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I came across an interesting pattern in the Pascal triangle, and thought I would post it as a problem here.

Given three consecutive binomial coefficients $$\binom n{r-1},\binom nr,\binom n{r+1}$$ which are in AP, where $$\begin{align} \binom n{r-1}&=1\cdot 10^m+1\\ \binom nr &=2\cdot 10^m+2\\ \binom n{r+1} &=3\cdot 10^m+3 \end{align}$$ where $m, n, r$ are integers, find $n, r$.

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    $\begingroup$ Hint: $$\binom n{r-1} = \frac{1}{2}\binom n{r} = \frac{1}{3}\binom n{r+1} = 10^m+1$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 12 '14 at 11:40
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It is well known (to those who know such things well) that $1001,2002,3003$ occur consecutively in the $14$th row of Pascal's triangle.

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  • $\begingroup$ Thank you for sharing your observation. You certainly know the Pascal triangle very well! I was hoping that someone on MSE might attempt a derivation based on the given problem statement, and in the process also identify if a solution exists for other values of $m$. Have upvoted nevertheless. $\endgroup$ – hypergeometric Nov 12 '14 at 14:34
  • $\begingroup$ I suspect it would be a very difficult problem to find any other solutions. But maybe somebody will be able to do it. $\endgroup$ – David Nov 12 '14 at 21:02
  • $\begingroup$ We'll see. But will accept your answer anyway as it's the one I had in mind. It's a surprisingly nice pattern to discover in the Pascal triangle. $\endgroup$ – hypergeometric Nov 13 '14 at 2:23

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