Wow, well done! One question before I try to remember how to invert this thing: in Dirichlet series, the index starts at 1, whereas in the Taylor series, the index starts at zero. Does that operator shift the sequence or drop a term, or what? Thanks, Mike ----- Original Message Follows -----

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=Mike Stay ...How does one find the sequence a(n) for a Dirichlet generating function f(s)=sum(n=1 to inf) a(n)/n^s ?

This isn't directly the answer you seek, but I recalled Z. A. Melzak gives a transform to "Dirichletize" a Taylor series:

If f(x) has Taylor series coefficients An then the operator Ds, defined by

Ds f(x) = integral z^(s-1) f(x e^-z) dz / Gamma(s), z=0..inf

gives the function with Taylor coefficients An/n^s.

Then you let x=1 and you get the corresponding Dirichlet series function, call it g(s).

Now if you can manage to invert that operator you could presumably start with g(s), transform it into the corresponding Taylor series, and then pick the coefficients off f(x) with the usual derivative operators. The compositions of the inverse with those operators would give you the operators yielding the Dirichlet sequence.

Probably not much help, but maybe a start...

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http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039