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Nov 26, 2016 at 6:02 history edited user78249 CC BY-SA 3.0
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Nov 26, 2016 at 5:17 vote accept CommunityBot
Nov 26, 2016 at 5:16 comment added user78249 @JulianRosen I feel like such a tool right now. I can't believe I missed that there was a $k$ in the upper argument. I can't even understand how I missed that. Everything is as clear as day now.
Nov 26, 2016 at 0:49 answer added Pat Devlin timeline score: 0
Nov 25, 2016 at 23:18 comment added Julian Rosen Should it be $1\times\ldots\text{($k$ times)}\ldots\times 1 = {n+k-1\choose k-1}$?
Nov 25, 2016 at 20:50 comment added Pat Devlin And this even fails for positive integers $s$ and $q$. Right?
Nov 25, 2016 at 20:45 comment added Pat Devlin This fails when $n=0,$ right?
Nov 25, 2016 at 19:58 comment added user78249 @PietroMajer I'm well aware of that, I guess I didn't make myself clear enough. I was more interested in the properties of the iterated sum, and then the complex iterated sum. If this construction exists in literature somewhere.
Nov 25, 2016 at 19:56 comment added Pietro Majer This is the discrete convolution, en.wikipedia.org/wiki/Convolution#Discrete_convolution In the particular case of sequences $u:\mathbb{Z}\to\mathbb{C}$ supported on $\mathbb{Z}_+$, it is also called "one-side convolution", and corresponds to the Cauchy product of power series.
Nov 25, 2016 at 19:42 history edited user78249 CC BY-SA 3.0
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Nov 25, 2016 at 19:37 history asked user78249 CC BY-SA 3.0