

A181297


Triangle read by rows: T(n,k) is the number of 2compositions of n having k even entries (0<=k<=n) 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.


3



1, 0, 2, 1, 0, 6, 0, 8, 0, 16, 3, 0, 35, 0, 44, 0, 28, 0, 132, 0, 120, 8, 0, 160, 0, 460, 0, 328, 0, 92, 0, 748, 0, 1528, 0, 896, 21, 0, 642, 0, 3117, 0, 4916, 0, 2448, 0, 290, 0, 3552, 0, 12062, 0, 15456, 0, 6688, 55, 0, 2380, 0, 17119, 0, 44318, 0, 47760, 0, 18272, 0, 888, 0
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OFFSET

0,3


COMMENTS

The sum of entries in row n is A003480(n).
T(2n1,0)=0.
T(2n,0)=A000045(2n) (Fibonacci numbers).
T(n,k)=0 if n and k have opposite parities.
T(n,n)=A002605(n+1).
Sum(k*T(n,k),k=0..n)=A181298.
For the statistics "number of odd entries" see A181295.


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..68.


FORMULA

G.f.=G(t,z)=(1z^2)^2/(13z^2+z^42sz2s^2*z^2+s^2*z^4).
The g.f. H(t,s,z), where z marks the size of the 2composition and t (s) marks the number of odd (even) entries, is H=1/(1h), where h=z(t+sz)(2s+tzsz^2)/(1z^2)^2.


EXAMPLE

T(2,2)=6 because we have (0 / 2), (2 / 0), (1,0 / 0,1), (0,1 / 1,0), (1,1 / 0,0), (0,0 / 1,1) (the 2compositions are written as (top row / bottom row).
Triangle starts:
1;
0,2;
1,0,6;
0,8,0,16;
3,0,35,0,44;


MAPLE

G := (1z^2)^2/(13*z^2+z^42*s*z2*s^2*z^2+s^2*z^4): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 11 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 11 do seq(coeff(P[n], s, k), k = 0 .. n) end do; # yields sequence in triangular form


CROSSREFS

Cf. A003480, A000045, A181295, A181296, A181298.
Sequence in context: A266904 A299198 A137477 * A196776 A157982 A119275
Adjacent sequences: A181294 A181295 A181296 * A181298 A181299 A181300


KEYWORD

nonn,tabl


AUTHOR

Emeric Deutsch, Oct 12 2010


STATUS

approved



