Hi.

Not sure if this is a "good" question for this forum or if it'll get panned, but here goes anyway...

Consider this problem. I've been trying to find a formula to expand the "regular iteration" of "exp". Regular iteration is a special kind of complex function that is a solution of the equation

$$f(z+1) = \exp(f(z))$$.

This regular iteration in particular is an entire function. To get it, we take a fixed point $L$ of $\exp$ and expand a solution in powers of $L^z$. The result is to obtain a Fourier series

$$f(z) = \sum_{n=0}^{\infty} a_n L^{nz}$$

where

$$a_0 = L$$
$$a_1 = 1$$
$$a_n = \frac{B_n(1! a_1, 2! a_2, ..., (n-1)! a_{n-1}, 0)}{n!(L^{n-1} - 1)}$$

with $B_n$ being the nth "complete" Bell polynomial. This <i>recursive</i> formula yields the following expansions:

$$a_2 = \frac{1}{2L - 2}$$
$$a_3 = \frac{L + 2}{6L^3 - 6L^2 - 6L + 6}$$
$$a_4 = \frac{L^3 + 5L^2 + 6L + 6}{24L^6 - 24L^5 - 24L^4 + 24L^2 + 24L - 24}$$
$$a_5 = \frac{L^6 + 9L^5 + 24L^4 + 40L^3 + 46L^2 + 35L + 24}{120L^{10} - 120L^9 - 120L^8 + 240L^5 - 120L^2 - 120L + 120}$$
...

It *appears* that, by pattern recognition and factoring the denominators,

$$a_n = \frac{\sum_{j=0}^{\frac{(n-1)(n-2)}{2}} mag_{n,j} L^j}{\prod_{j=2}^{n} j(L^{j-1} - 1)}$$

where $\mathrm{mag}_{n,j}$ is a sequence of "magic" numbers (integers) that looks like this (with the leftmost column being $j = 0$):

n = 1: 1  
n = 2: 1  
n = 3: 2, 1  
n = 4: 6, 6, 5, 1  
n = 5: 24, 36, 46, 40, 24, 9, 1  
n = 6: 120, 240, 390, 480, 514, 416, 301, 160, 64, 14, 1  
n = 7: 720, 1800, 3480, 5250, 7028, 8056, 8252, 7426, 5979, 4208, 2542, 1295, 504, 139, 20, 1  
n = 8: 5040, 15120, 33600, 58800, 91014, 124250, 155994, 177220, 186810, 181076, 163149, 134665, 102745, 71070, 44605, 24550, 11712, 4543, 1344, 265, 27, 1  
n = 9: 40320, 141120, 352800, 695520, 1204056, 1855728, 2640832, 3473156, 4277156, 4942428, 5395818, 5561296, 5433412, 5021790, 4391304, 3625896, 2820686, 2056845, 1398299, 879339, 504762, 260613, 117748, 45178, 13845, 3156, 461, 35, 1  
n = 10: 362880, 1451520, 4021920, 8769600, 16664760, 28264320, 44216040, 64324680, 88189476, 114342744, 141184014, 166279080, 187614312, 202901634, 210825718, 210403826, 201934358, 186191430, 164980407, 140216446, 114231817, 88934355, 66047166, 46576620, 31071602, 19460271, 11365652, 6112650, 2987358, 1298181, 488878, 153094, 37692, 6705, 749, 44, 1  
...

But what is the simplest (or at least "reasonably" simple) *non-recursive* formula for these numbers, or perhaps the numerators in general? Like a sum formula, or something like that.

And regardless of the formula for the "mag", can one <i>prove</i> from the recurrence formula that the $a_n$ have the form given, and if so, how? Especially, how can one <i>prove</i> the numerator has degree $\frac{(n-1)(n-2)}{2}$? Perhaps that might provide insight into how to find the formula for the "mag".

The ultimate goal here is to try and obtain a series expansion for the "tetration" function $^z e$, more specifically, Kneser's tetrational function, described in Kneser's papers on solutions of $f(f(x)) = \exp(x)$ and related equations (paper is in German, I only saw the translations.). Though this may not be the best way to go, since after constructing this regular iteration function, we then need a special mapping derived from a Riemann mapping to "distort" it so it becomes real-valued at the real axis, and I don't know if there's any good way to construct Riemann mappings even as "non-closed" infinite expansions. But I'm still curious to see if at least a formula for this function is possible.

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<b>Justification for thinking a formula exists</b>

Why do I think this even exists, when there's no guarantee that this kind of really non-trivial recurrence relation should even <i>have</i> a non-recursive solution in the first place? Well, for one, the fact that so much of it could be put in simple form as given, and also I *did* manage to come up with an explicit formula from a very roundabout way but this formula is *excessively* complicated and based on very *general* techniques.

It is difficult to describe that formula here, but the outline of the process to construct it is this, for all its worth:

1. A general recurrence of the form

$$A_1 = r_{1, 1}$$
$$A_n = \sum_{m=1}^{n-1} r_{n,m} A_m$$

has a non-recursive solution formula. This I found myself, but it is hideous and involves binary bit operations. This kind of recurrence is very general, and it also includes the recurrence for the Bernoulli numbers and other kinds of recurrences.

2. The "regular Schroder function" of $F(z) = e^{uz} - 1$, i.e. the function satisfying $\mathrm{RSF}(e^{uz} - 1) = K \mathrm{RSF}(z)$, can be given as a Taylor series

$$\mathrm{RSF}(z) = \sum_{n=1}^{\infty} A_n z^n$$

where $A_n$ is given by the recurrence-solving formula with $r_{1,1} = 1$ and $r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} S(n, m)$ (here, $S(n, m)$ is a Stirling number of the 2nd kind). This is hideous due to the binary counting stuff. Not sure <i>at all</i> how this could be simplified.

3. Invert the regular Schroder function using the Lagrange-Burmann formula. This can be expanded, apparently with a formula involving sums over partitions and a multinomial coefficient. Plug the huge $A_n$ formula into this. Horrific!

4. Now $U(z) = \mathrm{RSF}^{-1}(u^z)$ is a "regular iteration" of $e^{uz} - 1$, giveable as a Fourier series, or Taylor series in $u^z$.

5. Apply the topological conjugation to conjugate it to iteration of $e^{vz}$ by taking $v = ue^{-u}$ thus $u = -W(-v)$ (Lambert's W-function). Take $H(z) = z + 1$ then find $H^{-1} o U o H$. This gives a regular iteration of $e^{vz}$, thus set $u = 1$, though there may be a constant displacement of some kind offsetting it from the one with the $a_n$-formula.

So by this, I think an explicit formula <i>exists</i> (though that constant-shift at the end may be a little problem, but not much, since it is immaterial to the structure of the function). I'm just interested in something simpler than this, preferably something to "fill out" the "mag" formula I gave...

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