
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
i≥1
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 ...