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

Author

Vladimir Joseph Stephan Orlovsky, Nov 18 2009

Keywords

• nonn
• easy

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).

Programs

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 *)

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