A050155 Triangle T(n,k), k>=0 and n>=1, read by rows defined by: T(n,k) = (2k+3)*binomial(2n,n-k-1)/(n+k+2).
1, 3, 1, 9, 5, 1, 28, 20, 7, 1, 90, 75, 35, 9, 1, 297, 275, 154, 54, 11, 1, 1001, 1001, 637, 273, 77, 13, 1, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1, 41990, 48450, 38760, 23256, 10659, 3705, 950, 170, 19, 1
Offset: 1
Examples
1; 3, 1; 9, 5, 1; 28, 20, 7, 1; 90, 75, 35, 9, 1; 297, 275, 154, 54, 11, 1; ...
Links
- Alois P. Heinz, Rows n = 1..141, flattened
- R. K. Guy, Catwalks, Sansteps and Pascal Pyramids, J. Integer Seq., Vol. 3 (2000), #00.1.6
- V. E. Hoggatt, Jr. and M. Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., 14 (1976), 395-405.
- A. Papoulis, A new method of inversion of the Laplace transform, Quart. Appl. Math 14 (1957), 405-414.
- J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222.
- _Zoran Sunic_, Self-Describing Sequences and the Catalan Family Tree, Electronic Journal of Combinatorics, 10 (2003) #N5.
Crossrefs
Programs
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Maple
T:= (n, k)-> (2*k+3)*binomial(2*n, n-k-1)/(n+k+2): seq(seq(T(n, k), k=0..n-1), n=1..10); # Alois P. Heinz, Jan 19 2013
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Mathematica
T[n_, k_] := (2*k + 3)*Binomial[2*n, n - k - 1]/(n + k + 2); Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, May 21 2016 *)
Formula
Sum_{ k = 0, .., n-1} T(n, k) = binomial(2n, n-1) = A001791(n).
G.f. of column k: x^(k+1)*C^(2*k+3) where C = (1-(1-4*x)^(1/2))/(2*x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Feb 03 2004
T(n, k) = A039599(n, k+1) = A009766(n+k+1, n-k-1) = A033184(n+k+2, 2k+3) . - Philippe Deléham, May 28 2005
T(n, k)=(2k+3)binomial(2n+2, n+k+2)/(n+k+3)=C(2n+2, n+k+2)-C(2n+2, n+k+3) [offset (0, 0)]. - Paul Barry, Jul 06 2005
Extensions
Edited by Philippe Deléham, May 22 2005
Comments