Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$

Question

Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ (other than the trivial case $p=0$).

$$p=0\colon\,\sum_{k=0}^n\binom{n}kk^0=2^n$$

I know that $\sum_{k=0}^n\binom{n}k=2^n$ and $\sum_{k=0}^nk^n=\frac{k^{n+1}-1}{k-1}$ but I am unsure of whether these would be of much use now.

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Additionally, what about the similar series $\sum_{k=0}^n\binom{n}kk^n$ where $p=n$?

Answer 1

If you consider $p$ as fixed, then the below can be considered as closed form I suppose:

$$\sum_{k=0}^{n} \binom{n}{k} k^p = \sum_{k=1}^{p} S(p,k) n(n-1)\dots(n-k+1) 2^{n-k} \quad \quad (1)$$

where $S(k,p)$ is a Stirling number of the second kind.

If you denote the operator of differentiating and multiplying by $x$ as $D_{x}$

Then we have that

$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} S(n,k) f^{k}(x) x^{k}$$

where $S(n,k)$ is the Stirling number of the second kind and $f^k(x)$ is the $k^{th}$ derivative of $f(x)$.

This can easily be proven using the identity $$S(n,k) = S(n-1,k-1) + k \cdot S(n-1,k)$$

To prove $(1)$ above, we apply $D_x$, $p$ times to $(1+x)^n$, and set $x=1$.

Answer 2

Let $$ f(x)=(e^x+1)^n=\sum_{k=0}^n \binom{n}{k}e^{kx}. $$

Then $$ \left(\frac{d}{dx}\right)^p f(x)=\sum_{k=0}^n\binom{n}{k}k^pe^{kx}.$$

Plug in $x=0$.

Answer 3

We know that $(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$.Differentiate this, to get $n(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k} k x^{k-1}$. Multiply by $x$ to get $nx(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k} k x^k$. Take $x=1$ to get the first sum, And repeat this process for sums involving higher powers of $k$.

Answer 4

Use the identity $k\dbinom{n}k=n\dbinom{n-1}{k-1}$: for $p=1$ you get

$$\sum_k\binom{n}kk=n\sum_k\binom{n-1}{k-1}=n\sum_k\binom{n-1}k=n2^{n-1}\;.$$

For $p=2$:

$$\begin{align*} \sum_k\binom{n}kk^2&=n\sum_k\binom{n-1}{k-1}k\\ &=n\sum_k\binom{n-1}k(k+1)\\ &=n\sum_k\binom{n-1}kk+n\sum_k\binom{n-1}k\\ &=n(n-1)\sum_k\binom{n-2}{k-1}+n2^{n-1}\\ &=n(n-1)\sum_k\binom{n-2}k+n2^{n-1}\\ &=n(n-1)2^{n-2}+n2^{n-1}\;. \end{align*}$$

If you carry out the same sort of computation with $p=3$, you get

$$\sum_k\binom{n}kk^3=n(n-1)(n-2)2^{n-3}+2n(n-1)2^{n-2}+n2^{n-1}\;,$$

which can be written with falling factorials as $$\sum_k\binom{n}kk^3=n^{\underline3}2^{n-3}+3n^{\underline2}2^{n-2}+n^{\underline1}2^{n-1}\;.$$

After a little more experimentation one may conjecture and prove by induction that

$$\sum_k\binom{n}kk^p=\sum_{k=1}^p{p\brace k}n^{\underline k}2^{n-k}\;,$$

where $p\brace k$ is the Stirling number of the second kind.

Answer 5

A relate problem. Try this formula $$ \sum_{k=0}^n\binom{n}kk^p= 2^n\sum_{k=0}^{p}\begin{Bmatrix} p\\k \end{Bmatrix} {n\choose k}2^{-k}k!, $$

where $p \in \mathbb{N}$ and $\begin{Bmatrix} p\\k \end{Bmatrix}$ is the Stirling numbers of the second kind. You can plug in $p=n$ in the above formula.

Answer 6

Suppose we seek to evaluate $$\sum_{k=0}^n {n\choose k} k^p.$$

Introduce $$k^p = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(kz) \; dz.$$

This yields for the sum $$\frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \sum_{k=0}^n {n\choose k} \exp(kz) \; dz \\ = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} (1+\exp(z))^n \; dz \\ = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} (2+\exp(z)-1)^n \; dz \\ = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \sum_{q=0}^n {n\choose q} (\exp(z)-1)^q 2^{n-q} \; dz \\ = \sum_{q=0}^n {n\choose q} \times q! \times 2^{n-q} \times \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \frac{(\exp(z)-1)^q}{q!} \; dz.$$

Recall the species equation for labelled set partitions: $$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$

which yields the bivariate generating function of the Stirling numbers of the second kind $$\exp(u(\exp(z)-1)).$$

Substitute this into the sum to get $$\sum_{q=0}^n {n\choose q} \times q! \times 2^{n-q} \times {p\brace q}$$

Now observe that when $n\gt p$ the Stirling number is zero for the values $p\lt q \le n$ so we may replace $n$ by $p.$ On the other hand when $n\lt p$ the binomial coefficient is zero for the values $n\lt q \le p$ so we may again replace $n$ by $p.$ This finally yields

$$\sum_{q=0}^p {n\choose q} \times q! \times 2^{n-q} \times {p\brace q}$$

as observed by the other contributors.

Answer 7

We have:

$$\sum_{k=0}^n {n\choose k} k^p = \sum_{k=0}^n {n\choose k} \left[\sum_{j=0}^p j! {k\choose j} {p\brace j}\right] = \sum_{j=0}^p j! \left[\sum_{k=j}^n {n\choose k} {k\choose j} \right] {p\brace j} = \sum_{j=0}^p j! 2^{n-j}{n\choose j} {p\brace j} ,$$

with $p\brace j$ being the Stirling number of the second kind.