)) = +- 0.
Factorization of S(3,x) in terms of Psi polynomials:
S(3,x) = (2^3)*Psi(4,x/2)*Psi(8,x/2) = x*(x^2-2).
(End)
From _Wolfdieter Lang_, Nov 04 2011: (Start)
A- and Z- sequence recurrence:
T(4,0) = - (C(0)*T(3,1) + C(1)*T(3,3)) = -(-2 + 1) = +1,
T(5,3) = -3 - 1*1 = -4.
(End)
Boas-Buck recurrence for column k = 2, n = 6: S(6, 2) = (3/4)*(0 - 2* S(4 ,2) + 0 + 2*S(2, 2)) = (3/4)*(-2*(-3) + 2) = 6. - _Wolfdieter Lang_, Aug 11 2017
From _Wolfdieter Lang_, Apr 12 2018: (Start)
Factorization into C polynomials (see the Apr 12 2018 comment):
S(4, x) = 1 - 3*x^2 + x^4 = (-1 + x + x^2)*(-1 - x + x^2) = (-C(5, -x)) * C(5, x); the number of factors is 2 = 2*A095374(2).
S(5, x) = 3*x - 4*x^3 + x^5 = x*(-1 + x)*(1 + x)*(-3 + x^2) = C(2, x)*C(3, x)*(-C(3, -x))*C(6, x); the number of factors is 4 = A302707(2). (End)

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T. D. Noe, Rows 0 to 100 of the triangle, flattened.
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Index entries for sequences related to Chebyshev polynomials.
Index entries for "core" sequences

Cf. A000005, A000217, A000292, A000332, A000389, A001227, A007318, A008611, A008615, A101455, A010892, A011973, A053112 (without zeros), A053117, A053119 (reflection), A053121 (inverse triangle), A055034, A097610, A099774, A099777, A100258, A112552 (first column clipped), A127672, A168561 (absolute values), A187360. A194960, A232624, A255237.

Triangles of coefficients of Chebyshev's S(n,x+k) for k = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967.

A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
[A049310(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310 := proc(n,k): binomial((n+k)/2,(n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2 end: seq(seq(A049310(n,k), k=0..n),n=0..11); # Johannes W. Meijer, Aug 08 2011
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> ifelse(irem(n, 2) = 0, 0, (-1)^iquo(n-1, 2))); # Peter Luschny, Oct 06 2022

t[n_, k_] /; EvenQ[n+k] = ((-1)^((n+k)/2+k))*Binomial[(n+k)/2, k]; t[n_, k_] /; OddQ[n+k] = 0; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]][[;; 86]] (* Jean-François Alcover, Jul 05 2011 *)
Table[Coefficient[(-I)^n Fibonacci[n + 1, - I x], x, k], {n, 0, 10}, {k, 0, n}] //Flatten (* Clark Kimberling, Aug 02 2011; corrected by Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[ChebyshevU[Range[0, 10], -x/2], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[Table[(-I)^n Fibonacci[n + 1, -I x], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)

{T(n, k) = if( k<0 || k>n || (n + k)%2, 0, (-1)^((n + k)/2 + k) * binomial((n + k)/2, k))} /* Michael Somos, Jun 24 2002 */

@CachedFunction
def A049310(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A049310(n-1,k-1) - A049310(n-2,k)
for n in (0..9): [A049310(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) := 0 if n < k or n+k odd, otherwise ((-1)^((n+k)/2+k))*binomial((n+k)/2, k); T(n, k) = -T(n-2, k)+T(n-1, k-1), T(n, -1) := 0 =: T(-1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; g.f. k-th column: (1 / (1 + x^2)^(k + 1)) * x^k. - Michael Somos, Jun 24 2002
T(n,k) = binomial((n+k)/2, (n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Nov 21 2005
Recurrence for the (unsigned) Fibonacci polynomials: F(1)=1, F(2)=x; for n > 2, F(n) = x*F(n-1) + F(n-2).
From Wolfdieter Lang, Nov 04 2011: (Start)
The Riordan A- and Z-sequences, given in a comment above, lead together to the recurrence:
T(n,k) = 0 if n < k, if k=0 then T(0,0)=1 and
  T(n,0)= -Sum_{i=0..floor((n-1)/2)} C(i)*T(n-1,2*i+1), otherwise T(n,k) = T(n-1,k-1) - Sum_{i=1..floor((n-k)/2)} C(i)*T(n-1,k-1+2*i), with the Catalan numbers C(n)=A000108(n).
(End)
The row polynomials satisfy also S(n,x) = 2*(T(n+2, x/2) - T(n, x/2))/(x^2-4) with the Chebyshev T-polynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x) - S(n-2, x) (see the Dec 02 2010 comment above) and the S-recurrence several times. This is a formula which expresses the S- in terms of the T-polynomials. - Wolfdieter Lang, Aug 07 2014
From Tom Copeland, Dec 06 2015: (Start)
The non-vanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n-1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1-x)^(-(n+1)). E.g., Diag_4 contains D(2,3) = D(1,0) + D(1,1) + D(1,2) + D(1,3) = 1 + 2 + 3 + 4 = 10 = binomial(5,2). Diag_4 is shifted A000217; Diag_6, shifted A000292: Diag_8, shifted A000332; and Diag_10, A000389.
The non-vanishing antidiagonals are signed rows of the Pascal triangle A007318.
For a reversed, unsigned version with the zeros removed, see A011973. (End)
The Boas-Buck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(n-k))*Sum_{p=0..n-1-k} (1 - (-1)^p)*(-1)^((p+1)/2) * S(n-1-p, k), for n > k >= 0 and input S(k, k) = 1. - Wolfdieter Lang, Aug 11 2017
The m-th row consecutive nonzero entries in order are (-1)^c*(c+b)!/c!b! with c = m/2, m/2-1, ..., 0 and b = m-2c if m is even and with c = (m-1)/2, (m-1)/2-1, ..., 0 with b = m-2c if m is odd. For the 8th row starting at a(36) the 5 consecutive nonzero entries in order are 1,-10,15,-7,1 given by c = 4,3,2,1,0 and b = 0,2,4,6,8. - Richard Turk, Aug 20 2017
O.g.f.: exp( Sum_{n >= 0} 2*T(n,x/2)*t^n/n ) = 1 + x*t + (-1 + x^2)*t^2 + (-2*x + x^3)*t^3 + (1 - 3*x^2 + x^4)*t^4 + ..., where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Aug 15 2022

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050, 9127651499, 90354434940, 894416697901, 8853812544070, 87643708742799, 867583274883920, 8588189040096401, 85014307126080090, 841554882220704499, 8330534515080964900, 82463790268588944501, 816307368170808480110

Offset: 0

N. J. A. Sloane

Indices of square numbers which are also generalized pentagonal numbers.
If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008
The n-th term is a(n) = ((5+sqrt(24))^n - (5-sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, May 31 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 10's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) (A001079) are the nonnegative proper solutions of the Pell equation b(n)^2 - 6*(2*a(n))^2 = +1. See the cross reference to A001079 below. - Wolfdieter Lang, Jun 26 2013
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,9}. - Milan Janjic, Jan 25 2015
For n > 1, this also gives the number of (n-1)-decimal-digit numbers which avoid a particular two-digit number with distinct digits. For example, there are a(5) = 9701 4-digit numbers which do not include "39" as a substring; see Wikipedia link. - Charles R Greathouse IV, Jan 14 2016
All possible solutions for y in Pell equation x^2 - 24*y^2 = 1. The values for x are given in A001079. - Herbert Kociemba, Jun 05 2022
Dickson on page 384 gives the Diophantine equation "(20)  24x^2 + 1 = y^2" and later states "... three consecutive sets (x_i, y_i) of solutions of (20) or 2x^2 + 1 = 3y^2 satisfy x_{n+1} = 10x_n - x_{n-1}, y_{n+1} = 10y_n - y_{n-1} with (x_1, y_1) = (0, 1) or (1, 1), (x_2, y_2) = (1, 5) or (11, 9), respectively." The first set of values (x_n, y_n) = (A001079(n-1), a(n-1)). - Michael Somos, Jun 19 2023

			a(2)=10 and (3(-8)^2-(-8))/2=10^2, a(3)=99 and (3(81)^2-(81))/2=99^2. - _Michael Somos_, Sep 05 2006
G.f. = x + 10*x^2 + 99*x^3 + 980*x^4 + 9701*x^5 + 96030*x^6 + ...

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 12.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
D. Fortin, B-spline Toeplitz inverse under corner perturbations, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118. - From _N. J. A. Sloane_, Oct 22 2012
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=10, q=-1.
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
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=12.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Wikipedia, Curse of 39
Jianqiang Zhao, Finite Multiple zeta Values and Finite Euler Sums, arXiv:1507.04917 [math.NT], 2015.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A001079, A001109, A001353, A001906, A004187, A004254, A018913, A046173, A049310, A054320, A072256, A101950, A105138, A108741 (squares).
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), this sequence (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=5;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[ n eq 1 select 0 else n eq 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..20] ]; // Vincenzo Librandi, Aug 19 2011

A004189 := proc(n)
    option remember;
    if n <= 1 then
        n ;
    else
        10*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A004189(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
seq( simplify(ChebyshevU(n-1, 5)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n-1,1,5], {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008; modified by G. C. Greubel, Jun 06 2019 *)
LinearRecurrence[{10, -1}, {0, 1}, 20] (* Jean-François Alcover, Nov 15 2017 *)
ChebyshevU[Range[21] -2, 5] (* G. C. Greubel, Dec 23 2019 *)

{a(n) = subst(poltchebi(n+1) - 5*poltchebi(n), 'x, 5) / 24}; /* Michael Somos, Sep 05 2006 */

a(n)=([9,1;8,1]^(n-1)*[1;1])[1,1] \\ Charles R Greathouse IV, Jan 14 2016

vector(21, n, n--; polchebyshev(n-1, 2, 5) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number1(n,10,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,5) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = S(2*n-1, sqrt(12))/sqrt(12) = S(n-1, 10); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
A001079(n) = sqrt(24*(a(n)^2)+1), that is a(n) = sqrt((A001079(n)^2-1)/24).
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ( (5+2*sqrt(6))^n - (5-2*sqrt(6))^n )/(4*sqrt(6)).
G.f.: x/(1-10*x+x^2). (End)
a(-n) = -a(n). - Michael Somos, Sep 05 2006
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 10*a(n-1) - a(n-2). (End)
a(n+1) = Sum_{k=0..n} A101950(n,k)*9^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/2*(2 + sqrt(6)).
Product {n >= 2} (1 - 1/a(n)) = 1/5*(2 + sqrt(6)). (End)
a(n) = (A054320(n-1) + A072256(n))/2. - Richard R. Forberg, Nov 21 2013
a(2*n - 1) = A046173(n).
E.g.f.: exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*8^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*12^k. - Peter Bala, Jul 18 2023

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

Original entry on oeis.org

1, 10, 89, 791, 7030, 62479, 555281, 4935050, 43860169, 389806471, 3464398070, 30789776159, 273643587361, 2432002510090, 21614379003449, 192097408520951, 1707262297685110, 15173263270645039, 134852107138120241, 1198495700972437130, 10651609201613813929

Offset: 0

Wolfdieter Lang, Aug 04 2000

This is the m=11 member of the m-family of sequences S(n,m-2)+S(n-1,m-2) = S(2*n,sqrt(m)) (for S(n,x) see Formula). The m=4..10 instances are A005408, A002878, A001834, A030221, A002315, A033890 and A057080, resp. The m=1..3 (signed) sequences are: A057078, A057077 and A057079, resp.
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim_{n->oo} a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 89, 389806471, 192097408520951, 7477414486269626733119, ... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 10, 0, 89, 0, 791, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -7, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=11.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A005408, A002878, A001834, A030221, A002315, A033890, A057080.
Cf. A057078, A057077, A057079.
Cf. A018913, A049310.
Cf. A100047.

A057081 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,10]);
    else
        9*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1 + x)/(1 - 9*x + x^2), {x,0,50}], x] (* or *) LinearRecurrence[{9,-1}, {1,10}, 50] (* G. C. Greubel, Apr 12 2017 *)

Vec((1+x)/(1-9*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,9,1)-lucas_number2(n-1,9,1))/7 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

a(n) = 9*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 9) + S(n-1, 9) = S(2*n, sqrt(11)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 9) = A018913(n).
G.f.: (1+x)/(1-9*x+x^2).
Let q(n, x) = Sum{i=0..n} x^(n-i)*binomial(2*n-i, i), a(n) = (-1)^n*q(n, -11). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-9)*(-1)^n, where L is defined as in A108299; see also A070998 for L(n,+9). - Reinhard Zumkeller, Jun 01 2005
From Peter Bala, Jun 08 2025: (Start)
a(n) = (1/sqrt(7)) * ( ((sqrt(11) + sqrt(7))/2)^(2*n+1) - ((sqrt(11) - sqrt(7))/2)^(2*n+1) ).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/11 (telescoping series: 11/(a(n) - 1/a(n)) = 1/A018913(n+1) + 1/A018913(n)).
Conjecture: for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(11/7) [telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (1 - 4/b(n+1))/(1 - 4/b(n)), where b(n) = 2 + A056918(n)]. (End)

A070998 a(n) = 9*a(n-1) - a(n-2) for n > 0, a(0)=1, a(-1)=1.

Original entry on oeis.org

1, 8, 71, 631, 5608, 49841, 442961, 3936808, 34988311, 310957991, 2763633608, 24561744481, 218292066721, 1940066856008, 17242309637351, 153240719880151, 1361924169284008, 12104076803675921, 107574767063799281, 956068826770517608

Offset: 0

Joe Keane (jgk(AT)jgk.org), May 18 2002

A Pellian sequence.
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,9), where L is defined as in A108299; see also A057081 for L(n,-9). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(7)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Positive values of x (or y) satisfying x^2 - 9xy + y^2 + 7 = 0. - Colin Barker, Feb 09 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A057081, A056918.
Row 9 of array A094954.
Cf. similar sequences listed in A238379.

I:=[1,8]; [n le 2 select I[n] else 9*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014

CoefficientList[Series[(1 - x)/(1 - 9 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
LinearRecurrence[{9,-1},{1,8},30] (* Harvey P. Dale, Sep 24 2015 *)

[lucas_number1(n, 9, 1) - lucas_number1(n-1, 9, 1) for n in range(1, 19)]  # Zerinvary Lajos, Nov 10 2009

a(n) ~ (1/11)*sqrt(11)*((1/2)*(sqrt(11) + sqrt(7)))^(2*n+1).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 7) = a(n). - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 63 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = (-1)^n*U(2n, i*sqrt(7)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
G.f.: (1-x)/(1-9*x+x^2). - Philippe Deléham, Nov 03 2008
a(n) = A018913(n+1) - A018913(n). - R. J. Mathar, Jun 07 2013

More terms from Vincenzo Librandi, Feb 10 2014

A004190 Expansion of 1/(1 - 11*x + x^2).

Original entry on oeis.org

1, 11, 120, 1309, 14279, 155760, 1699081, 18534131, 202176360, 2205405829, 24057287759, 262424759520, 2862615066961, 31226340977051, 340627135680600, 3715672151509549, 40531766530924439, 442133759688659280, 4822939590044327641, 52610201730798944771, 573889279448744064840

Offset: 0

N. J. A. Sloane

Chebyshev or generalized Fibonacci sequence.
This is the m=13 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..12 (nonnegative) sequences are A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913 and A004189. The m=1..3 (signed) sequences are A049347, A056594, A010892.
All positive integer solutions of Pell equation b(n)^2 - 117*a(n)^2 = +4 together with b(n+1)=A057076(n+1), n >= 0. - Wolfdieter Lang, Aug 31 2004
For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 11's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,10}. - Milan Janjic, Jan 25 2015

			G.f. = 1 + 11*x + 120*x^2 + 1309*x^3 + 14279*x^4 + 155760*x^5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..900
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12
S. Falcon, Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence, Journal of Mathematics Research Vol. 4, No. 2; April 2012. - From _N. J. A. Sloane_, Sep 22 2012
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=11, q=-1.
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
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=13.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (11,-1).

Cf. A000027, A001090, A001109, A001353, A001906, A004187, A004189, A004254, A006190, A010892, A018913, A049310, A049347, A056594, A057076, A075835, A101950.

with(combinat):seq(fibonacci(2*n+2, 3)/3, n=0..20); # Zerinvary Lajos, Apr 20 2008

Join[{a=1,b=11},Table[c=11*b-a; a=b; b=c, {n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 20 2011 *)
CoefficientList[Series[1/(1-11*x+x^2),{x,0,30}],x] (* Vincenzo Librandi, Jun 13 2012 *)
Table[Fibonacci[2n + 2, 3]/3, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
a[ n_] := ChebyshevU[n, 11/2]; (* Michael Somos, Jul 14 2018 *)

Vec(1/(1-11*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

{a(n) = polchebyshev(n, 2, 11/2)}; /* Michael Somos, Jul 14 2018 */

[lucas_number1(n,11,1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008

Recursion: a(n) = 11*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(13))/sqrt(13) = S(n, 11); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
G.f.: 1/(1 - 11*x + x^2).
a(n) = ((11+3*sqrt(13))^(n+1) - (11-3*sqrt(13))^(n+1))/(2^(n+1)*3*sqrt(13)). - Rolf Pleisch, May 22 2011
a(n) = Sum_{k=0..n} A101950(n,k)*10^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n>=0} (1 + 1/a(n)) = 1/3*(3 + sqrt(13)).
Product_{n>=1} (1 - 1/a(n)) = 3/22*(3 + sqrt(13)). (End)
a(n) = sqrt((A057076(n+1)^2 - 4)/117).
a(n) = A075835(n+1)/3 = A006190(2*n+2)/3. - Vladimir Reshetnikov, Sep 16 2016
a(n) = -a(-2-n) for all n in Z. - Michael Somos, Jul 14 2018
E.g.f.: exp(11*x/2)*(39*cosh(3*sqrt(13)*x/2) + 11*sqrt(13)*sinh(3*sqrt(13)*x/2))/39. - Stefano Spezia, Aug 07 2024

Wolfdieter Lang, Oct 31 2002

A065100 a(n+2) = 9*a(n+1) - a(n), a(0) = 3, a(1) = 27.

Original entry on oeis.org

3, 27, 240, 2133, 18957, 168480, 1497363, 13307787, 118272720, 1051146693, 9342047517, 83027280960, 737903481123, 6558104049147, 58285032961200, 518007192601653, 4603779700453677, 40916010111481440, 363640311302879283

Offset: 0

N. J. A. Sloane, Nov 12 2001

Original definition: a(0) = c, a(1) = p*c^3; a(n+2) = p*c^2*a(n+1) - a(n), for p = 1, c = 3.

The sequence could have started with a(0) = 0, then a(1) = 3 etc. Any of the family of sequences x = (0, c, c^3, c^5 - c, ...) with x(n+1) = c^2*x(n) - x(n-1), c >= 1, provides a subset of solutions to A115169 (for n >= 2), their union yields all the solutions. See A052530 for c = 2. - M. F. Hasler, Jun 12 2019

			From _Vincenzo Librandi_, Aug 07 2010: (Start)
a(2) = 9*27 - 3 = 240;
a(3) = 9*240 - 27 = 2133;
a(4) = 9*2133 - 240 = 18957. (End)

Harry J. Smith, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
J.-P. Ehrmann et al., Problem POLYA002, Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A052530 (analog for c = 2).

a[0] = c; a[1] = p*c^3; a[n_] := a[n] = p*c^2*a[n - 1] - a[n - 2]; p = 1; c = 3; Table[ a[n], {n, 0, 20} ]
LinearRecurrence[{9,-1},{3,27},30] (* Harvey P. Dale, Sep 22 2016 *)

polya002(1,3,20) \\ See A052530 for definition of function polya002().

{ p=1; c=3; k=p*c^2; for (n=0, 100, if (n>1, a=k*a1 - a2; a2=a1; a1=a, if (n, a=a1=k*c, a=a2=c)); write("b065100.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 07 2009

G.f.: 3/(1 - 9*x + x^2). - Floor van Lamoen, Feb 07 2002
a(n) = 3*A018913(n+1). - R. J. Mathar, Oct 26 2009
a(n) = 9*a(n-1) - a(n-2) (with a(0)=3, a(1)=27). - Vincenzo Librandi, Aug 07 2010
E.g.f.: 3*exp(9*x/2)*(77*cosh(sqrt(77)*x/2) + 9*sqrt(77)*sinh(sqrt(77)*x/2))/77. - Stefano Spezia, Feb 23 2025

Definition simplified by M. F. Hasler, Jun 12 2019

A078362 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 13, 168, 2171, 28055, 362544, 4685017, 60542677, 782369784, 10110264515, 130651068911, 1688353631328, 21817946138353, 281944946167261, 3643466354036040, 47083117656301259, 608437063177880327

Offset: 0

Wolfdieter Lang, Nov 29 2002

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 165*a^2 = +4 with companion sequence b(n)=A078363(n+1), n >= 0.
This is the m=15 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..14 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190 and A004191. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 13's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,12}. - Milan Janjic, Jan 23 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..900
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=13, q=-1.
M. 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
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=15.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (13,-1).

Cf. A078363.

a:=[1,13,168];; for n in [4..20] do a[n]:=13*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 25 2019

I:=[1, 13, 168]; [n le 3 select I[n] else 13*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

CoefficientList[Series[1/(1 - 13 x + x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
LinearRecurrence[{13,-1},{1,13},20] (* Harvey P. Dale, Feb 07 2019 *)

my(x='x+O('x^20)); Vec(1/(1-13*x+x^2)) \\ G. C. Greubel, May 25 2019

[lucas_number1(n,13,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = 13*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(15))/sqrt(15) = S(n, 13), where S(n, x) = U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (13+sqrt(165))/2 and am = (13-sqrt(165))/2.
G.f.: 1/(1 - 13*x + x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*12^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = (1/11)*(11 + sqrt(165)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = (1/26)*(11 + sqrt(165)). - Peter Bala, Dec 23 2012
For n >= 1, a(n) = U(n-1,13/2), where U(k,x) represents Chebyshev polynomial of the second order.
a(n) = sqrt((A078363(n+1)^2 - 4)/165), n>=0, (Pell equation d=165, +4).

A092420 a(n+2) = 9*a(n+1) - a(n) + 1, with a(1)=1, a(2)=10.

Original entry on oeis.org

1, 10, 90, 801, 7120, 63280, 562401, 4998330, 44422570, 394804801, 3508820640, 31184580960, 277152408001, 2463187091050, 21891531411450, 194560595612001, 1729153829096560, 15367823866257040, 136581260967216801

Offset: 1

M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), Apr 04 2004

Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 9/11.

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,-10,1).

Cf. A092521. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

Cf. A018913 (first differences).

a[1] = 1; a[2] = 10; a[n_] := a[n] = 9a[n - 1] - a[n - 2] + 1; Table[ a[n], {n, 20}] (* Robert G. Wilson v, Apr 05 2004 *)
LinearRecurrence[{10,-10,1},{1,10,90},20] (* Harvey P. Dale, May 21 2023 *)

G.f.: x/(1-10*x+10*x^2-x^3) = x/((1-x)*(1-9*x+x^2)).
a(n) = 10*a(n-1) - 10*a(n-2) + a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=10.
a(n) = (S(n,9) - S(n-1,9) - 1)/7, n >= 1.
a(n+1) = Sum_{k=0..n} S(n,9), n >= 0, with S(n,9) = U(n,9/2) = A018913(n+1). (Partial sums of Chebyshev sequence A018913.)

More terms from Robert G. Wilson v, Apr 05 2004

Chebyshev comments from Wolfdieter Lang, Aug 31 2004

A316269 Array T(n,k) = n*T(n,k-1) - T(n,k-2) read by upward antidiagonals, with T(n,0) = 0, T(n,1) = 1, n >= 2.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 4, 8, 4, 0, 1, 5, 15, 21, 5, 0, 1, 6, 24, 56, 55, 6, 0, 1, 7, 35, 115, 209, 144, 7, 0, 1, 8, 48, 204, 551, 780, 377, 8, 0, 1, 9, 63, 329, 1189, 2640, 2911, 987, 9, 0, 1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10

Offset: 2

Jianing Song, Jun 28 2018

Define {x(k)} to be an integer sequence satisfying all the following conditions:
(i) {x(k)} satisfies second-order linear recursion, that is, there exists two integers P, Q such that x(k+2) = P*x(k+1) + Q*x(k) holds for all k >= 0.
(ii) {x(k)} is (not necessarily strictly) increasing. ({A000035(k)} satisfies condition (i), but it doesn't satisfy this.)
(iii) All terms in {x(k)} do not share a common factor. ({A024023(k)} satisfies both conditions (i) and (ii), but all terms share a common factor 2.)
(iv) {x(k)} satisfies strong divisibility, that is, gcd(x(m),x(n)) = x(gcd(m,n)) holds for all m, n >= 0. ({A093131(k)} satisfies all conditions (i) to (iii), but 5 = gcd(A093131(2),A093131(3)) != A093131(gcd(2,3)) = 1.)
(v) For all positive integers n, there eventually exists some m > 0 such that n divides x(m). ({A002275(k)} satisfies all conditions (i) to (iv), but 2, 5 and 10 never divide any term.)
Then it's easy to show that the only solutions to {x(k)} are x(k) = A172236(n,k) or x(k) = T(n,k), i.e., x(0) = 0, x(1) = 1, P >= 1, Q = 1 or P >= 2, Q = -1.
The case n = 0 is not included since it gives the period-4 signed sequence 0, 1, 0, -1, 0, 1, 0, -1, ..., the g.f. of which is the inverse of the 4th cyclotomic polynomial.
The case n = 1 is not included since it gives the period-6 signed sequence 0, 1, 1, 0, -1, -1, ..., the g.f. of which is the inverse of the 6th cyclotomic polynomial.
The congruence property: let p be an odd prime which is not divisible by n^2 - 4, then T(n,(p-1)/2) == 1/2(((n-2)/p) - ((n+2)/p)) (mod p), T(n,(p+1)/2) == 1/2(((n-2)/p) + ((n+2)/p)) (mod p). Here ((n-2)/p) is the Legendre symbol. Or equivalently:
((n-2)/p)...((n+2)/p)...T(n,(p-1)/2) mod p...T(n,(p+1)/2) mod p
.....1...........1...............0....................1
....-1..........-1...............0...................-1
.....1..........-1...............1....................0
....-1...........1..............-1....................0
To prove this, rewrite (n +- sqrt(n^2-4))/2 as ((sqrt(n+2) +- sqrt(n-2))/2)^2.
Let E(n,m) be the smallest number l such that m divides T(n,l), we have: E(n,p) divides (p - ((n^2-4)/p))/2 for odd primes p that are not divisible by n^2 - 4. E(n,p) = p for odd primes p that are divisible by n^2 - 4. E(n,2) = 2 for even n and 3 for odd n.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 - 4, p^2 is never divisible by T(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Given n, the largest possible value of E(n,m)/m is 1 for even n and 3/2 for odd n. It can be obtained by the value of E(n,2) described above.
Let pi(n,m) be the Pisano period of T(n,k) modulo m, i.e, the smallest number l such that T(n,k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p + ((n-2)/p) if ((n+2)/p) = -1 and (p - ((n-2)/p))/2 if ((n+2)/p) = 1. pi(n,p) = p for odd primes p that are divisible by n - 2 and 2p for odd primes p that are divisible by n + 2. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide T(n,E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 - 4, p^2 is never divisible by T(n,p), so pi(n,p^e) = p^e or 2p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1),pi(n,m_2)) if gcd(m_1,m_2) = 1.
Given n, the largest possible value of pi(n,m)/m is:
Parity of n...n + 2 is a power of 2 or 3...max{pi(n,m)/m}.....obtained by
....even..........yes (even exponent)............1...........pi(n,2^e) = 2^e
....even...........yes (odd exponent)...........4/3............pi(n,3) = 4
....even...................no....................2.............pi(n,p) = 2p (p >= 3 is any prime factor of n + 2)
.....odd..................yes....................2..........pi(n,3^e) = 2*3^e
.....odd...................no....................3..........pi(n,2p^e) = 6p^e (p >= 5 is any prime factor of n + 2)
The largest possible value of pi(n,m)/m is obtained by infinitely many m except for the case n = 10, in which we have pi(10,3) = 6, pi(10,7) = 8, pi(10,21) = 24 and pi(10,m)/m <= 14/13 for all other m. [Corrected by Jianing Song, Nov 04 2018]
Let z(n,m) be the number of zeros in a period of T(n,k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: for odd primes p that are not divisible by n^2 - 4, z(n,p) = 2 if ((n+2)/p) = -1; 1 or 2 if ((n+2)/p) = 1. z(n,p) = 1 for odd primes p that are divisible by n - 2 and 2 for odd primes p that are divisible by n + 2. z(n,2) = 1.
For all odd primes p, z(n,p) = 2 if and only if pi(n,p) is even, z(n,p) = 1 if and only if pi(n,p) is odd. For all odd primes p, if E(n,p) is even then z(n,p) = 2 (the converse is not necessarily true). [Comment revised by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p. z(n,4) = 1 if n == 2, 3 (mod 4) and 2 if n == 0, 1 (mod 4). z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
By induction it is easy to show that T(n,k) = T(n,m+1)*T(n,k-m) - T(n,m)*T(k-m-1). Let k = 2m we have T(n,2m) = T(n,m)*(T(n,m+1)-T(m-1)); let k = 2m+1 we have T(n,2m+1) = T(n,m+1)^2 - T(n,m)^2 = (T(n,m+1)+T(n,m))*(T(n,m+1)-T(n,m)). So T(n,k) is composite if n >= 3, k >= 3. - Jianing Song, Jul 06 2019

			The array starts in row n = 2 with columns k >= 0 as follows:
  0      1      2      3      4      5      6
  0      1      3      8     21     55    144
  0      1      4     15     56    209    780
  0      1      5     24    115    551   2640
  0      1      6     35    204   1189   6930
  0      1      7     48    329   2255  15456
  0      1      8     63    496   3905  30744
  0      1      9     80    711   6319  56160
  0      1     10     99    980   9701  96030
  0      1     11    120   1309  14279 155760

Jianing Song, Antidiagonals n = 2..101, flattened

Cf. A172236.
Sequences with g.f. 1/(1-k*x+x^2): A001477 (k=2), A001906 (k=3), A001353 (k=4), A004254 (k=5), A001109 (k=6), A004187 (k=7), A001090 (k=8), A018913 (k=9), A004189 (k=10).
Cf. A005563 (4th column), A242135 (5th column), A057722 (6th column).

Table[If[# == 2, k, Simplify[(((# + Sqrt[#^2 - 4])/2)^k - ((# - Sqrt[#^2 - 4])/2)^k)/Sqrt[#^2 - 4]]] &[n - k + 2], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jul 19 2018 *)

T(n, k) = if (k==0, 0, if (k==1, 1, n*T(n,k-1) - T(n,k-2)));
tabl(nn) = for(n=2, nn, for (k=0, nn, print1(T(n,k), ", ")); print); \\ Michel Marcus, Jul 03 2018

T(n, k) = ([n, -1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018

T(2,k) = k; T(n,k) = (((n+sqrt(n^2 - 4))/2)^k - ((n - sqrt(n^2 - 4))/2)^k)/sqrt(n^2 - 4), n >= 3, k >= 0.
T(n^2+2,k) = A172236(n,2k); T(n^4+4n^2+2,k) = A172236(n,4k)/A172236(n,4).
For n >= 2, Sum_{i=1..k} 1/T(n,2^i) = 2/n - ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/T(n,2^k)), where u = (n + sqrt(n^2 - 4))/2, v = (n - sqrt(n^2 - 4))/2 are the two roots of the polynomial x^2 - n*x + 1. As a result, Sum_{i=>1} 1/T(n,2^i) = (n - sqrt(n^2 - 4))/2. - Jianing Song, Apr 21 2019

cp's OEIS Frontend

A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

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A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

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A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

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A070998 a(n) = 9*a(n-1) - a(n-2) for n > 0, a(0)=1, a(-1)=1.

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A004190 Expansion of 1/(1 - 11*x + x^2).

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A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.