%I
%S 2,6,1,22,2,78,3,1,272,6,2,940,13,2,1,3232,28,2,2,11080,58,3,2,1,
%T 37920,118,6,2,2,129648,239,12,2,2,1,443008,484,22,2,2,2,1513248,979,
%U 37,3,2,2,1,5168000,1976,60,6,2,2,2,17647552,3980,97,12,2,2,2,1,60258304,8004
%N Triangle read by rows: T(n,k) is the number of 2compositions of n having least entry equal to k (n >= 1; 0 <= k <= floor(n/2)). A 2composition 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 Row n contains 1 + floor(n/2) entries.
%C The sum of entries in row n is A003480(n).
%C T(n,1) = A181367(n).
%C Sum_{k >= 0} k*T(n,k) = A181366.
%H Alois P. Heinz, <a href="/A181365/b181365.txt">Rows n = 1..200, 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 Lconvex polyominoes</a>, European J. Combin. 28 (2007), no. 6, 17241741.
%F G.f. for 2compositions with all entries >= k is h(k,z)=(1z)^2/(12z+z^2z^{2k}) if k>0 and h(0,z)=(1z)^2/(14z+2z^2) if k=0.
%F G.f. for 2compositions with least entry k is f(k,z)=h(k,z)h(k+1,z) (these are the column g.f.'s).
%F G.f.: G(t,z) = f(0,z) + Sum_{k>=1} f(k,z)*t^k.
%e T(4,1) = 3 because we have (1/3), (3/1), and (1,1/1,1) (the 2compositions are written as (top row / bottom row).
%e Triangle starts:
%e 2;
%e 6, 1;
%e 22, 2;
%e 78, 3, 1;
%e 272, 6, 2;
%e 940, 13, 2, 1;
%p h := proc (k) if k = 0 then (1z)^2/(14*z+2*z^2) else (1z)^2/(12*z+z^2z^(2*k)) end if end proc: f := proc (k) options operator, arrow: h(k)h(k+1) end proc; G := f(0)+sum(f(k)*t^k, k = 1 .. 30): Gser := simplify(series(G, z = 0, 20)): for n to 15 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 15 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
%p # second Maple program:
%p A:= proc(n, k) option remember; `if`(n=0, 1, add(add(
%p `if`(i=0 and j=0, 0, A(nij, k)), i=k..nj), j=k..n))
%p end:
%p T:= (n, k)> A(n, k) A(n, k+1):
%p seq(seq(T(n, k), k=0..n/2), n=1..15); # _Alois P. Heinz_, Mar 16 2014
%t A[n_, k_] := A[n, k] = If[n == 0, 1, Sum[Sum[If[i == 0 && j == 0, 0, A[nij, k]], {i, k, nj}], {j, k, n}]]; T[n_, k_] := A[n, k]  A[n, k+1]; Table[Table[T[n, k], {k, 0, n/2}], {n, 1, 15}] // Flatten (* _JeanFrançois Alcover_, May 28 2015, after _Alois P. Heinz_ *)
%Y Cf. A003480, A181366, A181367.
%K nonn,tabf
%O 1,1
%A _Emeric Deutsch_, Oct 15 2010
