How to get asymptotic approximations for complex summations like $S_n = \sum\limits_{i=0}^n {n \choose i}^2 {2(n-i)\choose (n-i)}$

Question

I asked a question about this summation (in the process of proving that a 3-d random walk is transient):

$$S_n = \sum\limits_{i=0}^n {n \choose i}^2 {2(n-i)\choose (n-i)}$$

in this post: Summation coming about in the process of solving a 3d random walk $\sum\limits_{i=0}^n {n \choose i}^2 {2(n-i)\choose (n-i)}$.

In the answer, Claude mentioned this sequence where it is buried the following approximation:

$$S_n \sim \frac{3^{2 n+\frac{3}{2}}}{4 \pi n}$$

This totally solved my problem and the desired result immediately popped out.

But since I was meaning it to be a complete end to end proof, the question remains how to come up with the approximation.

Answer 1

To save you the calculation of the hypergeometric function, take list of numbers given in $OEIS$, take their logarithms and use a linear regression , the prdictors being $n$ and $\log(n)$ $$\log(S_n)=a+ b\, n + c\, \log(n)$$ With an $R^2=0.99999984$, the resultsa are $$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & -1.06943 & 0.00713 & \{-1.08448,-1.05438\} \\ b & +2.18941 & 0.00100 & \{+2.18730,+2.19153\} \\ c & -0.89294 & 0.00729 & \{-0.90832,-0.87756\} \\ \end{array}$$

Now, search in the $ISC$ your prefered numbers for $$1.06943239808173345453079354467\cdots$$ $$2.18941203440648792537395963290\cdots$$ $$0.89293825070137998724963846596\cdots$$

Answer 2

I prefer to add a separate answer.

With a bit of heuristics plus the help of the $ISC$ for one of the constants

$$A_n=\log\Bigg(\, _3F_2\left(-n,-n,-n;1,\frac{1}{2}-n;\frac{1}{4}\right)\Bigg)$$ is $$A_n \sim -\frac{15 (2 \gamma -1)}{2+3 e^{\gamma }}+\Bigg(\frac{\pi }{4}+\frac{1}{100} \left(\Omega+e^{\frac{1}{e}+\frac{1}{\pi }}\right)\Bigg)\,n-\frac 12 \log(n)$$

For $n=456$, the above gives $e^A=1.3455\times 10^{159}$ while the value of the hypergeometric function is $1.3511\times 10^{159}$.

If the three parameter linear regression was made for $10\leq n \leq 10^4$ (step of $10$), the adjusted correlation coefficient would be $R^2=0.9999999999999943$.

So $$S_n \sim \binom{2 n}{n}e^{A_n}$$