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Iterate the following equation to obtain an explicit formula for $f( z)$: $$ \begin{align*} f( z) = 1 + z f\left( \frac{z}{1 + z}\right) .\end{align*} $$

Iterating this equation one obtains $$ \begin{align*} f( z) = \sum_{n \geq 0}^{} \frac{z^{n}}{\prod_{k = 0}^{n - 1} ( 1 + kz)} .\end{align*} $$ I wonder know whether this is already "explicit" enough or whether it can be simplified even further.

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    $\begingroup$ You have to study convergence of the series. $\endgroup$
    – geetha290krm
    Commented Sep 1, 2022 at 11:40
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    $\begingroup$ In terms of the first Kummer function $$ f\left( {\frac{1}{w}} \right) = 1 + \sum\limits_{n = 0}^\infty {\frac{1}{{w(w + 1) \cdots (w + n)}}} = 1 + \sum\limits_{n = 0}^\infty {\frac{{\Gamma (w)}}{{\Gamma (w + n + 1)}}} = 1 + \frac{{M(1,w + 1,1)}}{w}. $$ $\endgroup$
    – Gary
    Commented Sep 1, 2022 at 11:51
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    $\begingroup$ @Gary see my answer. $\endgroup$
    – Parcly Taxel
    Commented Sep 1, 2022 at 16:57
  • $\begingroup$ I think we need to assume continuity at most points, since the equation can only force values for algebraic $z$. $\endgroup$
    – aschepler
    Commented Sep 1, 2022 at 17:11

1 Answer 1

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Substituting $z\mapsto\frac1z$ in the functional equation and iterating yields $$f(1/z)=1+(1/z)f(1/(z+1))=\sum_{n=0}^\infty\frac1{z^{\overline n}}$$ which DLMF 8.11.4 simplifies to $$f(z)=e\gamma(1/z,1)+1$$ where $\gamma$ is the lower incomplete gamma function. $f$ has simple poles at the negative reciprocals of the positive integers; this can be seen by attempting to expand the functional equation at those points.

Taking the one-sided derivatives at $0$ from above reveals that $f$ is the ordinary generating function for a shifted, sign-alternated version of the Rao Uppuluri–Carpenter numbers A000587: $$f(z)=1+z+z^2-z^4+z^5+2z^6-9z^7+9z^8+50z^9-267z^{10}+\cdots$$

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