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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 $n, r$.

Hypergeometricx
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2 Answers2

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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.

David
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  • 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. – Hypergeometricx Nov 12 '14 at 14:34
  • I suspect it would be a very difficult problem to find any other solutions. But maybe somebody will be able to do it. – David Nov 12 '14 at 21:02
  • 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. – Hypergeometricx Nov 13 '14 at 02:23
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We prove the more general resukt that the system

$\dfrac{\binom{n}{r}}{\binom{n}{r-1}}=2$ Eq. 1

$\dfrac{\binom{n}{r+1}}{\binom{n}{r}}=\dfrac32$ Eq. 2

has a unique solution, which is the $\{1001,2002,3003\}$ set of binomial coefficients already identified.

To wit, from the expanded-product formula for the binomial coefficient

$\binom{n}{s}=\dfrac{n(n-1)(n-2)\cdot\cdot\cdot(n-s+1)}{(1)(2)(3)\cdot\cdot\cdot(s)}$

Eq. 1 renders

$\dfrac{n-r+1}{r}=2\implies n=3r-1$

Then Eq. 2 gives

$\dfrac{n-r}{r+1}=\dfrac{2r-1}{r+1}=\dfrac32\implies r=5$

Then $n=14$ giving the claimed result.

Oscar Lanzi
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