Editing Generating function (section)

===Dirichlet series generating functions (DGFs)===

[[Formal Dirichlet series]] are often classified as generating functions, although they are not strictly formal power series. The ''Dirichlet series generating function'' of a sequence {{math|''a''<sub>''n''</sub>}} is:<ref name="W56">{{harvnb|Wilf|1994|p=56}}</ref>
<math display="block">\operatorname{DG}(a_n;s)=\sum _{n=1}^\infty \frac{a_n}{n^s}.</math>

The Dirichlet series generating function is especially useful when {{math|''a''<sub>''n''</sub>}} is a [[multiplicative function]], in which case it has an [[Euler product]] expression<ref name="W59">{{harvnb|Wilf|1994|p=59}}</ref> in terms of the function's Bell series:
<math display="block">\operatorname{DG}(a_n;s)=\prod_{p} \operatorname{BG}_p(a_n;p^{-s})\,.</math>

If {{math|''a''<sub>''n''</sub>}} is a [[Dirichlet character]] then its Dirichlet series generating function is called a [[Dirichlet L-series|Dirichlet {{mvar|L}}-series]]. We also have a relation between the pair of coefficients in the [[Lambert series]] expansions above and their DGFs. Namely, we can prove that:
<math display="block">[x^n] \operatorname{LG}(a_n; x) = b_n</math>if and only if
<math display="block">\operatorname{DG}(a_n;s) \zeta(s) = \operatorname{DG}(b_n;s),</math>where {{math|''ζ''(''s'')}} is the [[Riemann zeta function]].<ref>{{cite book |last1=Hardy |first1=G.H. |last2=Wright |first2=E.M. |last3=Heath-Brown |first3=D.R |last4=Silverman |first4=J.H. |title=An Introduction to the Theory of Numbers|url=https://archive.org/details/introductiontoth00ghha_922|url-access=limited|publisher=Oxford University Press |page=[https://archive.org/details/introductiontoth00ghha_922/page/n357 339]|edition=6th |isbn=9780199219858 |year=2008}}</ref>

The sequence {{mvar|a<sub>k</sub>}} generated by a [[Dirichlet series]] generating function (DGF) corresponding to:<math display="block">\operatorname{DG}(a_k;s)=\zeta(s)^m</math>has the ordinary generating function:<math display="block">\sum_{k=1}^{k=n} a_k x^k = x + \binom{m}{1} \sum_{2 \leq a \leq n} x^{a} + \binom{m}{2}\underset{ab \leq n}{\sum_{a = 2}^\infty \sum_{b = 2}^\infty} x^{ab} + \binom{m}{3}\underset{abc \leq n}{\sum_{a = 2}^\infty \sum_{c = 2}^\infty \sum_{b = 2}^\infty} x^{abc} + \binom{m}{4}\underset{abcd \leq n}{\sum_{a = 2}^\infty \sum_{b = 2}^\infty \sum_{c = 2}^\infty \sum_{d = 2}^\infty} x^{abcd} + \cdots</math>