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%I #11 Nov 26 2013 11:26:04
%S 1,1,1,3,3,1,8,10,5,1,21,32,21,7,1,55,99,80,36,9,1,144,299,286,160,55,
%T 11,1,377,887,978,650,280,78,13,1,987,2595,3236,2482,1275,448,105,15,
%U 1,2584,7508,10438,9054,5377,2261,672,136,17,1,6765,21526,32991,31882
%N Triangle read by rows: T(n,k) is the number of 2-compositions of n having k 0's in the top row 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,0) = A000045(2n) (n>=1), Fibonacci numbers.
%C T(n,1) = A038731(n-1) (n>=1).
%C Sum(k*T(n,k), k>=0) = A181331.
%C For the statistic "number of nonzero entries in the top row" see A181332.
%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,x) = (1-x)^2/(1-3*x+x^2-t*x(1-x)).
%F The g.f. of column k is x^k*(1-x)^(k+2)/(1-3*x+x^2)^(k+1) (we have a Riordan array).
%F T(n,k) = 3*T(n-1,k) +T(n-1,k-1) -T(n-2,k) -T(n-2,k-1), with T(0,0)=T(1,0)=T(1,1)=T(2,2)=1, T(2,0)=T(2,1)=3, T(n,k)=0 if k<0 or if k>n. - _Philippe Deléham_, Nov 26 2013
%e T(2,1)=3 because we have (0/2), (1,0/0,1), and (0,1/1,0) (the 2-compositions are written as (top row / bottom row)).
%e Triangle starts:
%e 1;
%e 1,1;
%e 3,3,1;
%e 8,10,5,1;
%e 21,32,21,7,1;
%e 55,99,80,36,9,1;
%p G := (1-z)^2/(1-3*z+z^2-t*z*(1-z)): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 10 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 10 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form
%Y Cf. A003480, A000045, A038731, A181331, A181332.
%K nonn,tabl
%O 0,4
%A _Emeric Deutsch_, Oct 13 2010
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