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I'm trying work on my ability to break complex patterns down,
and in this case I'm trying to model the denominators of Lacsap's Fractions:

enter image description here

I managed to get the sequence that represents the denominators:

enter image description here

But now I want to convert it to a series in order to get rid of the tedious term D_n-1.
I guess I'm just not quite sure how to apply the sum identity to a parametric sequence: enter image description here

The red is the answer that I've found works, but I don't know how to get to it.
When I simply substitute k, D_1, and D_k with snippets from the sequence, I get it wrong.

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This is essentially OEIS A144216 –  Henry Aug 16 '12 at 7:24

1 Answer 1

up vote 2 down vote accepted

I think you wanted to write $$D_{n,r}=D_{n-1,r}+(n-r)$$ though you could have written $$D_{n,r}=D_{n-1,r-1}+r$$ with the combined effect that (assuming that you start from $D_{0,0}=0$) $$D_{n,r}=\sum_{i=1}^{n-r} i + \sum_{j=1}^{r} j = \frac{(n-r)(n-r+1)}{2} + \frac{r(r+1)}{2}$$ and hence your result.

Another way to look at it is to note that $D_{n,0}=D_{n,n}=\frac{n(n+1)}{2}$ and that $D_{n,r}=D_{n-r,0}+D_{r,r}$. For the numerator you have $N_{n,r}=\frac{n(n+1)}{2}$.

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Thanks for sharing your insight, but I'm still a bit confused. I thought you'd have to take D(n-1,r) out of the equation in order to perform summation since it merely acts as a symbol for recursive addition of all preceding terms (which is taken care of by the summation). According to your answer, SUM(n-r) is equal to SUM(D(n-1,r-1)) while SUM(r) is equal to SUM(D(n-1,r)). I don't see how D(n-1,r) and D(n-1,r-1) could have any direct value considering they are just indicators that the sequence needs to be summed. –  Griffin Aug 16 '12 at 10:44
    
Also, if I may prick your brain =), how did you so quickly realize that D(n,r) equaled D(n-r,0) + D(r,r) or that D(n,n) equaled n(n+1)/2 or that D(n-1, r) + (n-r) equaled D(n-1,r-1) + r? Are there some generic rules I'm not aware of? –  Griffin Aug 16 '12 at 10:47
    
@Griffin: If you look at your triangle, each time you go down to the left you add the next integer $(1+2+\cdots)$ and every time you go down to the right you add the next integer $(1+2+\cdots)$ so the final figure is the sum of two triangle numbers. –  Henry Aug 16 '12 at 11:58
    
Ok, but that still doesn't answer my first question. How EXACTLY did you get from D(n,r) = D(n−1,r) +(n−r) to the addition of those two summations (i and j)? –  Griffin Aug 17 '12 at 0:30
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$D(n,r) = D(n−1,r) +(n−r) = D(n−2,r) +(n-r-1) + (n−r) = \cdots $ $= D(r,r) + 1 +2 +\cdots + (n-r-1) + (n−r) = D(r,r) + \frac{(n-r)(n-r+1)}{2}$ while $D(r,r) = D(r-1,r-1) +r = D(r-2,r-2) +(r-1) +r = \cdots $ $= D(0,0) + 1 +2 +\cdots + (r-1) + r = D(0,0) + \frac{r(r+1)}{2}$. –  Henry Aug 17 '12 at 0:33

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