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In a problem I'm working on, the following two-index sequence keeps popping up:

$a(1,m) = 1$ for all $m$.

$a(n,m) = 1$ if $n = m$.

$a(n,m) = n(a(n,m-1) + a(n-1,m-1))$, $1 < n < m$.

And if $n > m$, $a(n,m)$ is not needed to be defined.

The sequence is almost like a messed-up Pascal's triangle. I would like to have a closed form relation for this sequence. I've tried messing around with 2D generating functions, but the results become messy very quickly.

Any help or hints would be appreciated!

EDIT: If anyone is interested in the math problem I'm working on, I am interested in finding a closed form of the sum $$\sum_{n=0}^{\infty}{n^pr^n},$$ For all $p\in\mathbb{N}$ and with $|r|<1$. So far, I've proven that the following relation holds: $$\sum_{n=0}^{\infty}{n^pr^n}=\frac{1}{(1-r)^{p+1}}\sum_{k=1}^{p}{a(k,p)(1-r)^{p-k}},$$ Where $a(k,p)$ is the sequence in question. Thus, you can see why finding a closed form for this sequence could be of some interest. Perhaps knowledge of the problem can help solve the enigma.

(NOTE: I'm a high school senior, so if there is already a closed-form for this sum that I just haven't seen due to lack of exposure, I'd love to see it!)

Bob Knighton
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  • Is the intent for $a_{n,m}$ with $n \ge m$? Otherwise it seems $a_{2,1} = -1/2$? In other words, if you don't want to define it for $n > m$, the first initial condition only gives $a_{1,1}$, no? – gt6989b Mar 17 '14 at 21:10

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At least the first couple of rows seem computable: $$a_{1,m} \equiv 1 \forall m$$ and for the second row one can see $$a_{2,m} = 3 \cdot 2^{m-2} - 2$$ and for the third row $$a_{3,m} = 16 \cdot 3^{m-3} - 9 \cdot 2^{m-2} + 3$$ and because of the propagation rule, it seems plausible that you get $$a_{n,m} = \sum_{k=1}^n c_{n,k} k^n$$ for some constants $\{c_{n,k}\}$...

gt6989b
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