A018913 a(n) = 9*a(n - 1) - a(n - 2) for n>1, a(0)=0, a(1)=1.

0, 1, 9, 80, 711, 6319, 56160, 499121, 4435929, 39424240, 350382231, 3114015839, 27675760320, 245967827041, 2186034683049, 19428344320400, 172669064200551, 1534593233484559, 13638670037160480, 121213437100959761

Offset: 0

Author

R. K. Guy

Keywords

• nonn
• easy

Comments

Define the sequence L(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0. This is L(1,9).

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 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,...,8}. - Milan Janjic, Jan 25 2015

Not to be confused with the Pisot L(1,9) sequence, which is A001019. - R. J. Mathar, Feb 13 2016

Lim_{n->oo} a(n+1)/a(n) = (9 + sqrt(77))/2 = A092290 + 1 = 8.887482... - Wolfdieter Lang, Nov 16 2023

Examples

			G.f. = x + 9*x^2 + 80*x^3 + 711*x^4 + 6319*x^5 + 56160*x^6 + 499121*x^7 + ...
		

Links

• Vincenzo Librandi, 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.
• 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.
• D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
• E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
• A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=9, 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=11.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (9,-1).
• Index entries for Pisot sequences

Crossrefs

Cf. A000027, A001906, A001353, A004254, A001109, A004187, A001090.

Cf. A056918(n)=sqrt{77*(a(n))^2 +4}, that is, a(n)=sqrt((A056918(n)^2 - 4)/77).

Cf. A092290 + 1.

Programs

Magma

I:=[0, 1]; [n le 2 select I[n] else 9*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

Mathematica

CoefficientList[Series[x/(1 - 9*x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)

PARI

concat(0, Vec(x/(1-9*x+x^2) + O(x^30))) \\ Michel Marcus, Sep 06 2017

Sage

[lucas_number1(n,9,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

Formula

G.f.: x/(1-9*x+x^2).

a(n) = S(2*n-1, sqrt(11))/sqrt(11) = S(n-1, 9); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.

a(n) = (((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n)/sqrt(77). - Barry E. Williams, Aug 21 2000

a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*8^k. - Philippe Deléham, Feb 10 2012

From Peter Bala, Dec 23 2012: (Start)

Product {n >= 1} (1 + 1/a(n)) = 1/7*(7 + sqrt(77)).

Product {n >= 2} (1 - 1/a(n)) = 1/18*(7 + sqrt(77)). (End)

a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*7^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*11^k. - Peter Bala, Jul 17 2023

E.g.f.: 2*exp(9*x/2)*sinh(sqrt(77)*x/2)/sqrt(77). - Stefano Spezia, Feb 23 2025

Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = sqrt(11/7) [telescoping product: ((a(2*n) + 1)/(a(2*n) - 1))^2 = (11 - 4/(a(n+1) - a(n))^2)/(11 - 4/(a(n) - a(n-1))^2), leading to 11 - 7*Product_{k = 1..n} ((a(2*k) + 1)/(a(2*k) - 1))^2 = 4/A070998(n)^2]. - Peter Bala, May 18 2025

Extensions

G.f. adapted to the offset by Vincenzo Librandi, Dec 23 2012