A168082 Fibonacci 11-step numbers.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4093, 8184, 16364, 32720, 65424, 130816, 261568, 523008, 1045760, 2091008, 4180992, 8359937, 16715781, 33423378, 66830392, 133628064, 267190704, 534250592, 1068239616

Offset

1,13

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.

Kai Wang, Identities for generalized enneanacci numbers, Generalized Fibonacci Sequences (2020).

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

Formula

From Joerg Arndt, Sep 22 2020: (Start)

a(n) = Sum_{k=1..11} a(n-k).

G.f.: x^11/(1 - Sum_{k=1..11} x^k ).

a(n) = 2*a(n-1) - a(n-12). (End)

Another form of the g.f. f: f(z) = (z^(k-1)-z^(k))/(1-2*z+z^(k+1)) with k=11. a(n) = Sum_((-1)^i*binomial(n-10-11*i,i)*2^(n-10-12*i), i=0..floor((n-10)/12))-Sum_((-1)^i*binomial(n-11-11*i,i)*2^(n-11-12*i), i=0..floor((n-11)/12)) with Sum_(alpha(i),i=m..n) = 0 for m>n. - Richard Choulet, Feb 22 2010

Maple

a:= proc(n) option remember; `if`(n<11, 0,
      `if`(n=11, 1, add(a(n-j), j=1..11)))
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Sep 23 2020

Mathematica

With[{nn=11}, LinearRecurrence[Table[1, {nn}], Join[Table[0, {nn-1}], {1}], 50]] (* Harvey P. Dale, Aug 17 2013 *)

Crossrefs

Sequence in context: A145117 A172320 A234592 * A295081 A227843 A271482

Adjacent sequences: A168079 A168080 A168081 * A168083 A168084 A168085

Keyword

nonn,easy

Author

Vladimir Joseph Stephan Orlovsky, Nov 18 2009

Status

approved