I wanted to split the expression $r^k (r+n)!$ as a sum of factorials, where $k ,n \ \in \ \mathbb{Z} \ ;\ k>0$.
For example,
-
$r(r+n)! = (r+n+1)! – (n+1)(r+n)!$
-
$r^2(r+n)! = (r+n+2)! – (2n+3)(r+n+1)! + (n+1)^2 (r+n)!$
And in general,
If $$ r^k (r+n)! = \sum_{m=0}^{k} \lambda_{m} (r+n+m)! $$
where $ \lambda_{m} $ is independent of $r$, find a closed form expression for $\lambda_{m}$.
My Try :
1) I tried to form a two variable recurrence for the expression and solve it such that we can express it in a summation form.
Let $r^k (r+n)! = A(k,n)$
Since $ r^k(r+n)! = r^{k-1}(\overline{r+n+1} – \overline{n+1})(r+n)! = r^{k-1}(r+n+1)! – r^{k-1} (n+1)(r+n)! $
$ \implies A(k,n) = A(k-1 , n+1) – (n+1)A(k-1,n) $
we can fix the initial conditions as $A(k,0) = r^k r!$ and $A(0,n) = (r+n)!$. I tried solving the above using generating functions, but eventually we'll have to find a Mac Lauren Series expansion for $\dfrac{1}{\Gamma \left( n+1 + \frac{2}{x} \right)}$
2) The expression can be rewritten as $$ r^k = \sum_{m=0}^{k} \lambda_{m} (r+n+m)_{m} $$
where $(x)_{y}$ denotes the falling factorial. This resembles the identity $$ r^k = \sum_{m=0}^{k} {k\brace m} (r)_{m} $$
where $ \displaystyle {a\brace b} $ denotes the Stirling Numbers of the Second Kind. So maybe the the coefficients can be expressed in terms of Stirling Numbers.
3) I also tried plugging in values of $r$ and making a system of $k+1$ equations, but it became tedious to solve.
Update (26th November, 2016) : The user @Marko Riedel has discovered a remarkable closed form expression, namely
$$ \lambda_m = (-1)^{k+m}
\sum_{p=0}^{k-m} {k\choose p} {k+1-p\brace m+1} n^p $$
I have put a bounty since I feel that there might be alternate solutions to prove (or maybe even simplify) the above proposition. I'm also looking for any references/links to this problem.
Also, the great answers so far reveal that the identity is valid even when $n$ is not an integer.
Any help will be greatly appreciated.
Best Answer
We are given that
$$r^k (r+n)! = \sum_{m=0}^k \lambda_m (r+n+m)!$$
and seek to determine the $\lambda_m$ independent of $r.$ We claim and prove that
$$\lambda_m = (-1)^{k+m} \sum_{p=0}^{k-m} {k\choose p} {k+1-p\brace m+1} n^p.$$
With this in mind we re-write the initial condition as
$$r^k = \sum_{m=0}^k \lambda_m m! {r+n+m\choose m}.$$
We evaluate the RHS starting with $\lambda_m$ using the EGF of the Stirling numbers of the second kind which in the present case says that
$${k+1-p\brace m+1} = \frac{(k+1-p)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2-p}} \frac{(\exp(z)-1)^{m+1}}{(m+1)!} \; dz.$$
We obtain for $\lambda_m$
$$(-1)^{k+m} \sum_{p=0}^{k-m} n^p {k\choose p} \frac{(k+1-p)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2-p}} \frac{(\exp(z)-1)^{m+1}}{(m+1)!} \; dz.$$
The inner term vanishes when $p\ge k+2$ but in fact even better it also vanishes when $p\gt k-m$ which implies $m+1\gt k+1-p$ because $(\exp(z)-1)^{m+1}$ starts at $[z^{m+1}]$ and we are extracting the term on $[z^{k+1-p}].$
Hence we may extend $p$ to infinity without picking up any extra contributions to get
$$(-1)^{k+m} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} \frac{(\exp(z)-1)^{m+1}}{(m+1)!} \sum_{p\ge 0} (k+1-p) \frac{n^p z^p}{p!} \; dz.$$
This is
$$(-1)^{k+m} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} \frac{(\exp(z)-1)^{m+1}}{(m+1)!} ((k+1)-nz) \exp(nz) \; dz.$$
Substitute this into the outer sum to get
$$(-1)^{k} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(nz) \\ \times \sum_{m=0}^k {r+n+m\choose m} (-1)^m \frac{(\exp(z)-1)^{m+1}}{m+1} \; dz.$$
We have
$${r+n+m\choose m} \frac{1}{m+1} = {r+n+m\choose m+1} \frac{1}{r+n}$$
and hence obtain
$$\frac{(-1)^{k}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(nz) \\ \times \sum_{m=0}^k {r+n+m\choose m+1} (-1)^m (\exp(z)-1)^{m+1} \; dz.$$
We may extend $m$ to $m\gt k$ in the remaining sum because the term $(\exp(z)-1)^{m+1}$ as before starts at $[z^{m+1}]$ which would then be $\gt k+1$ but we are extracting the coefficient on $[z^{k+1}],$ which makes for a zero contribution.
Continuing we find
$$-\sum_{m\ge 0} {r+n+m\choose r+n-1} (-1)^{m+1} (\exp(z)-1)^{m+1} \\ = 1 - \frac{1}{(1-(1-\exp(z)))^{r+n}} = 1 - \exp(-(r+n)z).$$
We get two pieces on substituting this back into the main integral, the first is
$$\frac{(-1)^{k}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(nz) \; dz \\ = \frac{(-1)^{k}}{r+n} (k+1)! \frac{n^{k+1}}{(k+1)!} - \frac{(-1)^k}{r+n} k! n \frac{n^{k}}{k!} = 0.$$
and the second is
$$\frac{(-1)^{k+1}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(nz) \exp(-(r+n)z) \; dz \\ = \frac{(-1)^{k+1}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(-rz) \; dz \\ = \frac{(-1)^{k+1}}{r+n} (k+1)! \frac{(-r)^{k+1}}{(k+1)!} - \frac{(-1)^{k+1}}{r+n} k! n \frac{(-r)^k}{k!} \\ = \frac{1}{r+n} (k+1)! \frac{r^{k+1}}{(k+1)!} + \frac{1}{r+n} k! n \frac{r^k}{k!} \\ = \frac{1}{r+n} r^{k+1} + \frac{1}{r+n} n r^k = r^k.$$
This concludes the argument.
Addendum Nov 27 2016. Markus Scheuer proposes the identity
$$\lambda_m = (-1)^{m+k} \sum_{p=m}^k {p\brace m} {k\choose p} (n+1)^{k-p}.$$
To see that this is the same as what I presented we extract the coefficient on $[n^q]$ to get
$$(-1)^{m+k} \sum_{p=m}^k {p\brace m} {k\choose p} {k-p\choose q}.$$
Now we have
$${k\choose p} {k-p\choose q} = \frac{k!}{p! q! (k-p-q)!} = {k\choose q} {k-q\choose p}.$$
We get
$$(-1)^{m+k} {k\choose q} \sum_{p=m}^k {p\brace m} {k-q\choose p}.$$
We now introduce with deployment of the Egorychev method in mind
$${k-q\choose p} = {k-q\choose k-q-p} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-q-p+1}} (1+z)^{k-q} \; dz.$$
This certainly vanishes when $p\gt k-q$ so we may extend $p$ to infinity, getting for the sum
$$(-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-q+1}} (1+z)^{k-q} \sum_{p\ge m} {p\brace m} z^p \; dz.$$
Using the OGF of the Stirling numbers of the second kind this becomes
$$(-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-q+1}} (1+z)^{k-q} \prod_{l=1}^m \frac{z}{1-lz} \; dz.$$
Now put $z/(1+z) = w$ to get $z = w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$ to obtain
$$(-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q}} \frac{1-w}{w} \frac{1}{(1-w)^2} \prod_{l=1}^m \frac{w/(1-w)}{1-lw/(1-w)} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+1}} \frac{1}{1-w} \prod_{l=1}^m \frac{w}{1-w-lw} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+1}} \frac{1}{1-w} \prod_{l=1}^m \frac{w}{1-(l+1)w} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+2}} \frac{w}{1-w} \prod_{l=2}^{m+1} \frac{w}{1-lw} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+2}} \prod_{l=1}^{m+1} \frac{w}{1-lw} \; dw \\= (-1)^{m+k} {k\choose q} {k-q+1\brace m+1}.$$
This is the claim and we are done.