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Combinatorics of Permutations, 2nd Edition
book

Combinatorics of Permutations, 2nd Edition

by Miklos Bona
April 2016
Intermediate to advanced content levelIntermediate to advanced
478 pages
14h 44m
English
Chapman and Hall/CRC
Content preview from Combinatorics of Permutations, 2nd Edition
In Many Circles. Permutations as Products of Cycles. 125
SQ(x)=
n=0
SQ
n
n!
x
n
. Then we have
SQ(x)=
1+x
1 x
i1
cosh
x
2i
2i
.
PROOF Recall that our permutations must have an even number of
cycles of each even length. Therefore, repeated application of the method
seen in the proof of Example 3.64 yields
SQ(x)=(expx) · (cosh(x
2
/2)) · (exp x
3
/3) · (cosh(x
4
/4)) ···. (3.15)
Our claim is then proved recalling that we have already computed the power
series exp
%
x +
x
3
3
+ ···
&
in Example 3.58.
COROLLARY 3.69
For all positive integers n, we have SQ
2n
· (2n +1)=SQ
2n+1
.
This is an interesting result. It means that when we pass from 2n to 2n +1,
the number of permutations with ...
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Publisher Resources

ISBN: 9781439850527