A132883 -id:A132883 - cp's OEIS frontend

A132886 Triangle read by rows: T(n,k) is the number of paths in the right half-plane, from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k U steps (0 <= k <= floor(n/2)).

1, 1, 2, 2, 3, 6, 5, 18, 6, 8, 44, 30, 13, 102, 120, 20, 21, 222, 390, 140, 34, 466, 1140, 700, 70, 55, 948, 3066, 2800, 630, 89, 1884, 7770, 9800, 3780, 252, 144, 3672, 18780, 31080, 17850, 2772, 233, 7044, 43710, 91560, 72450, 19404, 924, 377, 13332, 98610

Offset: 0

Emeric Deutsch, Sep 03 2007

Row n has 1+floor(n/2) terms. T(n,0) = A000045(n+1) (the Fibonacci numbers). T(2n,n) = binomial(2n,n) = A000984(n) (the central binomial coefficients). Row sums yield A059345. Column k has g.f. = binomial(2k,k)*z^(2k)/(1-z-z^2)^(2k+1); accordingly, T(n,1) = 2*A001628(n-2), T(n,2) = 6*A001873(n-4), T(n,3) = 20*A001875(n-6). See A132883 for the same statistic on paths restricted to the first quadrant.

			Triangle starts:
   1;
   1;
   2,   2;
   3,   6;
   5,  18,   6;
   8,  44,  30;
  13, 102, 120,  20;
T(3,1)=6 because we have hUD, UhD, UDh, hDU, DhU and DUh.

Cf. A000045, A059345, A001628, A001873, A001875, A132883.

G:=1/sqrt((1-z-z^2)^2-4*t*z^2): Gser:=simplify(series(G,z=0,17)): 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,j),j=0..floor((1/2)*n)) end do; # yields sequence in triangular form

G.f.: G(t,z) = 1/sqrt((1-z-z^2)^2 - 4tz^2).

cp's OEIS Frontend

A132886 Triangle read by rows: T(n,k) is the number of paths in the right half-plane, from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k U steps (0 <= k <= floor(n/2)).

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