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I got this question in my mind when I was working on a solution to factorial recurrence and came up with this recurrence relation: $$(2n)!=\binom{2n}{n}(n!)^2$$ which made me wonder: is there also a recurrence relation for $\tbinom{4n}{2n}$ in terms of $\tbinom{2n}{n}$? Please use no factorials greater than $(2n)!$, preferably not greater than $n!$.

Brian J. Fink
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    I'm not sure that fits the definition of a recurrence relation (i.e. $a_n = c_1a_{n-1}+\cdots+c_{n-1}a_1$). – Patrick Feb 12 '14 at 21:39
  • @Patrick That's a linear recurrence relation. To me, a recurrence relation is roughly anything of the form $\text{next thing}=f(\text{previous things})$. I don't see how the OP's example fits that either, though. – Jack M Feb 12 '14 at 22:06
  • @Jack M, True. To the OP take my "i.e." to be an "e.g." then. – Patrick Feb 12 '14 at 22:33
  • @Patrick this is binary recurrence. – Brian J. Fink Feb 12 '14 at 22:51

3 Answers3

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Here is an estimate that gives a good approximation of $\binom{4n}{2n}$ in terms of $\binom{2n}{n}$.

Using the identity $$ (2n-1)!!=\frac{(2n)!}{2^nn!}\tag{1} $$ it is straightforward to show that $$ \frac{\binom{4n}{2n}}{\binom{2n}{n}}=\frac{(4n-1)!!}{(2n-1)!!^2}\tag{2} $$ Notice that $$ \begin{align} \frac{(2n-1)!!}{2^nn!} &=\frac{2n-1}{2n}\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\cdots\frac12\\ &=\frac{n-\frac12}{n}\frac{n-\frac32}{n-1}\frac{n-\frac52}{n-2}\cdots\frac{\frac12}{\;1}\\ &=\frac1{\sqrt\pi}\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}\tag{3} \end{align} $$ By Gautschi's Inequality, we have $$ \frac1{\sqrt{n+1}}\le\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}\le\frac1{\sqrt{n}}\tag{4} $$ Thus, $(3)$ and $(4)$ yield $$ \frac1{\sqrt{\pi(n+1)}}\le\frac{(2n-1)!!}{2^nn!}\le\frac1{\sqrt{\pi n}}\tag{5} $$ and $$ \frac1{\sqrt{\pi(2n+1)}}\le\frac{(4n-1)!!}{2^{2n}(2n)!}\le\frac1{\sqrt{\pi 2n}}\tag{6} $$ Dividing $(6)$ by $(5)$ gives $$ \sqrt{\frac{n}{2n+1}}\le\frac{(4n-1)!!}{(2n-1)!!^2}4^{-n}\le\sqrt{\frac{n+1}{2n}}\tag{7} $$ Combine $(2)$ and $(7)$ to get $$ 4^n\sqrt{\frac{n}{2n+1}}\le\frac{\binom{4n}{2n}}{\binom{2n}{n}}\le4^n\sqrt{\frac{n+1}{2n}}\tag{8} $$ or asymptotically $$ \binom{4n}{2n}\sim\frac{4^n}{\sqrt2}\binom{2n}{n}\tag{9} $$

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robjohn
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  • Sorry but I'm looking for something I can plug into an exact formula, not an approximate one. – Brian J. Fink Feb 12 '14 at 23:29
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    @BrianJ.Fink: It might help if you specified exactly what you want. That way, time would not be wasted on formulas that don't exist. – robjohn Feb 12 '14 at 23:32
  • I appreciate your effort. Since it turns out that the one combinatorial does not divide the other, it seems approximations may be my only way of implementing this. Well, back to the drawing board! – Brian J. Fink Feb 12 '14 at 23:37
  • Can you give me that for $\binom{2n}{n}$ in terms of $\binom{n}{n/2}$? – Brian J. Fink Feb 12 '14 at 23:56
  • @BrianJ.Fink: just substitute $n\mapsto n/2$: $$\binom{2n}{n}\sim\frac{2^n}{\sqrt2}\binom{n}{n/2}$$ – robjohn Feb 13 '14 at 00:05
  • About how accurate is this formula? – Brian J. Fink Feb 13 '14 at 00:16
  • I see. I tried $\binom{8}{4}=70\approx 67.88225$? Needs a little improvement, I suspect. – Brian J. Fink Feb 13 '14 at 00:25
  • @BrianJ.Fink: The error bounds are given in the previous line: $$\frac{4^n}{\sqrt2} \binom{2n}{n} \sqrt{\frac{2n}{2n+1}} \le\binom{4n}{2n} \le\frac{4^n}{\sqrt2} \binom{2n}{n}\sqrt{\frac{n+1}{n}}$$ Plug in $n=2$ and we get $$\frac{16}{\sqrt2}\binom{4}{2}\sqrt{\frac45} \le \binom{8}{4} \le\frac{16}{\sqrt2}\binom{4}{2}\sqrt{\frac32}$$ $$60.7157310752329\le70\le83.1384387633061$$ – robjohn Feb 13 '14 at 00:52
  • @BrianJ.Fink: As $n$ gets bigger, the error factors in the square roots get closer to $1$; e.g. plug in $n=10$ $$\frac{1048576}{\sqrt2}\binom{20}{10}\sqrt{\frac{20}{21}} \le \binom{40}{20} \le\frac{1048576}{\sqrt2}\binom{20}{10}\sqrt{\frac{11}{10}}$$ $$133686889002.415\le137846528820\le143674537953.916$$ – robjohn Feb 13 '14 at 00:52
  • @BrianJ.Fink: the formula is guaranteed to be between $\sqrt{\frac{2n}{2n+1}}$ and $\sqrt{\frac{n+1}{n}}$ times the actual value. – robjohn Feb 13 '14 at 00:58
  • Well, I wish there was an approximation that was accurate to the one's digit minimum, and got better from there. – Brian J. Fink Feb 13 '14 at 01:16
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I think this is what you are looking for (but this is just back substitution):

$$\binom{4n}{2n} = \frac{(4n)!}{((2n)!)^2} = \frac{(4n)!}{\binom{2n}{n}^2 (n!)^4}.$$

Please update me on whether this is what you're looking for...

Brian J. Fink
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Christopher K
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  • Actually I'm trying to take a number that's in the form $n=2^k$ and express $n!$ in terms of $\tfrac{n}{2}!$, so expressing it in terms of $(4n)!$ is incomplete. – Brian J. Fink Feb 12 '14 at 21:51
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There can't be any multiplicative formula of the sort you describe for $4n\choose 2n$ in terms of $2n\choose n$, because there are prime factors of the former that aren't factors of the latter. By Bertrand's Postulate there's a prime number between $k=2n$ and $2k$($=4n$); that prime will be a factor of $4n\choose 2n$, but can't be a factor of $2n\choose n$ or any factorial of the form $(2n)!$ or less.

Steven Stadnicki
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