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Atoning (I hope) for my silly previous question, are there infinitely many equalities of the form $\binom{n}{k}=\binom{n'}{k'}$ with $k,k' \geq 2$, $n \geq 2k$, $n' \geq 2k'$ and $n \neq n'$?

EDIT: According to the Wikipedia article on Singmaster's Conjecture, yes there are: there are infinitely many solutions to $\binom{n}{k+2}=\binom{n+1}{k+1}$.

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Alon Amit
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  • Perhaps you can answer the question yourself so that it does not remain unanswered. I think I first saw the conjecture here: http://mathoverflow.net/questions/75698/examples-of-seemingly-elementary-problems-that-are-hard-to-solve/75721#757217 – Eric Naslund Nov 25 '11 at 11:13
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    The link to the article about Singmaster's conjecture is probably as good an answer to this question as anybody knows. – Michael Hardy Nov 25 '11 at 16:10

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So as not to leave the question unanswered: Yes, there are infinitely many such coincidences. An infinite family is given by

$\binom{n}{k+2} = \binom{n+1}{k+1}$

with $n = F_{2i+2}F_{2i+3}-1$, $k = F_{2i}F_{2i+3}-1$ and $F_n$ is the Fibonacci sequence indexed such that $F_1=F_2=1$.

Alon Amit
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