A092186 a(n) = 2(m!)^2 for n = 2m and m!(m+1)! for n = 2m+1.
2, 1, 2, 2, 8, 12, 72, 144, 1152, 2880, 28800, 86400, 1036800, 3628800, 50803200, 203212800, 3251404800, 14631321600, 263363788800, 1316818944000, 26336378880000, 144850083840000, 3186701844480000, 19120211066880000, 458885065605120000, 2982752926433280000
Offset: 0
Keywords
References
- A. O. Munagi, Alternating subsets and permutations, Rocky Mountain J. Math. 40 (6) (2010) 1965-1977 doi:10.1216/RJM-2010-40-6-1965, Corollary 3.2.
- David Singmaster, Problem 1654, Mathematics Magazine 75 (October 2002). Solution in Mathematics Magazine 76 (October 2003).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..300
- David Singmaster, Problem 1654, Mathematics Magazine 75 p. 317 (October 2002). Solution, Mathematics Magazine 76 p. 321-322 (October 2003).
- Shinji Tanimoto, Alternate Permutations and Signed Eulerian Numbers, arXiv:math/0612135 [math.CO], 2006; Ann. Comb. 14 (2010), 355.
Crossrefs
Cf. A152877.
Row sums of A125300. - Alois P. Heinz, Nov 18 2013
Programs
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Maple
a:= proc(n) option remember; `if`(n<2, 2-n, (n*(3*n-1)*(n-1)*a(n-2) -4*a(n-1))/(12*n-16)) end: seq(a(n), n=0..30); # Alois P. Heinz, Nov 11 2013
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Mathematica
f[n_] := If[EvenQ[n], 2 (n/2)!^2, ((n + 1)/2)! ((n - 1)/2)!]; Table[ f[n], {n, 0, 25}] (* Geoffrey Critzer, Aug 24 2013 *)
Comments