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This is a generalization of my earlier question here posted recently, and is a more interesting one.

Three consecutive binomial coefficients $$\binom n{r-1},\binom nr, \binom n{r+1}$$ are in an AP (arithmetic progression) with positive common difference. Find possible pairs of $(n, r)$ where both $n, r$ are positive integers.

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Hypergeometricx
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1 Answers1

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By scaling the triple we just need that: $$ \frac{r}{n-r+1},1,\frac{n-r}{r+1} $$ are in arithmetic progression, or: $$ r(r+1), (r+1)(n-r+1), (n-r)(n-r+1). $$ This leads to the condition: $$ r(r+1)+(n-r)(n-r+1) = 2(r+1)(n-r+1) $$ that is equivalent to: $$ (2r-n)^2 = n+2.\tag{1} $$ This leads to $n=a^2-2$ and $|2r-n|=a$. Assuming $r\leq\frac{n}{2}$, we get: $$ n=a^2-2,\qquad r=\frac{n-a}{2}=\frac{a^2-a-2}{2}=\frac{(a+1)(a-2)}{2}.\tag{2}$$

Jack D'Aurizio
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  • Thank you. That's a very nice solution. I suppose we should also specify that $n\geq3$. – Hypergeometricx Nov 13 '14 at 02:12
  • Just to point out that after the first pair, $n$ increases by $7, 9, 11, 13,...$ whilst $r$ increases by $3, 4, 5, 6, 7, ...$. This is of course a result of the general formula. – Hypergeometricx Nov 13 '14 at 02:16