Series expansion for $x_n=x_{n-1}+\log \left(x_{n-1}\right)$

Question

$x_0=2,\ x_n=x_{n-1}+\log \left(x_{n-1}\right)\quad$ has a series expansion about $1.$

Since $x_0=2,$ $(x_0-1)^k=1,$ so the recurrence can be written, up to the first $5$ terms as

\begin{align} 1+\frac{773788}{604800}\ 2^n-\frac{247200}{604800}\ 2^{2 n}+\frac{118825}{604800}\ 2^{3 n}-\frac{54300}{604800}\ 2^{4 n}+\frac{13687}{604800}\ 2^{5 n}-\dots \end{align}

found with

Evaluate[Plus @@ (((FindSequenceFunction[With[{nn = #}, (Rest@(List @@ 
Series[Nest[# + Log[#] &, x, #], {x, 1, 20}] & /@ Range[0, 20]))
[[All, 3, nn + 1]]]] & /@ Range[0, 5] /. # -> n)[[All, 1]]))]

and then combining like terms. The problem is finding the limits of the coefficients of $2^{k n}.$ Theoretically, the recurrence relation should be rewriteable as $$1+\sum _{k=1}^{\infty } (-1)^{k+1} c_k 2^{k n}$$

but I don't know how to go about finding a general solution for $c_k.$

Added

Building on @AntonioVargas' comment below, the recurrence relation seems very close numerically to $$\sum _{k=1}^{\infty } (-1)^{k+1} \left(n c_k \log (n)-\frac{1}{2} n \log \left(c_k\right)\right)\approx \pi n/2 +n \log n$$

(based on the first $9$ terms of the series expansion).

Numerically, an approximation of the coefficients can be found with something like

Abs[With[{aa = SplitBy[SortBy[Flatten@Join[{1, 2^n}, Drop[List @@@ 
Expand[Evaluate[(((FindSequenceFunction[With[{nn = #}, (Rest@(List @@ 
Series[Nest[# + Log[#] &, x, #], {x, 1, 20(*LARGER THAN number of terms*)}] 
& /@ Range[0,20(*LARGER THAN number of terms*)]))[[All, 3, nn + 1]]]] & /@ 
Range[0, 9(*number of terms*)] /. # -> n)[[All, 1]]))]], 2]],  
Coefficient[PowerExpand[Log2[#]], n] &], Coefficient[PowerExpand[Log2[#]], 
n] &]}, (a /. Solve[Plus @@ (aa[[#]]) == a 2^( (# - 1) n), a][[1]] // N) & 
/@ Range[2, Length@aa]]]

I doubt it. Any random recurrence you write is unlikely to have a closed-form solution. Only those with a very special form will have a simple formula. — Jair Taylor, May 14 '15 at 17:33
For clarity, are we talking about common log or natural log? — Sean Henderson, May 14 '15 at 17:42
@martin - in retrospect, it doesn't actually matter all that much. — Sean Henderson, May 14 '15 at 17:50
@martin Hmm, don't know. I suppose it's possible there's some brilliant Ramanujan-esque series solution for the sequence :) — Jair Taylor, May 14 '15 at 17:59
I'm actually experimenting with Faà di Bruno's formula because with finite n it can be viewed as a composition series of the series produced by $log(x)+x$ — Histograms, May 15 '15 at 08:19
Even if it did hypothetically have a nonrecursive form, that form probably wouldn't be useful at all. — DanielV, May 17 '15 at 09:26
@DanielV I agree that it would have to have a rediculous number of terms to approach anything of any "use", but as a mental exercise, I find it quite interesting. — martin, May 17 '15 at 09:38
@Did setting $x_0=\pi/2$ agrees with setting $x_0=2$ for recurrence. I am aware that this is an anomoly, but unsure how to pose it. — martin, May 17 '15 at 09:52
@Did, would posting a plot help to clarify where my error in explanation is? — martin, May 17 '15 at 09:53
Not really, but explaining what it is you are asking, definitely would. — Did, May 17 '15 at 09:55
@Did I am doing my best!!! Do you have any suggestions, or specific questions that I could answer hopefully more eloquently that I currently am? — martin, May 17 '15 at 09:57
I understand that you are asking for coefficients $c^n_k$ such that, for every $n\geqslant0$, if $x_0=x$ is close enough to $1$ and if $x_i=x_{i-1}+\ln x_{î-1}$ for every $1\leqslant i\leqslant n$, then $x_n=\sum\limits_kc^n_k\cdot(x-1)^k$. Right? — Did, May 17 '15 at 10:11
Numerically the asymptotic series for large $n$ appears to be something like $an\log n + bn + c + \cdots$ — Antonio Vargas, May 17 '15 at 14:07
@AntonioVargas great - that gives me something to work with - thanks :) — martin, May 17 '15 at 14:13
Based on some further analysis I would guess that $x_n = an\log n + (-a+\log a)n + o(n)$. Numerics based on this indicate something like $a \approx 1.25$. — Antonio Vargas, May 17 '15 at 14:35
@AntonioVargas great :) If you get any further with this, and if you have time, please do post an answer. Meanwhile, I will pursue with your observations in mind :) — martin, May 17 '15 at 14:38
I have absolutely no idea what this question is asking. Where do those power of $2$ come from ???? and how did you obtain those coefficients ? ( I can't read whatever code that is) — mercio, May 20 '15 at 08:20
@Histograms - have only just seen your comment - I think 2 must've come at the same time & I missed it! Than ks for the link re Faà di Bruno's formula - have you managed to get anywhere with it so far? — martin, May 20 '15 at 09:16
@mercio code is Mathematica & utilises Series and FindSequenceFunction (although the first few are guessable of course) as its basic functions. Since the series expands as \begin{align} \ &1\quad +\ \ &\left(2\cdot 2^{n-1}\right)(x_0-1)\quad +\ \ &\left(2^{n-2}-2^{2 n-2}\right)(x_0-1)^2\quad +\ \ &\left(\frac{2}{9}\ 2^{n-3}-\frac{9}{9}\ 2^{2 n-3}+\frac{7}{9}\ 2^{3 n-3}\right)(x_0-1)^3\quad +\ \ &\left(\frac{6}{252}\ 2^{n-4}-\frac{119}{252}\ 2^{2 n-4}+\frac{294}{252}\ 2^{3 n-4}-\frac{181}{252}\ 2^{4 n-4}\right)(x_0-1)^4\quad +\ \ &\dots \ \end{align} — martin, May 20 '15 at 11:20
@merci continued ... it is possible to eliminate the $(x_0−1)k=1$ since $x_0=2.$ The rest is combining like terms (since $2^{n-1},2^{n-2},2^{n-3},$ etc. are all expressible as $a 2^{n}$ ). This is actually based on an observation Histograms made here, and in the comment immediately preceding it. — martin, May 20 '15 at 11:21
@mercio continued ... Numerically, the coefficients approximate something like $\small{1.27941, 0.40922, 0.20359, 0.12001, 0.07558, 0.04616, 0.02606, 0.01575\dots},$ based on the first $11$ terms (error of course increasing with term number). — martin, May 20 '15 at 11:22
@martin no unfortunately it's pretty messy. The only other interesting thing I noticed was that you can "reverse" the nest from a given starting point y0 with Nest[ProductLog[Exp[#]] &, y0, n] where ProductLog is the Lambert-W function, for instance Nest[ProductLog[Exp[#]] &, 8.481100676447268, 5]` gets you back to a starting point of 2. — Histograms, May 20 '15 at 12:47
@Histograms yes, I have been playing around with the product log too. Nice observation, though agreed, it is pretty messy - there seems be a strange behaviour in terms of convergence to a limit, as it fluctuates. — martin, May 20 '15 at 12:54
why do you think the sum $2/2 + 1/4 + 2/(93) + 6/(25216) + ...$ converges ? it looks like the sequence is slowly converging to $0$ but it seems like an accident to me. — mercio, May 20 '15 at 15:15

Series expansion for $x_n=x_{n-1}+\log \left(x_{n-1}\right)$

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