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I am trying to solve/approximate $x$ in the following equation: $$\binom{x\cdot \log_2n}{\frac{x\cdot \log_2n}{2}} = n,$$ where $n$ is a natural number. Obviously $x$ is a decreasing function of $n$ and for $n=2$, $x=2$. But besides that and some hardcoded trials, I do not know how can I proceed. Some help here would be really awesome. Thanks!

gt6989b
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geosnow
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  • Set $2N=\log_2 n$ giving the nicer looking formula $\binom{x2N}{xN}=e^{2N}$ then use the classical approximation for the central binomial coefficient :$\binom{2k}{k} \approx \dfrac{4^k}{\sqrt{\pi k}}$. – Jean Marie Dec 30 '22 at 16:29
  • What do you want to do with your equation? You won't get to solve it in closed form, but the standard numerical methods should work fine. – Robert Israel Dec 30 '22 at 16:35
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    ... sorry, not $e^{2N}$ but $2^{2N}$ which is (interestingly !) $4^N$. At the end of the calculations, take the logarithls of both sides... – Jean Marie Dec 30 '22 at 16:37
  • Please share with us why are you interested with such strange non-integer indices. This site is based on "exchanges" as indicated in its name... – Jean Marie Dec 30 '22 at 16:39
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    @JeanMarie thank you for your reply, I want to calculate the number of logn binary sequences with same 0s and 1s such that it is equal to n. I think for my purpose it is enough that x belongs to the interval (1, 2] – geosnow Dec 30 '22 at 17:46

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