1,8

Sum of terms in row n = binom(n, floor(n/2))=A001405(n). T(n,2)=A052551(n-2) (n>=2). T(n,3)=A005672(n)=Fibonacci(n+1)-2^floor(n/2). Sum(k*T(n,k),k=1..n)=A132891(n).

The g.f. g[k](z) for column k is given by g[2k-1]=z^(2k-1)/(L[k]M[k]), g[2k]=z^(2k)/(L[k+1]M[k]), where L[1]=1, L[2]=1-2z^2, L[n]=L[n-1]-z^2*L[n-2] for n>=3 (Lucas type polynomials) and M[0]:=1, M[1]=1-z, M[n]=M[n-1]-z^2*M[n-2] for n>=2.

The g.f. for column k is g[k](z)=G[k](z)-G[k-1](z), where G[0]=1, G[n]=H[n]*(1+z*G[n-1]), where H[0]=1, H[n]:=1/(1-z^2*H[n-1]) (H[k](z) is the g.f. of the Dyck paths of height at most k and G[k](z) is the g.f. of the left factors of Dyck paths of height at most k).

T(5,3)=4 because we have UDUUU, UUDUU, UUUDD and UUUDU, where U=(1,1) and D=(1,-1).

Triangle starts:

1;

1,1;

1,1,1;

1,3,1,1;

1,3,4,1;1;

1,7,5,5,1,1;

H[0]:=1: for n to 15 do H[n]:=simplify(1/(1-z^2*H[n-1])) end do: G[0]:=1: for n to 15 do G[n]:=simplify(H[n]+z*H[n]*G[n-1]) end do: g[0]:=G[0]: for n to 15 do g[n]:=simplify(G[n]-G[n-1]) end do: for n from 0 to 15 do gser[n]:=series(g[n], z=0, 25) end do: for n to 15 do seq(coeff(gser[k], z, n), k=1..n) end do; # yields sequence in triangular form

L[1]:=1: L[2]:=1-2*z^2: for n from 3 to 15 do L[n]:=expand(L[n-1]-z^2*L[n-2]) end do: M[0]:=1: M[1]:=1-z: for n from 2 to 15 do M[n]:=expand(M[n-1]-z^2*M[n-2]) end do: for k to 7 do g[2*k-1]:=z^(2*k-1)/(L[k]*M[k]): g[2*k]:=z^(2*k)/(L[k+1]*M[k]) end do: for n to 14 do gser[n]:=series(g[n], z=0, 25) end do: for n to 14 do seq(coeff(gser[k], z, n), k=1..n) end do; # yields sequence in triangular form

Cf. A001405, A052551, A005672, A132891.

Sequence in context: A050328 A030401 A166030 this_sequence A069290 A076476 A016733

Adjacent sequences: A132887 A132888 A132889 this_sequence A132891 A132892 A132893

nonn

Emeric Deutsch (deutsch(AT)duke.poly.edu), Sep 08 2007