Power of natural numbers and [method of the second class of Stirling numbers]

Power of natural numbers and written in front

1. Both the Stirling number and the s mentioned in this article refer to the second class of Stirling numbers.
2. \ (s_{k,j}\) means the number of schemes for putting K-differentiated balls into J-box-free boxes [There is no empty box]

Derivation

For a \ (i^k\)

Can be specifically understood as having \ (i\) a different box, put \ (k\) a different ball into the box of the number of scenarios. [Allow empty box]

Now the number of scenarios that allow empty boxes to be converted into one ball for all boxes

First enumerate the number of boxes with at least one ball set to $ j$

Then for the existence of a J empty box scheme is \ (s_{k,j}*c_{i,j}*j!\)

For a \ (i^k\) , can be represented as \[\sum \limits _{j=1}^{i} s_{k,j}*c_{i,j}*j! \]

Then you can represent the scheme of \ (\sum \limits ^{n}_{i=0} i^k\) : \[\sum \limits _{i=0}^{n}\sum\limits _{j=1}^{i} s_{k,j}*c_{i, j}*j! \]

Then discuss the coefficients of \ (s_{k,j}\) and: \ (\sum \limits ^{n}_{i=0} c_{i,j} *j!\) , i.e. \ (j!* \sum\limits^{n}_{i=0} c_{i, j}\)

Known: \ (\sum \limits_{i=0}^{n}c_{i,j}=c_{n+1,j+1}\)

So \ (s_{j,k}\) has a coefficient of \ (j!*c_{n+1,j+1}\)

Then you can know \ (\sum \limits ^{n}_{i=0} i^k= \sum\limits_{j=1}^{n}s_{k,j}*j!*c_{n+1,j+1}\)

Then the pre-processing of the Stirling number and the number of combinations can be solved.

\ (by\) \ (zerokei\)

Power of natural numbers and [method of the second class of Stirling numbers]