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A364317 Irregular triangle T read by rows: T(n, k) gives the number of permutations of [n] = {1, 2, ..., n} with a cycle of length m = floor(n/2) + k = A138099(n, k), for 1 <= k <= n - floor(n/2) = ceiling(n/2).

%I A364317 #21 Sep 25 2023 18:22:37
%S A364317 1,1,3,2,8,6,40,30,24,180,144,120,1260,1008,840,720,8064,6720,5760,
%T A364317 5040,72576,60480,51840,45360,40320,604800,518400,453600,403200,
%U A364317 362880,6652800,5702400,4989600,4435200,3991680,3628800
%N A364317 Irregular triangle T read by rows: T(n, k) gives the number of permutations of [n] = {1, 2, ..., n} with a cycle of length m = floor(n/2) + k = A138099(n, k), for 1 <= k <= n - floor(n/2) = ceiling(n/2).
%C A364317 The length of row n is ceiling(n/2) = A008619(n-1).
%C A364317 The numbers for these cycles of permutations of [n], appear in the solution of the Locker Problem. See the link, p. 25.
%C A364317 For the probability of failures with the strategy used in the locker problem with n lockers and opening of up to floor(n/2) lockers see A058313(n)/A058312(n), for n > = 1. For n = 1 the one team member is not allowed to open the one locker (with the member's wallet) because (n/2) = 0; so certainly a failure.
%C A364317 For the probability of success in this locker problem for n lockers see A119248(n)/A058312(n), for n >= 1.
%H A364317 Paolo Xausa, <a href="/A364317/b364317.txt">Table of n, a(n) for n = 1..1024</a> (rows 1..63 of the triangle, flattened)
%H A364317 Eugene Curtis and Max Warshauer, <a href="http://gato-docs.its.txstate.edu/jcr:dabe5ae5-c4b8-4009-a00f-5215bf010aeb/Locker_Puzzle.pdf">The Locker Puzzle</a>, The Mathematical Intelligencer 28 (2006), pp. 28-31.
%H A364317 Christoph Pöppe, <a href="https://www.spektrum.de/magazin/freiheit-fuer-die-kombinatoriker/848868">Freiheit für die Kombinatoriker</a>, Spektrum, Highlights 1.21, pp. 16-18 (in German).
%H A364317 Wikipedia, <a href="https://en.wikipedia.org/wiki/100_prisoners_problem">100 prisoners problem</a>.
%H A364317 Wikipedia (in German), <a href="https://de.wikipedia.org/wiki/Problem_der_100_Gefangenen">Problem der 100 Gefangenen</a>.
%F A364317 T(n, k) = binomial(n, m(n, k))*(m(n, k) - 1)!*(n - m(n, k))! = n!/m(n, k), with m(n, k) = floor(n/2) + k = A138099(n, k), for n >= 1 and k = 1, 2, ..., ceiling(n/2).
%e A364317 The irregular triangle begins:
%e A364317 n\k       1       2       3       4       5       6 ...
%e A364317 -------------------------------------------------------
%e A364317 1:        1
%e A364317 2:        1
%e A364317 3:        3       2
%e A364317 4:        8       6
%e A364317 5:       40      30      24
%e A364317 6:      180     144     120
%e A364317 7:     1260    1008     840     720
%e A364317 8:     8064    6720    5760    5040
%e A364317 9:    72576   60480   51840   45360   40320
%e A364317 10:  604800  518400  453600  403200  362880
%e A364317 11: 6652800 5702400 4989600 4435200 3991680 3628800
%e A364317 ...
%e A364317 T(5, 1) = 40 because m(5, 1) = 2 + 1 = 3, and for each of the binomial(5, 3) = 10 possibilities for choosing three numbers from [5] there are (3 - 1)! = 2 3-cycles if each starts with the smallest number, e.g., for {2, 3, 5} the cycles are (2, 3, 5) and (2, 5, 3). For the remaining 5-3 = 2 numbers there are 2! possible permutations; in the example permutations of {1, 4}, namely (1)(4) and (1,4). Thus T(5, 3) = binomial(5, 3)*2!*2! = 10*2*2 = 40 = 5!/3.
%t A364317 A364317row[n_]:=n!/(Floor[n/2]+Range[Ceiling[n/2]]);Array[A364317row,10] (* _Paolo Xausa_, Sep 25 2023 *)
%Y A364317 Cf. A008619, A058313/A058312, A119248/A058312, A126074, A138099.
%K A364317 nonn,tabf,easy
%O A364317 1,3
%A A364317 _Wolfdieter Lang_, Aug 12 2023