A181332 - OEIS

%I #11 Nov 26 2013 11:24:10

%S 1,1,1,2,4,1,4,12,7,1,8,32,31,10,1,16,80,111,59,13,1,32,192,351,268,

%T 96,16,1,64,448,1023,1037,530,142,19,1,128,1024,2815,3598,2435,924,

%U 197,22,1,256,2304,7423,11535,9843,4923,1477,261,25,1,512,5120,18943,34832

%N Triangle read by rows: T(n,k) is the number of 2-compositions of n having k nonzero entries 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,1) = A001787(n).

%C T(n,2) = A055580(n-2) (n>=2).

%C T(n,3) = A055586(n-3) (n>=3).

%C Sum(k*T(n,k), k>=0) = A054146(n).

%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 T(n,k) = sum(2^j*binomial(k+j,k)*binomial(n-2-j,k-2), j=0..n-k).

%F G.f.: G(t,x) = (1-x)^2/(1-3*x+2*x^2-t*x).

%F The g.f. of column k is x^k/((1-2*x)^(k+1)*(1-x)^(k-1)) (we have a Riordan array).

%F T(n,k) = 3*T(n-1,k)+T(n-1,k-1)-2*T(n-2,k), with T(0,0)=T(1,0)=T(1,1)=T(2,2)=1, T(2,0)=2, T(2,1)=4, T(n,k)=0 if k<0 or if k>n. - _Philippe Deléham, Nov 26 2013

%e T(2,1)=4 because we have (1/1), (2/0), (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 2,4,1;

%e 4,12,7,1;

%e 8,32,31,10,1;

%e 16,80,111,59,13,1;

%p T := proc (n, k) options operator, arrow: sum(2^j*binomial(k+j, k)*binomial(n-j-2, k-2), j = 0 .. n-k) end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form

%Y Cf. A003480, A001787, A055580, A055586, A181330, A181332.

%K nonn,tabl

%O 0,4

%A _Emeric Deutsch_, Oct 13 2010