

A181299


Triangle read by rows: T(n,k) is the number of 2compositions of n having k columns in which the top entry is equal to the bottom entry (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.


2



1, 2, 6, 1, 20, 4, 64, 17, 1, 206, 68, 6, 662, 261, 32, 1, 2128, 976, 152, 8, 6840, 3577, 675, 51, 1, 21986, 12912, 2860, 280, 10, 70670, 46049, 11704, 1406, 74, 1, 227156, 162628, 46632, 6632, 460, 12, 730152, 569705, 181877, 29866, 2570, 101, 1, 2346942
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OFFSET

0,2


COMMENTS

Row n contains 1+floor(n/2) entries.
The sum of entries in row n is A003480(n).
T(n,0)=A181301(n).
Sum(k*T(n,k),k>=0)=A181300.


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..49.


FORMULA

G.f.=G(t,z)=(1+z)(1z)^2/[(13zz^2+z^3t(1z)z^2].


EXAMPLE

T(3,1)=4 because we have (1,1/1,0),(1,0/1,1),(1,1/0,1),(0,1/1,1) (the 2compositions are written as (top row/bottom row).
Triangle starts:
1;
2;
6,1;
20,4;
64,17,1;


MAPLE

G := (1z)^2*(1+z)/(13*zz^2+z^3t*z^2*(1z)): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 13 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form


CROSSREFS

Cf. A003480, A181300, A181301.
Sequence in context: A175353 A181307 A008855 * A181365 A221913 A280370
Adjacent sequences: A181296 A181297 A181298 * A181300 A181301 A181302


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Oct 12 2010


STATUS

approved



