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

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Examples

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

Links

• 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).

Crossrefs

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

Programs

Maple

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

Mathematica

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 *)

PARI

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

PARI

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

Sage

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

Formula

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

Extensions

Wolfdieter Lang, Oct 31 2002