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%I #18 Aug 18 2023 16:41:56
%S 1,1,1,4,7,13,25,46,85,157,289,532,979,1801,3313,6094,11209,20617,
%T 37921,69748,128287,235957,433993,798238,1468189,2700421,4966849,
%U 9135460,16802731,30905041,56843233,104551006,192299281,353693521
%N a(n) = 1 + a(n-1) + a(n-2) + a(n-3) if n>=4; a(1) = a(2) = a(3) = 1.
%C 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.
%C The number of leaves in the Fibonacci ternary tree T(n) is the tribonacci number A000213(n-1).
%C 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
%H Reinhard Zumkeller, <a href="/A248098/b248098.txt">Table of n, a(n) for n = 1..1000</a>
%H D. K. Chang, <a href="http://www.fq.math.ca/Scanned/24-3/chang.pdf">On Fibonacci k-ary trees</a>, The Fibonacci Quarterly, Volume 24, Number 3, August 1986, 258-262.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,0,-1).
%F G.f. = z*(1-z-z^2+2*z^3)/((1-z)*(1-z-z^2-z^3)).
%p 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);
%p 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);
%o (Haskell)
%o a248098 n = a248098_list !! (n-1)
%o a248098_list = 1 : 1 : 1 : map (+ 1) (zipWith3 (((+) .) . (+))
%o a248098_list (tail a248098_list) (drop 2 a248098_list))
%o -- _Reinhard Zumkeller_, Dec 29 2014
%Y Cf. A000213.
%Y Cf. A213967.
%K nonn,easy
%O 1,4
%A _Emeric Deutsch_, Dec 28 2014
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