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Consider the following iterative process. We start with the function having all $1$'s in its Taylor series expansion: $$f_0(x)=\frac1{1-x}=1+x+x^2+x^3+x^4+O\left(x^5\right).\tag1$$ Then, at each step we apply the following transformation: $$f_{n+1}(x)=x^{-1}\log\left(\frac{f_n(x)}{f_n(0)}\right).\tag2$$ A few initial iterations give us: $$ \begin{array}{l} f_1(x)=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\frac{x^4}{5}+O\left(x^5\right), \\ f_2(x)=\frac{1}{2}+\frac{5 x}{24}+\frac{x^2}{8}+\frac{251 x^3}{2880}+\frac{19 x^4}{288}+O\left(x^5\right), \\ f_3(x)=\frac{5}{12}+\frac{47 x}{288}+\frac{2443 x^2}{25920}+\frac{5303 x^3}{82944}+\frac{412589x^4}{8709120}+O\left(x^5\right), \end{array}\tag3 $$ and their initial terms form the following sequence: $$1,\,\frac{1}{2},\,\frac{5}{12},\,\frac{47}{120},\,\frac{12917}{33840},\,\frac{329458703}{874222560},\,\dots,\tag4$$ whose denominators grow pretty quickly, but which appear to slowly converge to a value $$\lambda\stackrel{\color{#aaaaaa}?}\approx0.3678\dots\tag5$$

If we assume that the process with the iterative step $(2)$ converges to a fixed point, we can see that it must have a form: $$f_\omega(x)=-x^{-1}\,W(-c\,x)=c+c^2 x+\frac{3\, c^3\, x^2}{2}+\frac{8\, c^4\, x^3}{3}+\frac{125\, c^5\, x^4}{24}+O\left(x^5\right),\tag6$$ where $W(\cdot)$ is the Lambert -function, and $c$ is a coefficient that is not uniquely determined but depends on the choice of the initial function $f_0(x)$. In our case, $c=\lambda$.

Questions: Does this process actually converge to a fixed point? If yes, then what is a closed-form expression (or another useful description) for $\lambda$?

Update: An explicit recurrent formula for the coefficients (the parenthesized superscript $m$ in $a_n^{\small(m)}$ is just the second index of the coefficient; the sum $\sum_{\ell=1}^m$ is assumed to be $0$ when $m=0$): $$f_n(x)=\sum_{m=0}^\infty a_{n\vphantom{+0}}^{\small(m)}x^m,$$ where $$a_{0\vphantom{+0}}^{\small(m)}=1,\quad a_{n\vphantom{+0}}^{\small(m)}=\frac1{\,a_{n-1}^{\small(0)}\,}\left(a_{n-1}^{\small(m+1)}-\frac1{m+1} \sum_{\ell=1}^m\ell\;a_{n\vphantom{+0}}^{\small(\ell-1)}\,a_{n-1}^{\small(m-\ell+1)}\right).$$ The sequence of coefficients $(4)$ is $\big\{a_{n\vphantom{+0}}^{\small(0)}\big\}$.

Vladimir Reshetnikov
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    By the mean value theorem, taking $x\to 0,$ you get $$f_{n+1}(0)=\frac{f_n’(0)}{f_n(0)}.$$ Not sure if that helps in any way. – Thomas Andrews Dec 20 '21 at 04:13
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    Whatever the answer could be, this is a nice problem. – Claude Leibovici Dec 20 '21 at 05:35
  • is there a typo in your explicit formula? We are after $a_n^{(0)}$ and when I tried plug $m = 0,$ the sum starts at $l=1$ and ends at $l = m.$ – dezdichado Dec 23 '21 at 06:03
  • The formula is correct. We assume that $\sum_{\ell=1}^0(\dots)=0$, because there are no indices satisfying $1\le\ell\le0$, so the sum is empty . – Vladimir Reshetnikov Dec 23 '21 at 07:24
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    I think $\lambda =1/e=0.367879441\ldots$. A rough argument is as follows. It seems that each $f_n(x)$ is singular at $x=1$. Since the limit function is $-x^{-1}W(-\lambda x)$ and $W(z)$ is singular at $z=-1/e$, $\lambda$ should be $1/e$ in order for $f_\omega(x)$ to be singular at $x=1$. – Gary Dec 23 '21 at 07:45
  • I concur with Claude and will go further saying this is a very nice problem in every respect. – A rural reader Dec 23 '21 at 22:18
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    it says the limit is $1/e$ here. – mathworker21 Dec 23 '21 at 22:52
  • @mathworker21 Thanks! It would be nice to see a rigorous proof. I haven't been able to prove it myself so far. – Vladimir Reshetnikov Dec 23 '21 at 22:56
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    @mathworker21 Usually, OEIS keeps numerators and denominators as distinct sequences. This entry mixes them into one interleaved. Probably this is why I was not able to find the coefficients in OEIS after I computed a few of them. – Vladimir Reshetnikov Dec 23 '21 at 22:58
  • Originally I found this sequence when I tried to construct the continued exponent expansion of the Fibonacci numbers: https://pbs.twimg.com/media/FHAQfurVQAANsLx?format=png&name=large, but later realized I can get it simply from $1/(1-x)$. – Vladimir Reshetnikov Dec 23 '21 at 23:05
  • Another very interesting question is how fast do denominators grow (or would grow if we ignored occasional cancellations of some of their factors with numerators). Empirically, they appear to grow (very roughly) doubly exponentially: $2^{2^{n-0.97942\dots}}$. – Vladimir Reshetnikov Dec 24 '21 at 15:55
  • @VladimirReshetnikov It seems that, if $f_n(0)=a/b$ (in lowest terms), then the denominator of $f_{n+1}(0)$ is a small multiple ($2$ or $6$) of $ab$. Let $b_n$ be the denominator of $f_n(0)$. If $f_n(0)$ converges to a limit $\lambda$, then this would imply that $b_{n+1}$ is multiplicatively not far off from $\lambda b_n^2$, which gives double-exponential growth. – Carl Schildkraut Dec 24 '21 at 23:09
  • Interestingly enough, Caylay briefly considered this problem in a note called "On Some Numerical Expansions" from the Quarterly Journal of Pure and Applied Mathematics, vol. III. (1860), pp. 366-369, but he didn't go much farther than you. What stroke me is that he got the same first few terms as you, but he was probably wrong for the last term "$\log\left(-\frac2x\log\left(\frac1x\log(1+x)\right)\right)=-\frac5{12}x+\frac{47}{288}x^2+\frac{2443}{25920}x^3-\frac{5303}{82944}x^4+\frac{19631}{580608}x^5+&c$". – Nolord Aug 23 '24 at 14:46
  • He also gives the first terms of the next power series in the sequence. Then, he writes the first ever recorded insights on the formula for iteration of power series $f \circ \ldots \circ f(x)$, which then insipired Schröder in his 1872 paper, in which he first considered, what now call, the "Schröder equation". This is one of the first paper on iteration theory, which unfortunately doesn't seem to apply to your case. – Nolord Aug 23 '24 at 14:55

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