A015456 Generalized Fibonacci numbers.

1, 1, 11, 111, 1121, 11321, 114331, 1154631, 11660641, 117761041, 1189271051, 12010471551, 121293986561, 1224950337161, 12370797358171, 124932923918871, 1261700036546881, 12741933289387681, 128681032930423691, 1299552262593624591, 13124203658866669601

Offset

0,3

Comments

For n>=1, row sums of triangle for numbers 10^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012

For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,10} containing no subwords ii, (i=0,1,...,9). - Milan Janjic, Jan 31 2015

a(n) equals the number of sequences over the alphabet {0,1,...,9,10} such that no two consecutive terms have distance 6. - David Nacin, Jun 02 2017

Links

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

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

Index entries for linear recurrences with constant coefficients, signature (10,1).

Formula

a(n) = 10*a(n-1) + a(n-2).

a(n) = Sum_{k=0..n} 9^k*A055830(n,k). - Philippe Deléham, Oct 18 2006

G.f.: (1-9*x)/(1-10*x-x^2). - Philippe Deléham, Nov 20 2008

For n>=2, a(n) = F_(n)(10) + F_(n+1)(10), where F_n(x) is Fibonacci polynomial (cf.A049310): F_n(x) = Sum_{i=0,...,floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012

Mathematica

LinearRecurrence[{10, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-9*x)/(1-10*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

Prog

(Magma) [n le 2 select 1 else 10*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012
(PARI) x='x+O('x^30); Vec((1-9*x)/(1-10*x-x^2)) \\ G. C. Greubel, Dec 19 2017

Crossrefs

Row m=10 of A135597.

Sequence in context: A134732 A166747 A337844 * A343355 A240367 A338679

Adjacent sequences: A015453 A015454 A015455 * A015457 A015458 A015459

Keyword

nonn,easy

Author

Olivier Gérard

Status

approved