Dear Ralf,

sorry, I cannot see the mistake. I don't see any permutations in my code, only
cycle types. 

Ralf Hemmecke <[EMAIL PROTECTED]> writes:

> OK, before I go further, just propose the code for what you have
> described. And I guess it will be wrong in any computer language.

But I did write code in my last email: two functions, one computes the cycle
type of the power of a permutation, only given its cycle type:

CYCLETYPE ==> List Integer
cycleTypePower(p: CYCLETYPE, k: Integer): CYCLETYPE  == {
    cycleType: CYCLETYPE := [];
    divisorsK: List Integer := divisors k;
    for m in #p..1 by -1 repeat {
        sum: Integer := 0;
        for d in divisorsK | gcd(m, k/d)=1 repeat {
            sum := sum + d * p.(d*m)
        }
        cylceType := cons(sum, cycleType)
    }
    cycleType
}

The second one computes the cycle type of F[\sigma], only given the cycle type
of \sigma and the cycleIndexSeries of F.

cycleTypeSpecies(p: CYCLETYPE): CYCLETYPE == {
    cycleType: CYCLETYPE := [];
    s := cycleIndexSeries;
    n := lcm([i for i in 1..#p | p.i ~= 0]);
    for k in n..1 by -1 repeat {
        divisorsK: List Integer := divisors k;
        sum: Integer := 0;
        for d in divisorsK repeat {
            sum := sum + moebiusMu(k/d)
                        *coefficient(s, cycleTypePower(p, d));
-- this is very rough... is there such a function coefficient? Or count?
        }
        cycleType := cons(sum/k, cycleType);
    }
    cycleType;
}


To get the cycleIndexSeries of F \square G, you have to compute 

\begin{equation*}
  Z_F \square Z_G = \sum_{n_1+2n_2+\dots}
                    \fix(F\square G)[n_1, n_2, \dots]
                    \frac{x_1^{n_1}x_2^{n_2}\dots}
                         {1^{n_1}n_1! 2^{n_2}n_2!\dots},
\end{equation*}

where

\fix(F\square G)[n_1, n_2, \dots]

is

coefficient(cycleIndexSeries$F, cycleTypeSpecies [n1, n2,...])

(I'm not sure what coefficient(cycleIndexSeries$F, ...) does exactly. If you
include the denominator 1^n1 n1! 2^n2 n2! ... in the coefficient you have to
multiply by it again... The same remark holds for the computation of
cycleTypeSpecies)


Does this answer your question? Did I miss something?

Martin


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