I am currently working on a problem related to graph theory an I came across this sum.

$ {\displaystyle a(n)=\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {n!}{2^{k}(n-2k)!k!}}.} $

Can someone tell me if it is possible to get a formula for finding the nth term of this series?


Computing for the first terms, the numbers are $$\{1,2,4,10,26,76,232,764,2620,9496,35696,140152,568504,2390480,10349536\}$$ This is sequence $A000085$ at $OEIS$. They seem to be somehow related to restricted Stirling numbers of the second kind.

Looking at the page, you will find soem interesting asymptotics.

  • $\begingroup$ OEIS, the greatest site when trying to find a closed form of a sequence or any sort of information about it! $\endgroup$ – Jakobian Jun 26 at 11:38
  • $\begingroup$ There is no closed form for this sequence on OEIS $\endgroup$ – user677887 Jun 28 at 10:28

While there's a lot of ways to find a closed form for combinatoric sums (including the most practical one of looking the terms up in OEIS, as Claude Leibovici had), I'd like to remind the more general way of using generalized hypergeometric functions.

Even though the sum is finite, we can easily treat it as a particular case of hypergeometric series.

$$a_n(x)=\sum _{k=0}^{\lfloor n/2\rfloor} \frac{n!}{(n-2k)!} \frac{x^k}{k!}$$

Naming the general term $c_k$ we consider the ratio:

$$\frac{c_{k+1}}{c_k}=(n-2k)(n-2k-1) \frac{x}{k+1}=\left(k-\frac{n}{2} \right)\left(k-\frac{n-1}{2} \right)\frac{4x}{k+1}$$


Using this ratio, we can immediately write:

$$a_n(x)={_2 F_0} \left(-\frac{n}{2},-\frac{n-1}{2};;4x \right)$$

We consider the case:

$$a_n \left( \frac12 \right) ={_2 F_0} \left(-\frac{n}{2},-\frac{n-1}{2};;2 \right)$$

In Mathematica or Wolfram Alpha, we can evaluate this function using the code:



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