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%I #35 Jun 22 2022 14:44:35
%S 1,1,4,25,210,2184,26796,376233,5920200,102816714,1947916100,
%T 39890416020,876478739164,20537052247300,510548782729680,
%U 13407568735200525,370553407586717490,10742998644116921160,325786278993936753300,10307990595756667951830
%N a(n) = B(n)*C(n), where B(n) are Bell numbers (A000110) and C(n) are Catalan numbers (A000108).
%C From _Joerg Arndt_, Oct 22 2012: (Start)
%C Number of strings of length 2*n of up to n different types t(k) of balanced parentheses, where the first appearance of type t(k) must precede the appearance of t(k+1) for all k<n.
%C For example, from the 5 parenthesis string of length 3
%C 1: ()()(); 2: ()(()); 3: (())(); 4: (()()); 5: ((())).
%C we obtain the B(3) * C(3) = 5 * 5 = 25 strings
%C 1: ()()(), ()()[], ()[](), ()[][], ()[]{};
%C 2: ()(()), ()([]), ()[()], ()[[]], ()[{}];
%C 3: (())(), (())[], ([])(), ([])[], ([]){};
%C 4: (()()), (()[]), ([]()), ([][]), ([]{});
%C 5: ((())), (([])), ([()]), ([[]]), ([{}]).
%C (End)
%H Alois P. Heinz, <a href="/A064299/b064299.txt">Table of n, a(n) for n = 0..400</a>
%H K. A. Penson and J.-M. Sixdeniers, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/Catalan.html">Integral Representations of Catalan and Related Numbers</a>, J. Integer Sequences, 4 (2001), #01.2.5.
%F Integral representation as n-th moment of a positive function on a positive half-axis, in Maple notation: a(n) = int(x^n*sum(sqrt((4*k-x)/x)*Heaviside(4*k-x)/(k*k!), k = 1..infinity)/(2*Pi*exp(1)), x = 0..infinity); this representation is unique.
%p with(combinat):
%p ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
%p a:= n-> bell(n)*ctln(n):
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Mar 23 2015
%t a[n_] := BellB[n]*CatalanNumber[n]; Table[a[n], {n, 0, 25}] (* _Jean-François Alcover_, Feb 25 2017 *)
%o (Sage) [bell_number(i)*catalan_number(i) for i in range(17)] # _Zerinvary Lajos_, Mar 14 2009
%o (Python)
%o from itertools import count, accumulate, islice
%o def A064299_gen(): # generator of terms
%o yield from (1,1)
%o blist, b, m = (1,2), 1, 1
%o for n in count(1):
%o blist = list(accumulate(blist, initial=(b:=blist[-1])))
%o yield b*(m := m*(4*n+2)//(n+2))
%o A064299_list = list(islice(A064299_gen(),20)) # _Chai Wah Wu_, Jun 22 2022
%Y Cf. A000108, A000110.
%Y Row sums of A253180.
%K nonn
%O 0,3
%A _Karol A. Penson_, Sep 05 2001
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