# Asymptotics of $\frac{\sum _{i=0}^{\lfloor n/2 \rfloor} {2(n-2i) \choose n-2i} {n \choose 2i} {4i \choose 2i}}{2^{3n - 1}}$, is it $\frac{2}{\pi n}$?

I am trying to work out the asymptotics of $$\frac{\sum _{i=0}^{\lfloor n/2 \rfloor} {2(n-2i) \choose n-2i} {n \choose 2i} {4i \choose 2i}}{2^{3n - 1}}.$$

My numerical experiments suggest it might be $\frac{2}{\pi n}$.

The only thing I know to do is to apply Stirling's formula. So we get ${4i \choose 2i} \sim 2^{4i+1/2}/\sqrt{2i\pi}$ I think. Similarly we get ${2(n-2i) \choose n-2i} \sim 2^{(n-2i)+1/2}/\sqrt{(n-2i)\pi}$ I think. Even assuming these are right, I not sure what to do with ${n \choose 2i}$.

• Summands $a_i$ attain their maximum at $i=n/4$. So the numerator can be estimated from above as $a_{n/4}\times \frac{n}{2}$. The lower bound is obviously $a_{n/4}$ – Norbert Jun 11 '15 at 13:45
• Numerical evaluations confirm your initial suspicions. – Lucian Jun 11 '15 at 16:21
• @Lucian Thank you very much for checking! – felipa Jun 11 '15 at 16:33
• @felipa: Have you tried writing $\displaystyle{n\choose2i}\le{n\choose[n/2]}$ ? – Lucian Jun 11 '15 at 16:55
• After applying Lucian's recommendation, Vandermonde's identity will simplify the summation, and what's left should go through easily after applying Stirling's Formula (I think). – TokenToucan Jun 11 '15 at 17:56

Here is an answer that evaluates part of the sum and locates the OEIS entry for the remaining one, thereby it is hoped facilitating further investigation.

Suppose we are interested in the asymptotics of $$\sum_{q=0}^{\lfloor n/2\rfloor} {n\choose 2q} {4q\choose 2q} {2n-4q\choose n-2q}.$$

Introduce $${4q\choose 2q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{4q}}{z^{2q+1}} \; dz$$

and $${2n-4q\choose n-2q} = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{2n-4q}}{w^{n-2q+1}} \; dw.$$

This gives for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{2n}}{w^{n+1}} \sum_{q=0}^{\lfloor n/2\rfloor} {n\choose 2q} \frac{(1+z)^{4q}}{z^{2q}} \frac{w^{2q}}{(1+w)^{4q}} \; dw \; dz$$

This has two pieces, namely $$A_1 = \frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{2n}}{w^{n+1}} \left(1+\frac{(1+z)^{2}}{z}\frac{w}{(1+w)^2}\right)^n \; dw \; dz$$ and $$A_2 = \frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{2n}}{w^{n+1}} \left(1-\frac{(1+z)^{2}}{z}\frac{w}{(1+w)^2}\right)^n \; dw \; dz.$$

Start with $A_2$ to get $$\frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{2n}}{w^{n+1}} \frac{(z(1+w)^2-w(1+z)^2)^n}{z^n(1+w)^{2n}} \; dw \; dz \\ = \frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (z(1+w)^2-w(1+z)^2)^n \; dw \; dz \\ = \frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (wz-1)^n (w-z)^n \; dw \; dz.$$

Extracting the inner residue we get $$\sum_{q=0}^n {n\choose q} z^q (-1)^{n-q} {n\choose n-q} (-1)^q z^q \\ = \sum_{q=0}^n {n\choose q}^2 z^{2q} (-1)^{n-2q}.$$

Extracting the outer residue from this we get zero when $n$ is odd and $${n\choose n/2}^2$$ when $n$ is even, giving the sequence (up to the $1/2$ multiplier) $$0, 4, 0, 36, 0, 400, 0, 4900, 0, 63504, 0, 853776, 0, 11778624,\ldots$$ which points us to OEIS A002894.

This can be treated with Stirling to get $$\frac{2^{2n+1}}{\pi n}.$$

We see that on dividing by $2^{3n-1}$ and including the $1/2$ factor we get $$\frac{1}{2^{n-1}\pi n}$$ which is an asymptotically lower order term.

Unfortunately there seems not to be a closed form expression for $A_1$ which is $$\frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (z(1+w)^2+w(1+z)^2)^n \; dw \; dz$$ because the central term does not factor. We get the sequence (factor $1/2$ not included) $$4, 20, 112, 676, 4304, 28496, 194240, 1353508, 9593104, 68906320,\ldots$$ which points us to OEIS A081085.

Among the references listed therein we find that this sequence is asymptotic to $$\frac{2^{3n+1}}{\pi n}$$ which on including the $1/2$ multiplier from $A_1$ and dividing by $2^{3n-1}$ does indeed yield $$\frac{2}{\pi n}$$ as conjectured.