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A015457

Generalized Fibonacci numbers.

1, 1, 12, 133, 1475, 16358, 181413, 2011901, 22312324, 247447465, 2744234439, 30434026294, 337518523673, 3743137786697, 41512034177340, 460375513737437, 5105642685289147, 56622445051918054, 627952538256387741, 6964100365872183205, 77233056562850402996

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OFFSET

0,3

COMMENTS

For n>=1, row sums of triangle for numbers 11^k*binomial(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,...,11} containing no subwords ii, (i=0,1,...,10). - Milan Janjic, Jan 31 2015

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..900

Tanya Khovanova, Recursive Sequences

Index entries for linear recurrences with constant coefficients, signature (11,1).

FORMULA

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

a(n) = Sum_{k=0..n} 10^k*A055830(n,k). - Philippe Deléham, Oct 18 2006

G.f.: (1-10*x)/(1-11*x-x^2). - Philippe Deléham, Nov 21 2008

For n>=2, a(n) = F_n(11)+F_(n+1)(11), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012

a(n) = (F(5*n-5) + F(5*n))/5 for F(n) the Fibonacci sequence A000045(n). - Greg Dresden, Aug 22 2021

MATHEMATICA

LinearRecurrence[{11, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)

CoefficientList[Series[(1-10*x)/(1-11*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

PROG

(Magma) [n le 2 select 1 else 11*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

(PARI) x='x+O('x^30); Vec((1-10*x)/(1-11*x-x^2)) \\ G. C. Greubel, Dec 19 2017

CROSSREFS

Row m=11 of A135597.

Cf. A000045, A049310, A055830.

Sequence in context: A244205 A016123 A353148 * A015469 A144785 A214994

Adjacent sequences: A015454 A015455 A015456 * A015458 A015459 A015460