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If we want a central binomial coefficient, for n greater than zero we have a ‘nice’ expression $${2n \choose n} =\prod_{k=1}^{n}{(4-\frac{2}{k})}$$ which is term-wise rational and produces each binomial coefficient matching ${2n \choose n}$ along the way.

But supposed we desired a similar expression for some other ‘linear pattern’ in Pascal’s triangle, like ${3n \choose n}$, or ${5n \choose 2n}$. Perhaps it is term-wise algebraic rather than rational, sans any ‘bespoke encoding of a sequence into a magic constant’ horsery. Can our desire be satisfied under any conditions? If so, how, and if not, why?

Edit:

Here is one other such expression I have been able to derive, in case anyone comes across this and is interested:

$${4n \choose 2n} = \prod_{k=1}^{n}{\frac{2(4k-1)(4k-3)}{k(2k-1)}}$$

And generally:

$${an \choose n} = \prod_{k=1}^{n}{\frac{a(k)_{a}}{(a-1)(k)_{(a-1)}}} $$

tuespetre
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    Something like ${3n\choose n}=\prod_{k=1}^n\frac{3n+1-k}{k} =\prod_{k=1}^n(\left(\frac{3n+1}{k}-1\right)$? – Hagen von Eitzen Apr 16 '19 at 03:15
  • That is a nice one, thank you for sharing! I am interested to know how that can be derived, not only for ${3n \choose n}$ but for any such ${an \choose bn}$ with $b < a$, especially if it is possible without the term depending on $n$. – tuespetre Apr 16 '19 at 04:11
  • I now see the general ‘multiplicative formula’ using a falling factorial, and how I could plug in the $a$ and $b$. I would still like to see if there is a way to avoid having the terms depend upon the upper bound. – tuespetre Apr 16 '19 at 05:25

2 Answers2

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From $${2n \choose n} =\prod_{k=1}^{n}{(4-\frac{2}{k})}$$ one can obviously get $${4n \choose 2n} =\prod_{k=1}^{2n}{(4-\frac{2}{k})}$$ and

$${6n \choose 3n} =\prod_{k=1}^{3n}{(4-\frac{2}{k})}$$ Which are not new formulas.

In general we have $$ {m \choose n} =\prod_{k=0}^{n-1}{(1+\frac{d}{n-k})} $$ where $d=m-n.$

Mohammad Riazi-Kermani
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It appears that for any $${an \choose bn}$$ with $0 \le b \le a$ we have the degree $(a-1)$ rational curve $$R_{a,b}(x) = \frac{(ax)^{\underline{a}}}{(bx)^{\underline{b}}((a-b)x)^{\underline{a-b}}}$$ such that $${a(n-1) \choose b(n-1)}R_{a,b}(n) = {an \choose bn} = \prod_{k=1}^{n} R_{a,b}(k).$$ I do not have a rigorous proof for this; I just discovered the pattern through computation.

tuespetre
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