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A052734 a(n) = 4^(n-1) * n! * Catalan(n-1) for n > 0, with a(0) = 0.

%I A052734 #55 Sep 08 2022 08:44:59
%S A052734 0,1,8,192,7680,430080,30965760,2724986880,283398635520,
%T A052734 34007836262400,4625065731686400,703009991216332800,
%U A052734 118105678524343910400,21731444848479279513600,4346288969695855902720000,938798417454304874987520000,217801232849398730997104640000,54014705746650885287281950720000,14259882317115833715842434990080000
%N A052734 a(n) = 4^(n-1) * n! * Catalan(n-1) for n > 0, with a(0) = 0.
%C A052734 For n>0, the number of fully-parenthesized expressions that you can form with n operands and 4 types of binary operators. - Yogy Namara (yogy.namara(AT)gmail.com), Mar 07 2010
%C A052734 a(n+1) is the number of square roots of any permutation in S_{16*n} whose disjoint cycle decomposition consists of 2*n cycles of length 8. - _Luis Manuel Rivera Martínez_, Feb 26 2015
%H A052734 Vincenzo Librandi, <a href="/A052734/b052734.txt">Table of n, a(n) for n = 0..200</a>
%H A052734 INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=690">Encyclopedia of Combinatorial Structures 690</a>.
%H A052734 Jesús Leaños, Rutilo Moreno and Luis Manuel Rivera-Martínez, <a href="http://arxiv.org/abs/1005.1531">On the number of mth roots of permutations</a>, arXiv:1005.1531 [math.CO], 2010-2011.
%H A052734 Jesús Leaños, Rutilo Moreno and Luis Manuel Rivera-Martínez, <a href="http://ajc.maths.uq.edu.au/pdf/52/ajc_v52_p041.pdf">On the number of mth roots of permutations</a>, Australas. J. Combin., Vol. 52 (2012), pp. 41-54 (Theorem 1).
%H A052734 W. van der Aalst, J. Buijs and B. van Dongen, <a href="https://hal.inria.fr/hal-01515548">Towards Improving the Representational Bias of Process Mining</a>, 2012.
%F A052734 E.g.f.: (1 - sqrt(1-16*x))/8.
%F A052734 Recurrence: a(1)=1, 8*(1 - 2*n)*a(n) + a(n+1) = 0.
%F A052734 a(n) = 16^n*Gamma(n+1/2)/sqrt(Pi).
%F A052734 a(0) = 0, a(1) = 1; a(n) = 4 * Sum_{k=1..n-1} binomial(n,k) * a(k) * a(n-k). - _Ilya Gutkovskiy_, Jul 09 2020
%F A052734 From _Amiram Eldar_, Jan 08 2022: (Start)
%F A052734 Sum_{n>=1} 1/a(n) = 1 + e^(1/16)*sqrt(Pi)*erf(1/4)/4, where erf is the error function.
%F A052734 Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - e^(-1/16)*sqrt(Pi)*erfi(1/4)/4, where erfi is the imaginary error function. (End)
%e A052734 Let's say the 4 types of binary operators are +, -, *, and /. Then, with 3 operands {a, b, c}, we can form expressions such as ((b+a)/c), (a-(c-b)), (c*(b+a)), etc. There are a(3)=192 such expressions. - Yogy Namara (yogy.namara(AT)gmail.com), Mar 07 2010
%p A052734 spec := [S,{B=Prod(C,C),S=Union(B,Z),C=Union(B,S,Z)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
%p A052734 seq((2*n)!/n! * 4^n, n = 0..10);
%t A052734 Join[{0}, Table[CatalanNumber[n-1] 4^(n-1) n!, {n, 1, 20}]] (* _Vincenzo Librandi_, Mar 11 2013 *)
%o A052734 (Magma) [0] cat [Catalan(n-1)*4^(n-1)*Factorial(n): n in [1..20]]; // _Vincenzo Librandi_, Mar 11 2013
%o A052734 (Sage) [0]+[4^(n-1)*factorial(n)*catalan_number(n-1) for n in (1..30)] # _G. C. Greubel_, Apr 02 2021
%Y A052734 Sequences of the form m^(n-1)*n!*Catalan(n-1): A001813 (m=1), A052714 (or A144828) (m=2), A221954 (m=3), this sequence (m=4), A221953 (m=5), A221955 (m=6).
%Y A052734 Equal to A000108 if all operands and all operators are indistinguishable.
%K A052734 easy,nonn
%O A052734 0,3
%A A052734 encyclopedia(AT)pommard.inria.fr, Jan 25 2000
%E A052734 Entry revised by _N. J. A. Sloane_, Feb 04 2013 and Feb 06 2013