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This question is motivated by a simple exercise in Peter Cameron's Combinatorics: Topics, Techniques, Algorithms:

A restaurant near Vancouver offered Dutch pancakes with ‘a thousand and one combinations’ of toppings. What do you conclude?

The intended solution (according to Cameron's website) is the following: since ${14 \choose 4}=1001$, most likely there were $14$ possible toppings and each serving of pancakes allowed the patron to choose $4$ toppings.

However, this begs a much more general question:

For a given positive integer $m$, how can we find all positive integers $n$ and $k$ such that ${n \choose k}=m$?

This seems like a very natural and obvious question to ask, but preliminary searches didn't yield much apart from specific values of $m$. Any insight to this question for more general classes of positive integers $m$, or links to relevant references, would be appreciated.

J. W. Tanner
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pyridoxal_trigeminus
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    See https://oeis.org/A003016 – jjagmath Mar 19 '24 at 02:42
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    We have $1001 = 7\cdot 11\cdot 13$. It is just the product of three relatively large primes, which means there are likely to be few ways to produce it as a binomial coefficient. The existence of $13$ means the top number is at least $13$, and the absence of $17$ almost means the top number is at most $16$. At this stage the existence of $7$ means the top number must acutally be at least $14$, since in $\binom{13}{k}$ the factor of $7$ always cancels out. And $\binom{16}k$ is always even (apart from $1$). Looking for combinations that leave the $11$ uncancelled makes the search go pretty fast. – Arthur Mar 19 '24 at 03:13
  • (This approach is very much not generalizable in a nice way, but I suspect that the intended solution to that particular problem would go along these lines.) – Arthur Mar 19 '24 at 03:14
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    It might be somewhat more reasonable to require at most $k$ toppings, rather than exactly $k$. However, it turns out the only solution to $\sum_{i=0}^k {n \choose i} = 1001$ turns out to be $n=1000$, $k=1$, and it would be hard for the restaurant to offer $1000$ toppings. – Robert Israel Mar 19 '24 at 03:27
  • For a given value of $~m,~$ my first step would be to determine the prime factorization of $~m.~$ My second step would be to focus on the highest power of prime $~p~$ dividing $~n!$. Then, I would try to determine under what circumstances $~\displaystyle \binom{n}{k}~$ would be divisible by $~p^\alpha,~$ but not divisible by $~p^{\alpha + 1}.~$ Personally, I have never attacked this problem, so my approach might well fail.See also Legendre's formula. – user2661923 Mar 19 '24 at 04:11
  • @Arthur I question your conclusion that the top number must be less than $~17.~$ Consider, for example $~\displaystyle \binom{34}{17},~$ or $~\displaystyle \binom{51}{34}.$ However, it may well be that your assertion is accurate, because (for example) $~\displaystyle \binom{34}{17}~$ is divisible by $~19, ~23, ~29, ~$ and $~31.$ So, setting $~r = \displaystyle \max[k,n-k],~$ apparently, the set $~{r+1, r+2, \cdots, n},~$ must contain no prime numbers absent from the prime factorization of $~m.$ – user2661923 Mar 19 '24 at 04:27
  • @user2661923 It isn't a full conclusion. Heck, $\binom{1001}1$ is obviously equal to $1001$, so it can't be. But it is a good liit on the initial search space. – Arthur Mar 19 '24 at 11:06

1 Answers1

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It turns out to be quite fast

If $k = 1$ then $n = m$

If $k > 1$ then $\frac{n^k}{k!} > m > \frac{(n-k)^k}{k!}$, therefore $(k!m)^{1/k} < n < (k!m)^{1/k}+k$

Notice that $\left(\begin{matrix}n\\k\end{matrix}\right) = \left(\begin{matrix}n\\n-k\end{matrix}\right)$ so we only need to check for $k \leq n/2$. This means $$2k < (k!m)^{1/k}+k$$ Therefore $$\frac{k^k}{k!} < m$$ As this grows quite fast for k (as shown in the below plot), only a handful of number of candidates need to be checked. Or one use Stirling approximation $$\frac{k^k}{k!} \approx \frac{e^k}{\sqrt{2\pi k}}$$

log plot of k^k/k!

UPDATE:

The range for possible values of k can be improved even further by directly using Stirling series:

$$ m = \left(\begin{matrix}n \\ k\end{matrix}\right) \geq \left( \begin{matrix}2k \\ k\end{matrix}\right) > \frac{4^k\left(1+\frac{1}{12k}\right)}{\sqrt{k\pi}\left(1+\frac{1}{12k}+\frac{1}{288k^2}\right)^2}$$

although it's probably faster to compute some values of $\left(\begin{matrix}2k\\k\end{matrix}\right)$ beforehand

log plot of gamma(2*n+1)/gamma(n+1)^2

ioveri
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    $+1$, nice approach! On the right you could use $-(k-1)$; that saves one check per $k$ value. And you could cut the number of checks roughly by half by using the AM/GM inequality: Since the correct numerator is the $k$-th power of the geometric mean of $n,n-1,\ldots,n-(k-1)$, the base on the left can be replaced by the arithmetic mean $n-\frac{k-1}2$. The search range for $n$ can thus be reduced to

    $$ (k!m)^{1/k}+\frac{k-1}2 < n < (k!m)^{1/k}+k-1;, $$

    with a corresponding slight improvement in the range for $k$,

    $$ \frac{(k+1)^k}{k!}\lt m;. $$

     – joriki Mar 19 '24 at 04:19
  • I've some further improvement using Stirling's formula. But the best strategy would be just precompute some values of $\left(\begin{matrix}2k\k\end{matrix}\right)$ – ioveri Mar 19 '24 at 05:43