A132885 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 H=(2,0) steps (0 <= k <= floor(n/2)).

1, 1, 3, 1, 7, 2, 19, 9, 1, 51, 28, 3, 141, 95, 18, 1, 393, 306, 70, 4, 1107, 987, 285, 30, 1, 3139, 3144, 1071, 140, 5, 8953, 9963, 3948, 665, 45, 1, 25653, 31390, 14148, 2856, 245, 6, 73789, 98483, 49815, 11844, 1330, 63, 1, 212941, 307836, 172645, 47160

Offset: 0

Author

Emeric Deutsch, Sep 03 2007

Keywords

• nonn
• tabf

Comments

Row n has 1+floor(n/2) terms. T(n,0)=A002426(n) (the central trinomial coefficients). T(n,1)=A109188(n-1). Row sums yield A059345. See A132280 for the same statistic on paths restricted to the first quadrant.

Examples

			T(4,1)=9 because we have hhH, hHh, Hhh, HUD, UDH, UHD, HDU, DUH and DHU.
Triangle starts:
                     1;
                     1;
                 3,      1;
                 7,      2;
            19,      9,      1;
            51,     28,      3;
       141,     95,     18,      1;
       393,    306,     70,      4;
  1107,    987,    285,     30,      1;
  3139,   3144,   1071,    140,      5;
		

Links

• G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Crossrefs

Columns k=0..3 give A002426, A109188(n-1), A373651(n-4), A375260(n-6).

Row sums gives A059345.

Cf. A132280.

Programs

Maple

G:=1/sqrt((1+z-t*z^2)*(1-3*z-t*z^2)): Gser:=simplify(series(G,z=0,18)): 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
A132885 := (n,k) -> binomial(n-k,k)*hypergeom([k-n/2,k-n/2+1/2], [1], 4): seq(print(seq(round(evalf(A132885(n,k))),k=0..iquo(n,2))),n=0..9); # Peter Luschny, Sep 18 2014

Mathematica

T[n_, k_] := Binomial[n - k, k]*Hypergeometric2F1[k - n/2, k - n/2 + 1/2, 1, 4]; Table[T[n, k], {n,0,10}, {k, 0, Floor[n/2]}] // Flatten  (* G. C. Greubel, Mar 01 2017 *)

Formula

G.f.: 1/sqrt((1+z-t*z^2)*(1-3*z-t*z^2)).

T(n,k) = C(n-k,k)*hypergeom([k-n/2,k-n/2+1/2], [1], 4). - Peter Luschny, Sep 18 2014