\( ^0(1+x) \,=\, \)\( {\color{Red} 1} \)\( + 0+ 0+0... \)

\( ^1(1+x) \,=\, \)\( {\color{Red} 1}+ x \)\( + 0+0+0... \)

\( ^2(1+x) \,=\, \)\( {\color{Red} 1}+ x+ x^2 \)\( +\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12}) \)

\( ^3(1+x) \,=\, \)\( {\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3} \)\( +\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12}) \)

\( ^4(1+x) \,=\, \)\( {\color{Red} 1}+x+x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4} \)\( +3x^{5}+\frac {163}{40}x^{6}+\frac {1861}{360}x^{7}+\frac {33641}{5040}x^{8}+\frac {8363}{1008}x^{9}+\frac {22391}{2160}x^{10}+\frac {7589}{600}x^{11}+O(x^{12}) \)

\( ^5(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5} \)\( +\frac {243}{40}x^{6}+\frac {3421}{360}x^{7}+\frac {71861}{5040}x^{8}+\frac {54371}{2520}x^{9}+\frac {69281}{2160}x^{10}+\frac {7200983}{151200}x^{11}+O(x^{12}) \)

\( ^6(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6} \)\( +\frac {4321}{360}x^{7}+\frac {102941}{5040}x^{8}+\frac {85871}{2520}x^{9}+\frac {61333}{1080}x^{10}+\frac {886763}{9450}x^{11}+O(x^{12}) \)

\( ^7(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7} \)\( +\frac {118061}{5040}x^{8}+\frac {106661}{2520}x^{9}+\frac {81583}{1080}x^{10}+\frac {10169449}{75600}x^{11}+O(x^{12}) \)

\( ^8(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8} \)\( +\frac {115481}{2520}x^{9}+\frac {93013}{1080}x^{10}+\frac {12169699}{75600}x^{11}+O(x^{12}) \)

\( ^9(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9} \)\( +\frac {97333}{1080}x^{10}+\frac {13165099}{75600}x^{11}+O(x^{12}) \)

\( ^{10}(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9}+\frac {98413}{1080}x^{10} \)\( +\frac {13505299}{75600}x^{11}+O(x^{12}) \)

with each integer power of n, the tetration \( \\[15pt]

{^n(x+1)} \) converges, one extra coefficient (highlighted in red), to the Taylor series of \( \\[15pt]

{\frac {lambertW(-ln(x+1))}{ln(x+1)} } \), which gives the known asymptotic limit for \( \\[15pt]

{e^{-e}<(x+1)<e^{e^{-1}}} \)

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

Some formulas for some of those tetrations are

\( \\[24pt]

{ ^1(x+1) \,=\, 1+

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \sum_{k=0}^{n} {{n} \choose {k}}*(n-k)^k

\\

} \)

\( \\[24pt]

{ ^2(x+1) \,=\, 1+

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \sum_{k=0}^{j} {{j} \choose {k}}*(j-k)^k

\\

} \)

\( \\[24pt]

{ ^\infty(x+1) \,=\,1+

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \sum_{k=0}^{j-1} {{j-1} \choose {k}}*{j^k}

\\

\\

} \)

where \( \\[24pt]

{ \begin{bmatrix}

& n & \\

& j &

\end{bmatrix}} \) 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 \( \\[15pt]

{b_j} \) of some relatively simple function (coefficients not including the factorial denominator in the Taylor series)

\( \\[24pt]

{ ^z(x+1) \,=\, ^0a +

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \,.\, b_j

\\

\\

} \)

Here \( \\[15pt]

{b_j} \) 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 \( \\[15pt]

{b_j} \), for any exponent n,it probably will be easier to generalize it to real exponents of tetration.

the \( \\[15pt]

{b_j} \) can be obtained by the direct Stirling transform of the coefficients shown at the start of the post.

For example, the coefficients \( c_j \) of the Taylor series of \( \\[15pt]

{ ^3(x+1)} \) 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 \( \\[15pt]

{b_j} \), we need to remove the n! denominators of the Taylor series, and apply the Stirling transform.

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

\( b_n=\sum_{j=1}^n \left\{\begin{matrix} n \\ j \end{matrix} \right\} c_j \)

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

Piece of code for calculation of \( b_j \)

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 \( b_j \) sequence for \( \\[15pt]

{^z(x+1)} \) seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725)

For n=3, \( \\[25pt]

{b_j=1+\sum_{k=0}^{j-1} {{j}\choose{k}} \sum_{i=1}^{j-k} {{j-k}\choose{i}}(i^{(j-k-i)}*k^i)} \)

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 \( \infty \)).

Can you make it simpler?

Can you identify a pattern?

Can you identify the exponent z in \( b_j \) from \( \\[24pt]

{ ^z(x+1) \,=\, ^0a +

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \,.\, b_j

\\

\\

} \) ? (I wrote here the formulas for n=1,2,3 and ∞)

\( ^1(1+x) \,=\, \)\( {\color{Red} 1}+ x \)\( + 0+0+0... \)

\( ^2(1+x) \,=\, \)\( {\color{Red} 1}+ x+ x^2 \)\( +\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12}) \)

\( ^3(1+x) \,=\, \)\( {\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3} \)\( +\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12}) \)

\( ^4(1+x) \,=\, \)\( {\color{Red} 1}+x+x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4} \)\( +3x^{5}+\frac {163}{40}x^{6}+\frac {1861}{360}x^{7}+\frac {33641}{5040}x^{8}+\frac {8363}{1008}x^{9}+\frac {22391}{2160}x^{10}+\frac {7589}{600}x^{11}+O(x^{12}) \)

\( ^5(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5} \)\( +\frac {243}{40}x^{6}+\frac {3421}{360}x^{7}+\frac {71861}{5040}x^{8}+\frac {54371}{2520}x^{9}+\frac {69281}{2160}x^{10}+\frac {7200983}{151200}x^{11}+O(x^{12}) \)

\( ^6(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6} \)\( +\frac {4321}{360}x^{7}+\frac {102941}{5040}x^{8}+\frac {85871}{2520}x^{9}+\frac {61333}{1080}x^{10}+\frac {886763}{9450}x^{11}+O(x^{12}) \)

\( ^7(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7} \)\( +\frac {118061}{5040}x^{8}+\frac {106661}{2520}x^{9}+\frac {81583}{1080}x^{10}+\frac {10169449}{75600}x^{11}+O(x^{12}) \)

\( ^8(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8} \)\( +\frac {115481}{2520}x^{9}+\frac {93013}{1080}x^{10}+\frac {12169699}{75600}x^{11}+O(x^{12}) \)

\( ^9(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9} \)\( +\frac {97333}{1080}x^{10}+\frac {13165099}{75600}x^{11}+O(x^{12}) \)

\( ^{10}(1+x) \,=\, \)\( {\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9}+\frac {98413}{1080}x^{10} \)\( +\frac {13505299}{75600}x^{11}+O(x^{12}) \)

with each integer power of n, the tetration \( \\[15pt]

{^n(x+1)} \) converges, one extra coefficient (highlighted in red), to the Taylor series of \( \\[15pt]

{\frac {lambertW(-ln(x+1))}{ln(x+1)} } \), which gives the known asymptotic limit for \( \\[15pt]

{e^{-e}<(x+1)<e^{e^{-1}}} \)

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

Some formulas for some of those tetrations are

\( \\[24pt]

{ ^1(x+1) \,=\, 1+

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \sum_{k=0}^{n} {{n} \choose {k}}*(n-k)^k

\\

} \)

\( \\[24pt]

{ ^2(x+1) \,=\, 1+

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \sum_{k=0}^{j} {{j} \choose {k}}*(j-k)^k

\\

} \)

\( \\[24pt]

{ ^\infty(x+1) \,=\,1+

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \sum_{k=0}^{j-1} {{j-1} \choose {k}}*{j^k}

\\

\\

} \)

where \( \\[24pt]

{ \begin{bmatrix}

& n & \\

& j &

\end{bmatrix}} \) 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 \( \\[15pt]

{b_j} \) of some relatively simple function (coefficients not including the factorial denominator in the Taylor series)

\( \\[24pt]

{ ^z(x+1) \,=\, ^0a +

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \,.\, b_j

\\

\\

} \)

Here \( \\[15pt]

{b_j} \) 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 \( \\[15pt]

{b_j} \), for any exponent n,it probably will be easier to generalize it to real exponents of tetration.

the \( \\[15pt]

{b_j} \) can be obtained by the direct Stirling transform of the coefficients shown at the start of the post.

For example, the coefficients \( c_j \) of the Taylor series of \( \\[15pt]

{ ^3(x+1)} \) 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 \( \\[15pt]

{b_j} \), we need to remove the n! denominators of the Taylor series, and apply the Stirling transform.

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

\( b_n=\sum_{j=1}^n \left\{\begin{matrix} n \\ j \end{matrix} \right\} c_j \)

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

Piece of code for calculation of \( b_j \)

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 \( b_j \) sequence for \( \\[15pt]

{^z(x+1)} \) seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725)

For n=3, \( \\[25pt]

{b_j=1+\sum_{k=0}^{j-1} {{j}\choose{k}} \sum_{i=1}^{j-k} {{j-k}\choose{i}}(i^{(j-k-i)}*k^i)} \)

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 \( \infty \)).

Can you make it simpler?

Can you identify a pattern?

Can you identify the exponent z in \( b_j \) from \( \\[24pt]

{ ^z(x+1) \,=\, ^0a +

\sum_{n=1}^{\infty}

\frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix}

& n & \\

& j &

\end{bmatrix} \,.\, b_j

\\

\\

} \) ? (I wrote here the formulas for n=1,2,3 and ∞)

I have the result, but I do not yet know how to get it.