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A168082
Fibonacci 11-step numbers.
3
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
Martin Burtscher, Igor Szczyrba, and RafaƂ 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).
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
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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