Search: id:A103770
Results 1-1 of 1 results found.

%I A103770
%S A103770 1,1,4,16,37,121,376,1072,3289,9889,29404,88672,265885,796537,2392240,
%T A103770 7174816,21520369,64574977,193709428,581117680,1743420757,5230158649,
%U A103770 15690480040,47071742800,141214610761,423644159521,1270933677004
%N A103770 A weighted tribonacci sequence, (1,3,9).
%C A103770 The weighted tribonacci (1,r,r^2) with g.f. 1/(1-x-r*x^2-r^2x^3) has 
               general term sum{k=0..n, T(n-k,k)r^k}.
%C A103770 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(n-1)+b(n-2)+b(n-3)) 
               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) suffices 
               A(x)=B(3*x)/(3*x)^2. Because b(n) converges to the limit lim (1-x)*B(x)=1/
               6*(b(0)+2*b(1)+3*b(2))=1/2 (for x-->1), 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^(p-1)) with natural p>0. Forming such sequences c(n):=c(n-1)+p^1*c(n-2)+...+p^(p-1)*c(n-p) 
               the limit of c(n)/p^n is 2/(p+1) (see also A001045). - Hieronymus 
               Fischer (Hieronymus.Fischer(AT)gmx.de), Feb 04 2006
%F A103770 G.f.: 1/(1-x-3x^2-9x^3); a(n)=sum{k=0..n, T(n-k, k)3^k}, T(n, k) = trinomial 
               coefficients (A027907).
%F A103770 a(n)=sum{k=0..n, sum{i=0..floor((n-k)/2), C(n-k-i, i)C(k, n-k-i)}*3^(n-k)}; 
               - Paul Barry (pbarry(AT)wit.ie), Apr 26 2005
%F A103770 a(n)/3^n converges to 1/2. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), 
               Feb 02 2006
%F A103770 a(0)=1, a(1)=1, a(2)=4 [For the matter of clearness only]; a(n)=a(n-1)+3*a(n-2)+9*a(n-3), 
               n>=3; - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Feb 04 
               2006
%F A103770 a(n) = 3^n + b(n) + b(n-1), with b(n) = (-1)^A121262(n+1)*A088137(n+1). 
               - Ralf Stephan, May 20 2007
%Y A103770 Cf. A000073, A102001.
%Y A103770 Cf. A071675.
%Y A103770 Sequence in context: A054246 A080709 A080855 this_sequence A121318 A152133 
               A110477
%Y A103770 Adjacent sequences: A103767 A103768 A103769 this_sequence A103771 A103772 
               A103773
%K A103770 easy,nonn
%O A103770 0,3
%A A103770 Paul Barry (pbarry(AT)wit.ie), Feb 15 2005

Search completed in 0.001 seconds