A145027 a(n) = a(n-1) + a(n-2) + a(n-3) with a(1) = 2, a(2) = 3, a(3) = 4.

2, 3, 4, 9, 16, 29, 54, 99, 182, 335, 616, 1133, 2084, 3833, 7050, 12967, 23850, 43867, 80684, 148401, 272952, 502037, 923390, 1698379, 3123806, 5745575, 10567760, 19437141, 35750476, 65755377, 120942994, 222448847, 409147218

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Sep 30 2008

Keywords

• nonn

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).

Crossrefs

Cf. A000073, A000213, A001590, A003265, A056816, A081172, A001644.

Programs

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1+x)*(2-x)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 22 2019

Mathematica

LinearRecurrence[{1,1,1},{2,3,4},33] (* Ray Chandler, Dec 08 2013 *)

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

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

Formula

From R. J. Mathar, Jan 30 2011: (Start)

a(n) = -A000073(n-1) + A000073(n) + 2*A000073(n+1).

G.f. x*(1+x)*(2-x)/(1-x-x^2-x^3). (End)