OFFSET
1,1
COMMENTS
If the conjectured recurrence in A000382 is correct, then a(n) = A000382(n+2) - A000382(n+1), n>=4. - R. J. Mathar, Jan 30 2011
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Index entries for linear recurrences with constant coefficients, signature (1,1,1).
FORMULA
From R. J. Mathar, Jan 30 2011: (Start)
G.f. x*(1+x)*(2-x)/(1-x-x^2-x^3). (End)
MATHEMATICA
LinearRecurrence[{1, 1, 1}, {2, 3, 4}, 33] (* Ray Chandler, Dec 08 2013 *)
PROG
(PARI) my(x='x+O('x^30)); Vec(x*(1+x)*(2-x)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 22 2019
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1+x)*(2-x)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 22 2019
(Sage) a=(x*(1+x)*(2-x)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Apr 22 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Joseph Stephan Orlovsky, Sep 30 2008
STATUS
approved