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%I #11 Jun 01 2015 07:17:35
%S 16796,3233230,354660460,29214542500,2013190058880,122762429039250,
%T 6850724997273300,357603651626578500,17726205673051976100,
%U 843509478504416874150,38843740303576863755100,1741683387026398566250500,76401095775145069217992560
%N Number of 2n-length strings of balanced parentheses of exactly 10 different types that are introduced in ascending order.
%C In general, column k>0 of A253180 is asymptotic to (4*k)^n / (k!*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Jun 01 2015
%H Alois P. Heinz, <a href="/A258398/b258398.txt">Table of n, a(n) for n = 10..600</a>
%F Recurrence: (n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*n*(n+1)*a(n) = 110*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*n*(2*n - 1)*a(n-1) - 5280*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(n-1)*(2*n - 3)*(2*n - 1)*a(n-2) + 145200*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(n-2)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-3) - 2524368*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(n-3)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-4) + 28865760*(n-8)*(n-7)*(n-6)*(n-5)*(n-4)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-5) - 218683520*(n-8)*(n-7)*(n-6)*(n-5)*(2*n - 11)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-6) + 1076416000*(n-8)*(n-7)*(n-6)*(2*n - 13)*(2*n - 11)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-7) - 3264915456*(n-8)*(n-7)*(2*n - 15)*(2*n - 13)*(2*n - 11)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-8) + 5441863680*(n-8)*(2*n - 17)*(2*n - 15)*(2*n - 13)*(2*n - 11)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-9) - 3715891200*(2*n - 19)*(2*n - 17)*(2*n - 15)*(2*n - 13)*(2*n - 11)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*a(n-10). - _Vaclav Kotesovec_, Jun 01 2015
%F a(n) ~ 40^n / (10!*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Jun 01 2015
%p ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
%p A:= proc(n, k) option remember; k^n*ctln(n) end:
%p T:= (n, k)-> add(A(n, k-i)*(-1)^i/((k-i)!*i!), i=0..k):
%p a:= n-> T(n, 10):
%p seq(a(n), n=10..25);
%Y Column k=10 of A253180.
%K nonn
%O 10,1
%A _Alois P. Heinz_, May 28 2015
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