A248098 - OEIS

A248098

a(n) = 1 + a(n-1) + a(n-2) + a(n-3) if n>=4; a(1) = a(2) = a(3) = 1.

1, 1, 1, 4, 7, 13, 25, 46, 85, 157, 289, 532, 979, 1801, 3313, 6094, 11209, 20617, 37921, 69748, 128287, 235957, 433993, 798238, 1468189, 2700421, 4966849, 9135460, 16802731, 30905041, 56843233, 104551006, 192299281, 353693521

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OFFSET

1,4

COMMENTS

Number of vertices in the Fibonacci ternary tree T(n). The Fibonacci ternary tree T(n) is defined inductively: T(1), T(2), T(3) consist of only a root node, while for n>=4, T(n) consists of a root node with 3 ordered children T(n-1), T(n-2), T(n-3) from left to right. See the Chang reference.

The number of leaves in the Fibonacci ternary tree T(n) is the tribonacci number A000213(n-1).

In general, adding a constant to each successive term of a third-order linear recurrence with signature (x,y,z) results in a fourth-order recurrence with signature (x+1, y-x, z-y, -z). - Gary Detlefs, Jul 20 2023

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

D. K. Chang, On Fibonacci k-ary trees, The Fibonacci Quarterly, Volume 24, Number 3, August 1986, 258-262.

Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

FORMULA

G.f. = z*(1-z-z^2+2*z^3)/((1-z)*(1-z-z^2-z^3)).

MAPLE

a[1]:=1: a[2]:=1: a[3]:=1: for n from 4 to 40 do a[n] := 1+a[n-1]+a[n-2]+a[n-3] end do: seq(a[n], n=1..40);

g:=z*(1-z^2+2*z^3-z)/((1-z)*(1-z-z^2-z^3)): gser:=series(g, z=0, 45): seq(coeff(gser, z, n), n=1..40);

PROG

(Haskell)

a248098 n = a248098_list !! (n-1)