|
%I #7 Dec 30 2016 02:30:25
%S 1,0,3,1,0,9,0,9,0,27,2,0,54,0,81,0,30,0,270,0,243,5,0,270,0,1215,0,
%T 729,0,105,0,1890,0,5103,0,2187,14,0,1260,0,11340,0,20412,0,6561,0,
%U 378,0,11340,0,61236,0,78732,0,19683,42,0,5670,0,85050,0,306180,0,295245,0
%N Triangle read by rows: T(n,k) is number of hex trees with n edges and k vertices of outdegree 1 (0<=k<=n).
%C A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper).
%C Sum of terms in row n = A002212(n+1).
%C Column 0 yields the aerated Catalan numbers (1,0,1,0,2,0,5,0,14,...).
%C T(n,n) = 3^n (see A000244).
%C Sum_{k=0..n} k*T(n,k) = 3*A026376(n) (n>=1).
%H F. Harary and R. C. Read, <a href="https://doi.org/10.1017/S0013091500009135">The enumeration of tree-like polyhexes</a>, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.
%F T(n,k) = [3^k/(n+1)]binomial(n+1,k)*binomial(n+1-k,(n-k)/2) (0<=k<=n).
%F G.f.: G=G(t,z) satisfies G=1+3tzG+z^2*G^2.
%e Triangle starts:
%e 1;
%e 0, 3;
%e 1, 0, 9;
%e 0, 9, 0, 27;
%e 2, 0, 54, 0, 81;
%p T:=proc(n,k) if n-k mod 2 = 0 then 3^k*binomial(n+1,k)*binomial(n+1-k,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 11 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%Y Cf. A000244, A002212, A026376.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Dec 19 2006
|
