A118720 -id:A118720 - cp's OEIS frontend

A132883 Triangle read by rows: T(n,k) is the number of paths in the first quadrant 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, 1, 3, 3, 5, 9, 2, 8, 22, 10, 13, 51, 40, 5, 21, 111, 130, 35, 34, 233, 380, 175, 14, 55, 474, 1022, 700, 126, 89, 942, 2590, 2450, 756, 42, 144, 1836, 6260, 7770, 3570, 462, 233, 3522, 14570, 22890, 14490, 3234, 132, 377, 6666, 32870, 63600, 52668

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)/(n+1) = A000108(n) (the Catalan numbers). Row sums yield A118720. Column k has g.f. = c(k)z^(2k)/(1-z-z^2)^(2k+1), where c(k) = binomial(2k,k)/(k+1) are the Catalan numbers; accordingly, T(n,1) = A001628(n-2), T(n,2) = 2*A001873(n-4), T(n,3) = 5*A001875(n-6). Sum_{k>=0} k*T(n,k) = A106050(n+1).

			Triangle starts:
   1;
   1;
   2,  1;
   3,  3;
   5,  9,  2;
   8, 22, 10;
  13, 51, 40,  5;
T(3,1)=3 because we have hUD, UhD and UDh.

Cf. A000045, A118720, A106050, A000108.

G:=((1-z-z^2-sqrt(1-2*z-z^2+2*z^3+z^4-4*t*z^2))*1/2)/(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 = G(t,z) satisfies G = 1 + zG + z^2*G + tz^2*G^2 (see explicit expression at the Maple program).

A132279 Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k doublerises (i.e., UU's) (0 <= k <= floor(n/2) - 1 for n >= 2).

Original entry on oeis.org

1, 1, 3, 6, 15, 1, 36, 4, 91, 17, 1, 232, 60, 5, 603, 206, 26, 1, 1585, 676, 110, 6, 4213, 2174, 444, 37, 1, 11298, 6868, 1687, 182, 7, 30537, 21446, 6196, 841, 50, 1, 83097, 66356, 22100, 3612, 280, 8, 227475, 203914, 77138, 14833, 1455, 65, 1

Offset: 0

Emeric Deutsch, Sep 03 2007

Row n contains floor(n/2) terms (n>=2). Row sums yield A118720. T(n,0) = A005043(n+2) (the Riordan numbers).

			Triangle starts:
    1;
    1;
    3;
    6;
   15,  1;
   36,  4;
   91, 17,  1;
  232, 60,  5;
T(5,1)=4 because we have UUhDD, UUDhD, hUUDD and UUDDh.

Cf. A005043, A118720.

G:=((1-z-2*z^2+z^2*t-sqrt((1+z-z^2*t)*(1-3*z-z^2*t)))*1/2)/(z^2*(t+z+z^2-z*t-z^2*t)): Gser:=simplify(series(G,z=0,18)): for n from 0 to 15 do P[n]:=sort(coeff(Gser,z,n)) end do: 1; 1; for n from 2 to 14 do seq(coeff(P[n],t,j),j= 0..floor((1/2)*n)-1) end do; # yields sequence in triangular form

G.f.: G = G(t,z) satisfies G = 1 + zG + z^2*G + z^2*(t(G-1-zG-z^2*G) + 1 + zG + z^2*G)G (see explicit expression at the Maple program).

G.f.: G = 2/(1-z-2*z^2+t*z^2+sqrt(1-2*z-3*z^2-2*t*z^2+2*t*z^3+t^2*z^4)). - Olivier Gérard, Sep 27 2007

cp's OEIS Frontend

A132883 Triangle read by rows: T(n,k) is the number of paths in the first quadrant 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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