%I #2 Mar 30 2012 17:36:24
%S 1,0,2,1,0,6,0,8,0,16,3,0,35,0,44,0,28,0,132,0,120,8,0,160,0,460,0,
%T 328,0,92,0,748,0,1528,0,896,21,0,642,0,3117,0,4916,0,2448,0,290,0,
%U 3552,0,12062,0,15456,0,6688,55,0,2380,0,17119,0,44318,0,47760,0,18272,0,888,0
%N Triangle read by rows: T(n,k) is the number of 2-compositions of n having k even entries (0<=k<=n) A 2-composition of n is a nonnegative matrix with two rows, such that each column has at least one nonzero entry and whose entries sum up to n.
%C The sum of entries in row n is A003480(n).
%C T(2n-1,0)=0.
%C T(2n,0)=A000045(2n) (Fibonacci numbers).
%C T(n,k)=0 if n and k have opposite parities.
%C T(n,n)=A002605(n+1).
%C Sum(k*T(n,k),k=0..n)=A181298.
%C For the statistics "number of odd entries" see A181295.
%D G. Castiglione, A. Frosini, E. Munarini, A. Restivo and S. Rinaldi, Combinatorial aspects of L-convex polyominoes, European Journal of Combinatorics, 28, 2007, 1724-1741.
%F G.f.=G(t,z)=(1-z^2)^2/(1-3z^2+z^4-2sz-2s^2*z^2+s^2*z^4).
%F The g.f. H(t,s,z), where z marks the size of the 2-composition and t (s) marks the number of odd (even) entries, is H=1/(1-h), where h=z(t+sz)(2s+tz-sz^2)/(1-z^2)^2.
%e T(2,2)=6 because we have (0 / 2), (2 / 0), (1,0 / 0,1), (0,1 / 1,0), (1,1 / 0,0), (0,0 / 1,1) (the 2-compositions are written as (top row / bottom row).
%e Triangle starts:
%e 1;
%e 0,2;
%e 1,0,6;
%e 0,8,0,16;
%e 3,0,35,0,44;
%p G := (1-z^2)^2/(1-3*z^2+z^4-2*s*z-2*s^2*z^2+s^2*z^4): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 11 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 11 do seq(coeff(P[n], s, k), k = 0 .. n) end do; # yields sequence in triangular form
%Y Cf. A003480, A000045, A181295, A181296, A181298.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Oct 12 2010