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%I #13 Feb 19 2015 05:11:48
%S 1,0,2,3,0,4,0,16,0,8,14,0,52,0,16,0,104,0,144,0,32,64,0,460,0,368,0,
%T 64,0,616,0,1624,0,896,0,128,292,0,3428,0,5056,0,2112,0,256,0,3456,0,
%U 14688,0,14528,0,4864,0,512,1332,0,23132,0,53920,0,39488,0,11008,0,1024,0
%N Triangle read by rows: T(n,k) is the number of 2-compositions of n having k columns with an odd sum (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(n,k) = 0 if n and k have opposite parities.
%C T(2n,0) = A060801(n).
%C Sum(k*T(n,k), k=0..n) = A181326(n).
%C For the statistic "number of column with an even sum" see A181327.
%H Alois P. Heinz, <a href="/A181308/b181308.txt">Table of n, a(n) for n = 0..140, flattened</a>
%H G. Castiglione, A. Frosini, E. Munarini, A. Restivo and S. Rinaldi, <a href="http://dx.doi.org/10.1016/j.ejc.2006.06.020">Combinatorial aspects of L-convex polyominoes</a>, European J. Combin. 28 (2007), no. 6, 1724-1741.
%F G.f.: G(t,z) = (1-z)^2*(1+z)^2/(1-5z^2+2z^4-2tz).
%F The g.f. of column k is (2z)^k*(1-z^2)^2/(1-5z^2+2z^4)^{k+1} (we have a Riordan array).
%F The g.f. H(t,s,z), where z marks size and t (s) marks number of columns with an odd (even) sum, is H=(1-z^2)^2/(1-2z^2+z^4-2tz-3sz^2+sz^4).
%e T(2,2) = 4 because we have (1,0/0,1), (0,1/1,0), (1,1/0,0), and (0,0/1,1) (the 2-compositions are written as (top row/bottom row)).
%e Triangle starts:
%e 1;
%e 0, 2;
%e 3, 0, 4;
%e 0, 16, 0, 8;
%e 14, 0, 52, 0, 16;
%p G := (1-z^2)^2/(1-5*z^2+2*z^4-2*t*z): 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], t, k), k = 0 .. n) end do; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n) option remember; `if`(n=0, 1,
%p expand(add(add(`if`(i=0 and j=0, 0, b(n-i-j)*
%p `if`(irem(i+j,2)=1, x, 1)), i=0..n-j), j=0..n)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
%p seq(T(n), n=0..15); # _Alois P. Heinz_, Mar 16 2014
%t b[n_] := b[n] = If[n == 0, 1, Expand[Sum[Sum[If[i == 0 && j == 0, 0, b[n-i-j]* If[Mod[i+j, 2] == 1, x, 1]], {i, 0, n-j}], {j, 0, n}]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n]]; Table[T[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, Feb 19 2015, after _Alois P. Heinz_ *)
%Y Cf. A003480, A060801, A181326, A181327.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Oct 13 2010
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