A007974 Number of permutations that are n-3 "block reversals" away from 12...n.
1, 6, 52, 389, 2539, 16604, 105365, 654030, 3900116, 22073230, 116855950, 567965207
Offset: 3
Programs
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Python
def a(n): perm = tuple(range(1, n+1)); reach = {perm}; frontier = {perm} for k in range(n-3): r1 = set() while len(frontier) > 0: q = frontier.pop() for i in range(n): for j in range(i+1, n+1): qp = list(q) qp[i:j] = qp[i:j][::-1] qp = tuple(qp) if qp not in reach: reach.add(qp) r1.add(qp) frontier = r1 return len(frontier) print([a(n) for n in range(3, 10)]) # Michael S. Branicky, Jan 26 2023
Formula
a(n) = A300003(n,n-3).
Extensions
a(9)-a(11) from Sean A. Irvine, Feb 22 2018
a(12) from Thomas Baruchel, Mar 26 2025
a(13)-a(14) from Bert Dobbelaere, Apr 12 2025