%I #13 Nov 29 2017 23:02:36
%S 1,1,1,3,5,2,12,28,21,5,55,165,180,84,14,273,1001,1430,990,330,42,
%T 1428,6188,10920,10010,5005,1287,132,7752,38760,81396,92820,61880,
%U 24024,5005,429,43263,245157,596904,813960,678300,352716,111384,19448,1430,246675
%N Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have k peaks of the form Ud.
%C Row sums yield A027307.
%C T(n,n) = A000108(n) (the Catalan numbers).
%C T(n,0) = A001764(n) = binomial(3n,n)/(2n+1).
%C Number of Ud peaks in all paths from (0,0) to (3n,0) is given by A108427.
%H G. C. Greubel, <a href="/A108426/b108426.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H Emeric Deutsch, <a href="http://www.jstor.org/stable/2589192">Problem 10658: Another Type of Lattice Path</a>, American Math. Monthly, 107, 2000, 368-370.
%F T(n,k) = (1/n)*binomial(n,k)*binomial(3*n-k,n-1).
%F G.f.: G = G(t,z) satisfies G=1+z(t+G)G^2.
%e Example T(2,1) = 5 because we have udUdd, uUddd, Uddud, Ududd and UUdddd.
%e Triangle begins:
%e 1;
%e 1,1;
%e 3,5,2;
%e 12,28,21,5;
%e ...
%p T:=(n,k)->binomial(n,k)*binomial(3*n-k,n-1)/n: print(1); for n from 1 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%t Table[If[n == 0, 1, (1/n)*Binomial[n, k]*Binomial[3 n - k, n - 1]], {n, 0, 10}, {k, 0, n}] // Flatten (* _G. C. Greubel_, Nov 29 2017 *)
%o (PARI) for(n=0,10, for(k=0,n, print1(if(n==0, 1, (1/n)*binomial(n,k) *binomial(3*n-k,n-1)), ", "))) \\ _G. C. Greubel_, Nov 29 2017
%Y Cf. A027307, A000108, A001764, A108425, A108427.
%K nonn,tabl
%O 0,4
%A _Emeric Deutsch_, Jun 03 2005