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Here given is a sequence from OEIS.

The sequence is triangle of coefficients from fractional iteration of $e^x - 1$. Few terms are:
1, 1, 3, 1, 13, 18, 1, 50, 205, 180, 1, 201, 1865, 4245, 2700, 1, 875, 16674

The expression of finding the sequence is also given as:
$A(n;x)$ for n-th row satisfies
$$A(n;x) = \sum_{k=0}^{n-1} \mathrm{Stirling}_2(n, k) * A(k;x)*x,\ A(1;x) = 1.$$

The tabular view shows the entries row wise. $\mathrm{Stirling}_2(n,k)$ is most probably stirling numbers of the second type

I am not able to get how above expression is resulting in the given sequence.
In summation, $k$ begins from 0, but nothing is mentioned about $A(0;x)$. I assumed it to be 0, but still can't get the above values.

Please explain how the first few terms are resulting from the expression.

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Isn't this the same as mathoverflow.net/questions/133271/… ? –  Gerry Myerson Jun 13 '13 at 6:02
    
The Mathematica function at OEIS works as desired. –  Fred Kline Jun 13 '13 at 8:36
    
@Fred Kline are you sure of that? Can you explain the the computation manually for the first few terms? I am getting wrong results. –  Nadia Jun 13 '13 at 9:02
    
I have changed the mathematical expressions to correct LaTeX. Hope this helps. –  rem Jun 26 '13 at 8:36

2 Answers 2

[update] Here -for possibly a nicer reference- I give a screenshot, where the identity by the OEIS-formulae (and kindly expanded by @Carlo Beenakker) is interpreted as matrix-multiplication. It indicates also, that the given formula of relations between the Stirling and A-numbers is near to an eigenvector/eigenmatrix-relation (in fact is a Jordan-decomposition) which I described in my first version of this post (which I kept below).

Here is the matrix-multiplication scheme:
here

[the original answer: Jordan-decomposition]
Another approach gave the numbers in a completely surprising context; I've no explanation so far. ([update]:It is clear now. Carlo's expansion gave the key hint!).

For the function $\exp(x)-1$ the matrix $S_2$ of Stirling numbers 2'nd kind is involved. I've just recently begun to exercise with the Jordan-decomposition, and for some exercise I took this matrix

$$ \large S_2 \qquad = \qquad \small{ \begin{bmatrix} 1 & . & . & . & . \\ 1 & 1 & . & . & . \\ 1 & 3 & 1 & . & . \\ 1 & 7 & 6 & 1 & . \\ 1 & 15 & 25 & 10 & 1 \end{bmatrix}} $$ and let WolframAlpha Jordan-decompose it such that $$S_2 = A \cdot J \cdot A^{-1}$$

This gave the three matrices:

$$ \begin{array} {} A &=&\small { \begin{bmatrix} 1 & . & . & . & . \\ 0 & 1 & . & . & . \\ 0 & 1 & 3 & . & . \\ 0 & 1 & 13 & 18 & . \\ 0 & 1 & 50 & 205 & 180 \end{bmatrix} } \\ J&=& \small {\begin{bmatrix} 1 & . & . & . & . \\ 1 & 1 & . & . & . \\ 0 & 1 & 1 & . & . \\ 0 & 0 & 1 & 1 & . \\ 0 & 0 & 0 & 1 & 1 \end{bmatrix} } \\ A^{-1}&=&\small {\begin{bmatrix} 1 & . & . & . & . \\ 0 & 1 & . & . & . \\ 0 & -1/3 & 1/3 & . & . \\ 0 & 5/27 & -13/54 & 1/18 & . \\ 0 & -301/2430 & 353/1944 & -41/648 & 1/180 \end{bmatrix}}\end{array}$$ (Note: I've transposed the input to Woframalpha and then also the output to keep in line with my usual conventions with that type of matrix-discussion)
The surprise is: that we find the coefficients in the matrix $A$

The relation can also be written as $$ S_2 \cdot A = A \cdot J $$ and if $J$ is decomposed in the sum of two (diagonal and subdiagonal) components $J=J_0+J_1$ where $J_0=I$ and $J_1$ the transpose of my $J$ in the update of this post above, then we can reformulate $$ S_2 \cdot A = A \cdot (I+J_1) \\ (S_2 - I) \cdot A = A \cdot J_1 \\ S_2^* \cdot A = A \cdot J_1 \\ $$ from which -in my opinion- we could formulate a slightly more convenient relation than that given in the OEIS.


Additional remark because there is a relation to the Schröder-function for fractional iteration: for the use for the function $f(x) = \exp(x)-1$ the matrix $S_2$ is factorially similarity-scaled and -when let unscaled- becomes the Bell/(transposed) Carleman-matrix for that function. In that view the scaling $F^{-1} \cdot S_2 \cdot F$ is a matrixoperator; and the matrices $A$ and $A^{-1}$ are in a very similar role as the operators for the Schröder-function of $f(x)$ which is indeed at the heart of the Schröder/Abel-type of fractional iteration...


This is the input for W/A's input field:
JordanForm({{1,1,1,1,1},{0,1,3,7,15},{0,0,1,6,25},{0,0,0,1,10},{0,0,0,0,1}})

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Stirling numbers of the second kind $S_2(n,k)$ vanish for $k=0$, so the $k=0$ term in the sum does not contribute. You ask for a manual computation of the first few terms. Here we go:

Stirling numbers: $S_2(n,1)=S_2(n,n)=1$, $S_2(3,2)=3$, $S_2(4,2)=7$, $S_2(4,3)=6$

$$a(2,x)=xa(1,x)=x$$

$$a(3,x)=xa(1,x)+3xa(2,x)=x+3x^2$$

$$a(4,x)=xa(1,x)+7xa(2,x)+6xa(3,x)=x+7x^2+6x^2+18x^3=x+13x^2+18x^3$$

This gives the first few coefficients: $1,1,3,1,13,18$.

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if x means the column-index, then all elements in the matrix $A$ in the first column are zero by $a(2,0)=0 \cdot a(1,0) = 0 $. Is that really meant? It would contradict the table-view in OEIS where the first column is 1. Are the indexes for S2 and A are thought differently (for S2 beginning at 0 and for A beginning at 1 ?) –  Gottfried Helms Dec 9 '13 at 14:54
    
@GottfriedHelms --- my understanding is that $x$ is a variable, not an index; the function $a(n,x)$ is a polynomial of degree $n-1$ in the variable $x$; you are seeking a triangle of coefficients, such that the $n$-th row of the triangle has at the $m$-th position the coefficient of the term $x^m$ in $a(n,x)$. As I worked it out above, you see that this indeed gives the series mentioned by the OP. In particular, the fourth row has elements $1,13,18$, which are the coefficients of $a(4,x)=x+13x^2+18x^3$. –  Carlo Beenakker Dec 9 '13 at 16:34
    
Carlo, see my accidental beautiful observation in my new answer... –  Gottfried Helms Dec 10 '13 at 23:55

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