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A000246 Number of permutations in the symmetric group S_n that have odd order.
(Formerly M2824 N1137)
+0
16
1, 1, 1, 3, 9, 45, 225, 1575, 11025, 99225, 893025, 9823275, 108056025, 1404728325, 18261468225, 273922023375, 4108830350625, 69850115960625, 1187451971330625, 22561587455281875, 428670161650355625 (list; graph; listen)
OFFSET

0,4

COMMENT

Michael Reid (mreid(AT)math.umass.edu) points out that the e.g.f. for the number of permutations of odd order can be obtained from the cycle index for S_n, F(Y; X1, X2, X3, ... ) := e^(X1 Y + X2 Y^2/2 + X3 Y^3/3 + ... ) and is F(Y, 1, 0, 1, 0, 1, 0, ... ) = sqrt((1 + Y)/(1 - Y)).

a(n)=number of permutations on [n] whose up-down signature has nonnegative partial sums. For example, the up-down signature of (2,4,5,1,3) is (+1,+1,-1,+1) with nonnegative partial sums 1,2,1,2 and a(3)=3 counts (1,2,3), (1,3,2), (2,3,1). - David Callan (callan(AT)stat.wisc.edu), Jul 14 2006

a(n) is the number of permutations of [n] for which all left-to-right minima occur in odd locations in the permutation. For example, a(3)=3 counts 123, 132, 231. Proof: For such a permutation of length 2n, you can append 1,2,..., or 2n+1 (2n+1 choices) and increase by 1 the original entries that weakly exceed the appended entry. This gives all such permutations of length 2n+1. But if the original length is 2n-1, you cannot append 1 (for then 1 would be a left-to-right min in an even location) so you can only append 2,3,..., or 2n (2n-1 choices). This count matches the given recurrence relation a(2n)=(2n-1)a(2n-1), a(2n+1)=(2n+1)a(2n). - David Callan (callan(AT)stat.wisc.edu), Jul 22 2008

REFERENCES

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

H.-D. Ebbinghaus et al., Numbers, Springer, 1990, p. 146.

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 87.

LINKS

T. D. Noe, Table of n, a(n) for n=0..100

J. Sondow, A faster product for Pi and a new integral for ln(Pi/2)

FORMULA

E.g.f.: sqrt(1-x^2)/(1-x). a(2n)=(2n-1)a(2n-1), a(2n+1)=(2n+1)a(2n).

Let b(1)=0, b(2)=1, b(k+2)=b(k+1)/k + b(k); then a(n+1)=n!*b(n+2). - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 03 2002

a(n) = sum((2k)! * C(n-1, 2k) * a(n-2k-1), k=0 to floor((n-1)/2)) for n>0. - Noam Katz (noamkj(AT)hotmail.com), Feb 27 2001

Also successive denominators of Wallis's approximation to pi/2 (unreduced): 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * ...

a(n)=a(n-1)+(n-1)*(n-2)*a(n-2) - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 30 2003

a(n) is asymptotic to (n-1)!*sqrt(2*n/Pi) - Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 19 2004

n! * C(n-1, [(n-1)/2]) / 2^(n-1), n>0. - R. Stephan, Mar 22 2004

PROGRAM

(PARI) a(n)=if(n<1, !n, a(n-1)*(n+n%2-1))

CROSSREFS

Cf. A001900, A059838.

Cf. A002867.

Bisections are A001818 and A079484.

Row sums of unsigned triangle A049218 and of A111594.

Sequence in context: A001902 A068100 A012821 this_sequence A103620 A138315 A038059

Adjacent sequences: A000243 A000244 A000245 this_sequence A000247 A000248 A000249

KEYWORD

nonn

AUTHOR

N. J. A. Sloane (njas(AT)research.att.com).

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Last modified November 21 21:21 EST 2009. Contains 167310 sequences.


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