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

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

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

Added 16 August 2024

In response to a question below I find that I no longer see how to carry out the suggested induction, if indeed I ever did, but the suggested calculations are still useful as a way to arrive at the conjecture $(1)$. The conjecture itself then has a fairly straightforward combinatorial proof.

Let $\mathscr{P}$ be the set of ordered pairs $\langle A,\varphi\rangle$ such that $A$ is a non-empty subset of $[n]=\{1,2,\ldots,n\}$ and $\varphi$ is a function from $[p]=\{1,2,\ldots,p\}$ to $A$; we’ll calculate $|\mathscr{P}|$ in two ways.

If $\varnothing\ne A\subseteq[n]$, and $|A|=k$, there are $k^p$ functions from $[p]$ to $A$. There are $\binom{n}k$ subsets of $[n]$ of cardinality $k$, so there are $\binom{n}kk^p$ pairs $\langle A,\varphi\rangle\in\mathscr{P}$ such that $|A|=k$, and it follows that

$$|\mathscr{P}|=\sum_{k=1}^n\binom{n}kk^p=\sum_k\binom{n}kk^p\,.$$

(The second equality holds because $0^p=0$ for $p\in\Bbb Z^+$.)

Alternatively, we can begin by noting that each $\varphi:[p]\to[n]$ induces a partition $\pi_\varphi$ of $[p]$ into the fibres of $\varphi$: $\pi_\varphi=\{\varphi^{-1}[\{a\}]:a\in[n]\}$. If $|\pi_\varphi|=k$, we let $\pi_\varphi=\{\pi_\varphi(1),\ldots,\pi_\varphi(k)\}$, where

$$\min\pi_\varphi(1)<\ldots<\min\pi_\varphi(k)\,.$$

At this point $\varphi$ is completely determined by the $k$-tuple $\langle\varphi(\min\pi_\varphi(1)),\ldots,\varphi(\pi_\varphi(k))\rangle$, which can be any $k$-tuple of distinct elements of $[n]$, so if $\pi$ is any partition of $[p]$ into $k$ parts, there are

$$n(n-1)\ldots(n-k+1)=n^{\underline k}$$

functions $\varphi:[p]\to[n]$ such that $\pi_\varphi=\pi$.

Let $\varphi$ be one of these $n^{\underline k}$ functions; clearly $\langle A,\varphi\rangle\in\mathscr{P}$ iff $\varphi\big[[p]\big]\subseteq A\subseteq[n]$, and there are $2^{n-k}$ such sets $A$, one for each subset of $[n]\setminus\varphi\big[[p]\big]$. And there are $n\brace k$ partitions of $[p]$ into $k$ parts, so there are

$$|\{\langle A,\varphi\rangle\in\mathscr{P}:|\varphi[A]|=k\}|={n\brace k}n^{\underline k}2^{n-k}\,,$$

and therefore

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

as desired.

Answer 5

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 combinatorial class equation for labelled set partitions: $$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z}))$$

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 6

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.

Answer 7

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.