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%I #20 Jan 26 2023 16:14:12
%S 1,6,52,389,2539,16604,105365,654030,3900116
%N Number of permutations that are n-3 "block reversals" away from 12...n.
%F a(n) = A300003(n,n-3).
%o (Python)
%o def a(n):
%o perm = tuple(range(1, n+1)); reach = {perm}; frontier = {perm}
%o for k in range(n-3):
%o r1 = set()
%o while len(frontier) > 0:
%o q = frontier.pop()
%o for i in range(n):
%o for j in range(i+1, n+1):
%o qp = list(q)
%o qp[i:j] = qp[i:j][::-1]
%o qp = tuple(qp)
%o if qp not in reach:
%o reach.add(qp)
%o r1.add(qp)
%o frontier = r1
%o return len(frontier)
%o print([a(n) for n in range(3, 10)]) # _Michael S. Branicky_, Jan 26 2023
%Y Cf. A007972, A007973, A007975, A300003.
%K nonn,more
%O 3,2
%A _J. H. Conway_
%E a(9)-a(11) from _Sean A. Irvine_, Feb 22 2018
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