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A102593 Triangle read by rows: T(n,k) is the number of noncrossing trees with n edges in which the maximum number of contiguous border edges starting from the root in counterclockwise direction is equal to k.

%I A102593 #15 Jul 06 2018 08:09:24
%S A102593 1,0,1,1,1,1,5,4,2,1,25,18,8,3,1,130,88,37,13,4,1,700,455,185,63,19,5,
%T A102593 1,3876,2448,973,325,97,26,6,1,21945,13566,5304,1748,518,140,34,7,1,
%U A102593 126500,76912,29697,9690,2856,775,193,43,8,1,740025,444015,169763,54967
%N A102593 Triangle read by rows: T(n,k) is the number of noncrossing trees with n edges in which the maximum number of contiguous border edges starting from the root in counterclockwise direction is equal to k.
%C A102593 Row n has n+1 terms. Row sums yield the ternary numbers (A001764). T(n,0)=A102893(n).
%H A102593 Andrew Howroyd, <a href="/A102593/b102593.txt">Table of n, a(n) for n = 0..1274</a>
%H A102593 P. Flajolet and M. Noy, <a href="http://dx.doi.org/10.1016/S0012-365X(98)00372-0">Analytic combinatorics of non-crossing configurations</a>, Discrete Math., 204, 203-229, 1999.
%H A102593 M. Noy, <a href="http://dx.doi.org/10.1016/S0012-365X(97)00121-0">Enumeration of noncrossing trees on a circle</a>, Discrete Math., 180, 301-313, 1998.
%F A102593 T(n, k) = (k+1)binomial(3n-2k, n-k)/(2n-k+1) - (k+2)binomial(3n-2k-2, n-k-1)/(2n-k) if n > 1, 0 <= k <= n; T(1, 1)=1; T(0, 0)=1; T(n, k)=0 if k > n.
%F A102593 G.f.: G(t, z) = g(1-zg)/(1-tzg), where g = 1+zg^3 is the g.f. for the ternary numbers (A001764).
%e A102593 T(2,0) = T(2,1) = T(2,2) = 1 because in _\, /\ and /_ the maximum number of contiguous border edges starting from the root in counterclockwise direction is 0,1 and 2, respectively.
%e A102593 Triangle starts:
%e A102593     1;
%e A102593     0,   1;
%e A102593     1,   1,   1;
%e A102593     5,   4,   2,  1;
%e A102593    25,  18,   8,  3,  1;
%e A102593   130,  88,  37, 13,  4, 1;
%e A102593   700, 455, 185, 63, 19, 5, 1;
%e A102593   ...
%p A102593 T:=proc(n,k) if n=0 and k=0 then 1 elif n=1 and k=1 then 1 elif k<=n then (k+1)*binomial(3*n-2*k,n-k)/(2*n-k+1)-(k+2)*binomial(3*n-2*k-2,n-k-1)/(2*n-k) else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%t A102593 T[n_ /; n>1, k_] /; 0 <= k <= n := (k + 1) Binomial[3n - 2k, n - k]/(2n - k + 1) - (k + 2) Binomial[3n - 2k - 2, n - k - 1]/(2n - k); T[1, 1] = T[0, 0] = 1; T[_, _] = 0;
%t A102593 Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 06 2018 *)
%o A102593 (PARI) T(n,k) = {if(n==0, k==0, if(k<=n, (k+1)*binomial(3*n-2*k, n-k)/(2*n-k+1)-(k+2)*binomial(3*n-2*k-2, n-k-1)/(2*n-k)))} \\ _Andrew Howroyd_, Nov 06 2017
%Y A102593 Cf. A001764, A102893, A006629.
%K A102593 nonn,tabl
%O A102593 0,7
%A A102593 _Emeric Deutsch_, Jan 22 2005