%I #49 Aug 18 2022 13:59:27
%S 1,1,0,3,1,0,19,10,1,0,193,119,23,1,0,2721,1806,466,46,1,0,49171,
%T 34017,10262,1502,87,1,0,1084483,770274,255795,47020,4425,162,1,0,
%U 28245729,20429551,7235853,1539939,193699,12525,303,1,0,848456353,621858526,230629024,54314242,8273758,755170,34912,574,1,0
%N Triangle read by rows: T(n,m) = Sum_{k=m+1..n} (n-1)!/(k-1)!*binomial(2*n-k-1, n-1)*E(k,m) where E(n,m) is Euler's triangle A173018, T(0,0) = 1, n >= m >= 0.
%C T(n,m) is the number of labeled binary trees of size n with m ascents on the left branch.
%H Michael De Vlieger, <a href="/A316773/b316773.txt">Table of n, a(n) for n = 0..11475</a> (rows 0 <= n <= 150, flattened)
%H Yuriy Shablya, Dmitry Kruchinin, Vladimir Kruchinin, <a href="https://doi.org/10.3390/math8060962">Method for Developing Combinatorial Generation Algorithms Based on AND/OR Trees and Its Application</a>, Mathematics (2020) Vol. 8, No. 6, 962.
%F E.g.f.: Sum_{n >= m >= 0} T(n, m)/n! * x^n * y^m = E(C(x),y) = (y-1)/(y-exp(C(x)*(y-1))), where E(x,y) is an e.g.f. for Euler's triangle A173018.
%F T(n,m) = Sum_{k = m+1..n} C(n,k)*E(k,m)*P(n,n-k), T(0,0)=1, where C(n,m) is the transposed Catalan's triangle A033184, E(n,m) is Euler's triangle A173018, and P(n,m) is the number of k-permutations of n A008279.
%e Triangle begins:
%e --------------------------------------------------------------------------
%e n\k| 0 1 2 3 4 5 6 7 8 9
%e ------+-------------------------------------------------------------------
%e 0 | 1
%e 1 | 1 0
%e 2 | 3 1 0
%e 3 | 19 10 1 0
%e 4 | 193 119 23 1 0
%e 5 | 2721 1806 466 46 1 0
%e 6 | 49171 34017 10262 1502 87 1 0
%e 7 | 1084483 770274 255795 47020 4425 162 1 0
%e 8 | 28245729 20429551 7235853 1539939 193699 12525 303 1 0
%e 9 | 848456353 621858526 230629024 54314242 8273758 755170 34912 574 1 0
%p T := (n,m) -> `if`(n=0, 1, add((n-1)!/(k-1)!*binomial(2*n-k-1, n-1)*
%p combinat:-eulerian1(k, m), k = m+1..n)):
%p for n from 0 to 6 do seq(T(n, k), k=0..n) od; # _Peter Luschny_, Sep 04 2020
%t Table[Boole[n == 0] + Sum[(n - 1)!/(k - 1)!*Binomial[2 n - k - 1, n - 1]*Sum[(-1)^j*(m + 1 - j)^k*Binomial[k + 1, j], {j, 0, m}], {k, m + 1, n}], {n, 0, 8}, {m, 0, n}] // Flatten (* _Michael De Vlieger_, Sep 04 2020 *)
%o (Maxima)
%o T(n,m):=if m>n then 0 else if n=0 then 1 else sum((n-1)!/(k-1)!*binomial(2*n-k-1,n-1)*sum((-1)^j*(m+1-j)^k*binomial(k+1,j),j,0,m),k,m+1,n);
%Y Cf. A033184, A008279, A173018.
%K nonn,tabl
%O 0,4
%A _Yuriy Shablya_, Sep 13 2018