01/30/2016, 09:06 PM

with each integer power of n, the tetration converges, one extra coefficient (highlighted in red), to the Taylor series of , which gives the known asymptotic limit for

the converged coefficients are 1, followed by the sequence http://oeis.org/A033917 (divided by n!)

Some formulas for some of those tetrations are

where are the Stirling numbers of the first kind.

From the definition of the inverse Stirling transform, it looks like each tetration can be written as (1 plus...) a polynomial whose coefficients are inverse Stirling transform of the coefficients of some relatively simple function (coefficients not including the factorial denominator in the Taylor series)

Here represents the last summation on the previous set of equations. The point of them being "relatively simple", compared to tetration, is that if we can find a general expression for , for any exponent n,it probably will be easier to generalize it to real exponents of tetration.

the can be obtained by the direct Stirling transform of the coefficients shown at the start of the post.

For example, the coefficients of the Taylor series of are 1 followed by [ 1, 1, 3/2, 4/3, 3/2, 53/40, 233/180, 5627/5040, 2501/2520, ...] (as shown at the start of this post).

To get the , we need to remove the n! denominators of the Taylor series, and apply the Stirling transform.

= [1, 2, 9, 32, 180, 954, 6524, 45016, 360144, 3023640, 27617832, 271481880, ...]

and we get = [1, 3, 16, 101, 756, 6607, 65794, 733833, 9046648, 121961051, 1782690174, 28055070397, ...]

Piece of code for calculation of

Code:

`tet=x+O(x^21)+1`

{

for (iter=1,10,

tet=(x+O(x^21)+1)^tet;

c_j=vector( 12,n,polcoeff(tet,n)*n! );

b_j=vector( 12,n,sum(j=1,n,stirling(n,j,2)*c_j[j]));

print("^"iter+1"(x+1): b_j="b_j);

); /*for iter*/

}

gives this output:

^2(x+1): b_j=[1, 3, 10, 41, 196, 1057, 6322, 41393, 293608, 2237921, 18210094, 157329097]

^3(x+1): b_j=[1, 3, 16, 101, 756, 6607, 65794, 733833, 9046648, 121961051, 1782690174, 28055070397]

^4(x+1): b_j=[1, 3, 16, 125, 1176, 12847, 160504, 2261289, 35464816, 612419291, 11539360944, 235469524237]

^5(x+1): b_j=[1, 3, 16, 125, 1296, 16087, 229384, 3687609, 66025360, 1303751051, 28151798544, 659841763957]

^6(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 257104, 4480569, 87238720, 1874561291, 44057589984, 1124459440117]

^7(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4742649, 96915520, 2197675691, 54640864224, 1476693931957]

^8(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 99637120, 2323474091, 59755204224, 1676301882037]

^9(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2354318891, 61498237824, 1760945456437]

^10(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2357947691, 61877447424, 1786651875637]

^11(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2357947691, 61917364224, 1791681392437]

According of OEIS, the sequence for seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725)

For n=3,

I made everything I could to simplify this last expression, to make it look as similar as possible to the others (for n=1, 2, and ).

Can you make it simpler?

Can you identify a pattern?

Can you identify the exponent z in from ? (I wrote here the formulas for n=1,2,3 and ∞)