Formula for $\sum_{k=0}^n {n \choose k}^4$

Question

May I know where I can read/find simpler formula for $$\sum_{k=0}^n {n \choose k}^4?$$

I tried to use generating function trick just like in the squared case, but I lost it as the multiplication is too much..and I feel like it is not gonna work the same but I tried it anyway xD

Any suggestion is appreciated. Thank you!

Answer 1

This is A005260 in the OEIS. The nicest expression they give (in my opinion) is this recursive one: $$ a_{n} = \frac{1}{n^3}\left((4n - 2)(3n^2 - 3n + 1)\cdot a_{n-1} + (4n - 3)(4n - 4)(4n - 5)\cdot a_{n-2}\right) $$ for $n \ge 2$ (with $a_0=1$ and $a_1=2$).

Answer 2

If you want an approximation, you could use $$\log\left( \sum_{k=0}^n {n \choose k}^4\right)\sim \frac{27608 }{9977}n+\frac{2 }{2669}n \log (n)-\frac{3919 }{2839}\log (n)-\frac{15533}{9047}$$

For $n=50$, the exact value is $$1141414024819789256187074524647238375882479333861915381768$$ while this approximation would give $$1150625550178599919625707223110892786861320701420884344305$$