Does this sequence of inverse-binomial numbers have a name? I was inspired by considering the following:
$$\left(\sum_{i=1}^n i\right)^2=\sum_{i=1}^n i^3$$
Are there exact formulas for the sums of the powers of the integers?  For example, we have:
$$\sum_{i=1}^n i={n(n+1)\over 2}$$
$$\sum_{i=1}^n i^2={n(n+1)(2n+1)\over 6}$$
Do we have more of these?  Then I began to consider them using binomial notation:
$$\sum_{i=1}^n i={n(n+1)\over 2}={n+1\choose 2}$$
$$\sum_{i=1}^n i^2={n(n+1)(2n+1)\over 6}={n+1\choose 3}+{n\choose 3}$$
Upon further searching, I managed to come up with the following:
$$\sum_{i=1}^n i^3={n+2\choose 4}+4{n+1\choose 4}+{n\choose 4}={n+1\choose 2}^2$$
$$\sum_{i=1}^n i^4={n+3\choose 5}+11{n+2\choose 5}+11{n+1\choose 5}+{n\choose 5}$$
$$\sum_{i=1}^n i^5={n+4\choose 6}+26{n+3\choose 6}+66{n+2\choose 6}+26{n+1\choose 6}+{n\choose 6}$$
I know that these could also be written in terms of decreasing choice (i.e., $a{n\choose 5}+b{n\choose 4}+c{n\choose 3}+\dots$), but I wonder about the particular coefficients displayed here.  Note that these coefficients are the same ones that would be used if the formula were to generate the sequence of "powers of n" instead of the sum over them due to the identity ${n+1\choose k+1}-{n\choose k+1}={n\choose k}$.  I have found a recursive formula for them given exponent $k$:
$$a_k(i)=i^k-\sum_{j=1}^{i-1}a_k(j){k+i-j\choose k}$$
The first few sequences are:
1 = 1!
1 1 = 2!
1 4 1 = 3!
1 11 11 1 = 4!
1 26 66 26 1 = 5!

So my questions are:


*

*Does the sequence of coefficients for a given $k$ have a name?  I see that for each $k$ the sum over the sequence is $k!$.  Note that the Bernoulli numbers are not the same as these since these are constructed individually for each $k$, and for all $i\gt k, a_k(i)=0$.

*These sequences are symmetrical and appear to have properties very similar to that of Pascal's Triangle.  Is there a simple rule to generate them along the same lines?

 A: Using $\def\falling#1{^{\underline#1}}x\falling k=k!\binom xk=x(x-1)\ldots(x-k+1)$ (also written as $(x)_k$, but I find is unpleasant to read), one has
$$
  \def\stirs{\genfrac\{\}0{}}x^n=\sum_{k=0}^n \stirs nk x\falling k,
  \qquad\text{and hence}\quad
  x^n=\sum_{k=0}^n k!\stirs nk \binom xk
$$
where $\stirs nk$ denote the Stirling numbers of the second kind. The inverse transformation uses (signed) Stirling numbers of the first kind $\def\stirf{\genfrac[]0{}}(-1)^k\stirf nk$:
$$
  x\falling k=\sum_{k=0}^n (-1)^k\stirf nk x^k
  \qquad\text{and hence}\quad
  \binom nk=\sum_{k=0}^n \frac{(-1)^k}{k!}\stirf nk x^k.
$$
These numbers have many interesting combinatorial properties, but no simple closed formula.
Your question seems to pursue another course, the numbers $a_k(i)$ appear to be the Eulerian numbers $\genfrac<>0{}ki$.
A: Have a look at $\sum_{i=1}^{n}i\left(i-1\right)\cdots\left(i-k+1\right)$.
They are easier to deal with and the $\sum_{i=1}^{n}i^{k}$ can be deduced
from them. It becomes something like $\left(n+1\right)n\cdots\left(n-k+1\right)$ preceded by a constant and it can easily be verified with induction.
