

A181336


Triangle read by rows: T(n,k) is the number of 2compositions of n having k even entries in the top row. 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.


2



1, 1, 1, 2, 4, 1, 5, 11, 7, 1, 11, 31, 29, 10, 1, 25, 83, 102, 56, 13, 1, 56, 217, 329, 245, 92, 16, 1, 126, 556, 1000, 938, 487, 137, 19, 1, 283, 1403, 2917, 3292, 2180, 855, 191, 22, 1, 636, 3498, 8247, 10865, 8740, 4406, 1376, 254, 25, 1, 1429, 8636, 22756, 34248
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OFFSET

0,4


COMMENTS

The sum of entries in row n is A003480(n).
T(n,0)=A006054(n+1) (n>=1).
Sum(k*T(n,k), k>=0)=A181337(n).
For the statistic "number of odd entries in the top row" see A181304.


REFERENCES

G. Castiglione, A. Frosini, E. Munarini, A. Restivo and S. Rinaldi, Combinatorial aspects of Lconvex polyominoes, European Journal of Combinatorics, 28, 2007, 17241741.


LINKS

Table of n, a(n) for n=0..58.


FORMULA

G.f.=G(s,z)=(1+z)(1z)^2/[12zz^2+z^3sz(1+zz^2)].
The g.f. of column k is z^k*(1+z)(1z)^2*(1+zz^2)^k/(12zz^2+z^3)^{k+1} (we have a Riordan array).
The g.f. H=H(t,s,z), where z marks size and t (s) marks odd (even) entries in the top row, is given by H = (1+z)(1z)^2/[(1+z)(1z)^2(t+s)zsz^2*(1z)].


EXAMPLE

T(2,1)=4 because we have (0/2), (2/0), (1,0/0,1), and (0,1/1,0) (the 2compositions are written as (top row / bottom row)).
Triangle starts:
1;
1,1;
2,4,1;
5,11,7,1;
11,31,29,10,1;
25,83,102,56,13,1;


MAPLE

G := (1+z)*(1z)^2/(12*zz^2+z^3s*z*(1+zz^2)): Gser := simplify(series(G, z = 0, 13)): 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], s, k), k = 0 .. n) end do; # yields sequence in triangular form


CROSSREFS

Cf. A003480, A006054, A181304, A181337,
Sequence in context: A080427 A118906 A085059 * A238731 A124037 A126126
Adjacent sequences: A181333 A181334 A181335 * A181337 A181338 A181339


KEYWORD

nonn,tabl


AUTHOR

Emeric Deutsch, Oct 14 2010


STATUS

approved



