For each $n \in \mathbb{N}$ define $b(n)$ to be the smallest $m \in \mathbb{N}$ such that $n = {m \choose k}$ for some $k$, i.e. $b(n)$ is the first row in Pascal's triangle containing $n$. Question, what are the $n$ such that $b(n) = n$?

Ideas. I think that for any prime $p$, $b(p) = p$. I think it follows from the fact that $p$ is prime if and only if ${p \choose k} \equiv 0 \pmod p$ for all $0 < k < p$. I also guess that powers of primes $n = p^k$ also satisfy $b(n) = n$ but not sure how to prove this. Other numbers satisfy the property by inspection, for instance $12$ and $14$ so I wonder what properties are shared by all these numbers. Perhaps it may be easier to consider for which $n$, $b(n) < n$.

Any references to related results would be greatly appreciated.

  • 1
    $\begingroup$ Numbers not satisfying $b(n)=n$ are $\{6,10,15,20,21,28,35,36,\cdots\}$. See oeis.org for some references. $\endgroup$ Apr 24 '17 at 9:36
  • $\begingroup$ Your guesses are correct, if $n$ appears before row $n$ it will have the form $\binom{m}{k}$ where $2\le m\le m-2$ so $n$ will be possible to factor into at least two distinct factors which excludes $p^l$ if $p$ is prime. $\endgroup$
    – skyking
    Apr 24 '17 at 9:41

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