

A103770


A weighted tribonacci sequence, (1,3,9).


1



1, 1, 4, 16, 37, 121, 376, 1072, 3289, 9889, 29404, 88672, 265885, 796537, 2392240, 7174816, 21520369, 64574977, 193709428, 581117680, 1743420757, 5230158649, 15690480040, 47071742800, 141214610761, 423644159521, 1270933677004
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

The weighted tribonacci (1,r,r^2) with g.f. 1/(1  x  r*x^2  r^2*x^3) has general term Sum_{k=0..n} T(nk,k)r^k.
Correspondence: a(n) = b(n+2)*3^n, where b(n) is the sequence of the arithmetic means of the previous three terms defined by b(n) = (1/3)*(b(n1) + b(n2) + b(n3)) with initial values b(0)=0, b(1)=0, b(2)=1; the g.f. for b(n) is B(x) := x^2/(1(x^1+x^2+x^3)/3), so the g.f. A(x) for a(n) satisfies A(x) = B(3*x)/(3*x)^2. Because b(n) converges to the limit lim_{x>1} (1x)*B(x) = (1/6)*(b(0) + 2*b(1) + 3*b(2)) = 1/2, it follows that a(n)/3^n also converges to 1/2. This correspondence is valid in general (with necessary changes) for weighted sequences of order (1, p, p^2, p^3, p^4, ..., p^(p1)) with integer p > 0. Forming such sequences c(n) := c(n1) + p^1*c(n2) + ... + p^(p1)*c(np) the limit of c(n)/p^n is 2/(p+1) (see also A001045).  Hieronymus Fischer, Feb 04 2006
a(n)/3^n equals the probability that n will occur as a partial sum in a randomlygenerated infinite sequence of 1s, 2s and 3s. The limiting ratio is 1/2.  Bob Selcoe, Jul 05 2013
Number of compositions of n into one sort of 1's, three sorts of 2's, and nine sorts of 3's.  Joerg Arndt, Jul 06 2013
Using the Markov Chain {{0, 1, 0}, {0, 0, 1}, {1/3, 1/3, 1/3}} and raising it to the nth power can generate this sequence when looking at the element in the third row and third column and reading the numerator.  Robert P. P. McKone, May 25 2021


LINKS

Table of n, a(n) for n=0..26.
Index entries for linear recurrences with constant coefficients, signature (1,3,9).


FORMULA

G.f.: 1/(1  x  3*x^2  9*x^3).
a(n) = Sum_{k=0..n} T(nk, k)*3^k, T(n, k) = trinomial coefficients (A027907).
a(n) = Sum_{k=0..n} 3^(nk) * Sum_{i=0..floor((nk)/2)} C(nki, i)*C(k, nki)).  Paul Barry, Apr 26 2005
a(n)/3^n converges to 1/2.  Hieronymus Fischer, Feb 02 2006
a(n) = a(n1) + 3*a(n2) + 9*a(n3), n >= 3; a(0)=1, a(1)=1, a(2)=4.  Hieronymus Fischer, Feb 04 2006
a(n) = 3^n + b(n) + b(n1), with b(n) = (1)^A121262(n+1)*A088137(n+1).  Ralf Stephan, May 20 2007


MATHEMATICA

LinearRecurrence[{1, 3, 9}, {1, 1, 4}, {1, 27}] (* Robert P. P. McKone, May 25 2021 *)


CROSSREFS

Cf. A000073, A102001.
Cf. A071675, A027907.
Cf. A121262, A088137.
Sequence in context: A080855 A203299 A198015 * A121318 A152133 A297361
Adjacent sequences: A103767 A103768 A103769 * A103771 A103772 A103773


KEYWORD

easy,nonn


AUTHOR

Paul Barry, Feb 15 2005


STATUS

approved



