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What function satisfies the following:

Let the matrix: $$\displaystyle T = \left(\begin{matrix} 1&0&0&0&0&0&0&\cdots \\ 1&1&0&0&0&0&0 \\ 1&1&1&0&0&0&0 \\ 1&1&1&1&0&0&0 \\ 1&1&1&1&1&0&0 \\ 1&1&1&1&1&1&0 \\ 1&1&1&1&1&1&1 \\ \vdots&&&&&&&\ddots \end{matrix}\right)$$

be defined by recurrence:

$$\text{If}\; n\geq k \; \text{then} \; T(n,k) = x \sum _{i=1}^{n-1} T(n-i,k-1)-y \sum _{i=1}^{n-1} T(n-i,k) \; \text{else} \; T(n,k) = 0$$

Then there exists a function $f(n,s)$ such that:

$$\lim_{s\to 1-y} f(n,s) = \sum\limits_{k=1}^{k=n} T(n,k)$$

This function then satisfies:

Harmonic numbers:

$$H_n=\lim_{s\to 0} \, \int f(-n,s) \, ds$$

Logarithmic integral:

$$\text{li}(k+1)-\gamma +k-\log (k) = \int_0^k \left(\int_1^2 f(n,s) \, dn\right) \, ds$$

Pascal triangle:

$$\binom{n-1}{1}=\lim_{s\to 1-1} \, f(n-1,s)$$ $$\binom{n-1}{2}=\lim_{s\to 1-1} \, \frac{1}{1} \frac{\partial f(n,s)}{\partial s}$$ $$\binom{n-1}{3}=\lim_{s\to 1-1} \, \frac{1}{2} \frac{\partial }{\partial s}\left(\frac{1}{1} \frac{\partial f(n,s)}{\partial s}\right)$$ $$\binom{n-1}{4}=\lim_{s\to 1-1} \, \frac{1}{3} \frac{\partial }{\partial s}\left(\frac{1}{2} \frac{\partial }{\partial s}\left(\frac{1}{1} \frac{\partial f(n,s)}{\partial s}\right)\right)$$ $$...$$

Polygamma:

$$\sum _{n=0}^{n=k} \frac{(-1)^{n+1} \psi ^{(n)}\left(c\right)}{n!}=\lim_{s\to c-1} \, \left(\sum _s f(-n,s)\right)$$

where $c$ is a complex number.

So again, what function am I thinking of?

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The function is: $$f(n,s)=\frac{(s+1)^{n-1}+s-1}{s}$$

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