New wrong recurrence formula for Bell numbers Bell numbers are the numbers counting the total partitions on a set with $n$ distinct elements.
Explanation:
Consider a set like $A:=\left\{x_{1},x_{2},...,x_{n}\right\}$
A partial equivalence relation is either reflexive or it is not, so the number of partial equivalence relation on a set with cardinality $n$ is $B_n$ plus the relations which are partial equivalence relation but not reflexive, to make these relations we can consider that a set with $n$ element cannot be reflexive if at least one of the ordered pairs $(x_i,x_i)$ is not in that relation,so at first we can remove one of the ordered pairs  $(x_i,x_i)$ $(0\le i\le n)$ and continue till we have $n-1$ ordered pair in the form $(x_i,x_i)$ removed (It cannot be $n$ because if we have $n$ ordered pairs then we are counting a relation which has been already counted by $B_n$)
Also given these kind of the partial equivalence relation, they can be combined with each other to make another   equivalence relation that is not reflexive.
For example for $A=\left\{1,2,3\right\}$ the total conditions that a partial equivalence relation is not reflexive is:
$$\left\{\right\}\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}\choose{0}}$$
$$\left\{\left(1,1\right)\right\}\left\{\left(2,2\right)\right\}\left\{\left(3,3\right)\right\}\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}\choose{1}}$$
$$\left\{\left(11\right),\left(22\right)\right\}\left\{\left(1,1\right),\left(3,3\right)\right\}\left\{\left(2,2\right),\left(3,3\right)\right\}\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}\choose{2}}$$
$$\left\{\left(11\right),\left(22\right),\left(12\right),\left(21\right)\right\}\left\{\left(1,1\right),\left(3,3\right),\left(13\right),\left(31\right)\right\}\left\{\left(2,2\right),\left(3,3\right),\left(23\right),\left(32\right)\right\}\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}\choose{2}}{{2}\choose{2}}$$
The relations in the last row are made form combining the second and third row, the number of choosing $2$ element form the set $A$ is ${{3}\choose{2}}$ ( indeed the number of way to choose $2$ of these elements to make an ordered pair in the form  $(x_i,x_i)$) and there is ${{2}\choose{2}}$ ways to make a new partial equivalence relation with the $2$ elements.
If we sum the number of these kind of partial equivalence relations that are not reflexive with those partial equivalence relations which are reflexive we get :
$$\color{blue}{9+B_3}=10+5=15=\color{blue}{B_4}$$ 
Which is indeed the number of partial equivalence relations on $A$.
I used this strategy and tried for $n=4$, finally could derive the recurrence formula for Bell numbers:
$$2^{n}-1+\sum_{k=2}^{n-1}\sum_{m=2}^{k}{{n}\choose{k}}{{k}\choose{m}}+B_n=B_{n+1}$$
Mapping $k-2\mapsto k$ and $m-2\mapsto m$  equivalently the formula can be rewritten as:

$$2^{n}-1+\sum_{k=0}^{n-3}\sum_{m=0}^{k+2}{{n}\choose{k+2}}{{k+2}\choose{m+2}}+B_n=B_{n+1}$$

With the initial value $B_0=1$ the formula gives:
$$0+B_{0}=0+1=\color{blue}{1}=\color{blue}{B_{1}}$$
$$1+B_{1}=1+1=\color{blue}{2}=\color{blue}{B_{2}}$$
$$3+B_{2}=3+2=\color{blue}{5}=\color{blue}{B_{3}}$$
$$10+B_{3}=10+5=\color{blue}{15}=\color{blue}{B_{4}}$$
$$37+B_{4}=37+15=\color{blue}{52}=\color{blue}{B_{5}}$$
$$136+B_{5}=136+52=\color{red}{188}\ne \color{red}{B_{6}}$$
The formula does not give the right number for $B_{6}$, but I'm sure the validity of the other Bell numbers is not accidental, so why the formula gives such a wrong number? where I was wrong?
Finally I should say that I came up with this calculation, because I cannot understand why the total number of partial equivalence relations on a set with cardinality $n$ is $B_{n+1}$, so It would be really appreciated if someone explain that with more details.
 A: First, the number of PER's on $\{1,\ldots,n\}$ is indeed $B_{n+1}$. To show this, we construct a bijection between PER's on $\{1,\ldots,n\}$ and equivalence relations on  $\{0,1,\ldots,n\}$: given a PER, keep the existing equivalence classes and put the non-reflexive elements in a new equivalence class with $0$ (if there are none, put $0$ in an equivalence class on its own).
So if your formula breaks down, it must be getting the number of non-reflexive PER's wrong, but only for $n$ at least $5$. In fact, it appears you are only counting the number of non-reflexive PER's which have at most one non-singleton class (in your formula, $k+2$ is the number of reflexive elements, and $m+2$ is the number of these in the large equivalence class). If $n\leq 4$ this covers all the non-reflexive PER's, but if $n=5$ you will miss out PER's with two classes of size $2$ and one irreflexive element, that is  $$\{(2,2),(2,3),(3,2),(3,3),(4,4),(4,5),(5,4),(5,5)\}$$
and isomorphic PER's. There are $15$ PER's of this form ($5$ ways to choose the irreflexive element and then $3$ ways to pair off the remaining four), which is exactly the discrepancy. For larger numbers, the number of missing PER's will grow very quickly.
