Surjections

(
n
)

{\displaystyle k\_{(n)}}

k

(
n
)

{\displaystyle k\_{(n)}}

  A008279 injections

O
# e

∑
n

(
y
x
y

)
n

{\displaystyle Oe= \sum\_n(y+xy)^n}

e
# e
e
x
p
(
y
y
x
)

{\displaystyle ee = exp(y+yx)}

e
# O
e
x
p
(
y
)

∑
k

(
x
y

)
k

{\displaystyle eO = exp(y)\sum\_k(xy)^k}

k
!

(

n
k

)

{\displaystyle k! \binom n k}

T
(
n
,
k
# )
n
∗
T
(
n
−
1
,
k
−
1
# )
k
∗
T
(
n
−
1
,
k
−
1
)
T
(
n
−
1
,
k
)

{\displaystyle T(n, k) = n\*T(n-1, k-1) = k\*T(n-1, k-1)+T(n-1, k)}

all functions

k
n

{\displaystyle k^n}

O
G
e
# f
l
i
n
(
k
.
e
x
p
(
n
)
)

{\displaystyle OGef = lin(k.exp(n)) }

e
e
g
# f
e
x
p
(
k
.
e
x
p
(
n
)
)

{\displaystyle eegf = exp(k.exp(n)) }

∑

# k
1

n

n
!

k
!

(

n
−
1

k
−
1

)

{\displaystyle \sum\_{k=1}^n \frac{n!}{k!}\binom{n-1}{k-1}}

**Ens(Lin\*)**;
Lah numbers

n
!

k
!

(

n
−
1

k
−
1

)

{\displaystyle \frac {n!} {k!} \binom {n-1} {k-1} }

(

n
−
1

k
−
1

)

n
!

{\displaystyle \binom{n-1}{k-1}n! }

  E.g.f. for column k is (x/(1-x))^k.

l
i
n
(
k
.
l
i

n
∗

(
n
)
)

{\displaystyle lin(k.lin^\*(n)) }

 missing

2

n
−
1

n
!

{\displaystyle 2^{n-1}n!}

 A000110 Bell  **Ens(Ens\*)** | A008277
Stirling II

{

n

k

}

{\displaystyle \left\{\begin{matrix} n \\ k \end{matrix}\right\}}

  Oe:

e
x
p
(
k
.
e
x

p
∗

(
n
)
)

{\displaystyle exp(k.exp^\*(n)) }

k
!

{

n

k

}

{\displaystyle k! \left\{\begin{matrix} n \\ k \end{matrix}\right\} }

Oe:

∑
k

(
y
.
e
x

p
∗

(
x
)

)
k

{\displaystyle \sum\_k(y.exp^\*(x))^k }

  ee:

e
x
p
(
y
.
e
x

p
∗

(
x
)
)

{\displaystyle exp(y.exp^\*(x))}

∑

# k
1

n

k
!

{

n

k

}

{\displaystyle \sum\_{k=1}^n k! \left\{\begin{matrix} n \\ k \end{matrix}\right\} }

 A000041
Partitions of n
partitions of n in k parts

(

n
k

)

=

(

n
−
1

k
−
1

)

(

n
−
k

k

)

{\textstyle \binom n k = \binom {n-1} {k-1} + \binom {n-k} k }

(

n
−
1

k
−
1

)

{\displaystyle \binom{n-1}{k-1} }

2

n
−
1

{\displaystyle 2^{n-1}}

### **B3: Onto functions**

A019538 onto functions

(

n
k

)

=
k

(

n
−
1

k

)

k

(

n
−
1

k
−
1

)

{\displaystyle {n \choose k} = k{n-1 \choose k} + k{n-1 \choose k-1}}

Put 5 pennies in a row, leaving a little gap between consecutive pennies. There are 4 interpenny gaps. CHOOSE any number of these gaps (0, 1, 2, 3, or 4) to put a grain of rice into. Any such choice gives rise to a unique ordered partition of 5, and all of them arise in this way. For example, the trivial partition 5 comes from using no grain. The partition 4+1 comes from putting a grain of rice after the 4th penny. And so on. So there are exactly as many ordered partitions of 5 as there are ways of choosing a SUBSET of the set of gaps. But a set of 4 elements has 2^4 subsets.

Or else one could attack the problem by induction. For example, let P(n)P(n) be the number of ordered partitions of nn. Now look at P(n+1)P(n+1). Ordered partitions of n+1 are of two types: (i) last element 1 and (ii) last element bigger than 1. You should be able to see that there are P(n) ordered partitions of n+1 of each type, meaning that P(n+1)=2P(n).
André Nicolas
https://math.stackexchange.com/questions/31562/number-of-ordered-partitions-of-integer

### **E3: Stirling II - partitions of n-sets**

A008277 - Triangle of Stirling numbers of the second kind

e
.
g
.
f
# .
A
(
x
,
y
# )
e
x
p
(
y
.

e
x

−
y
)

{\displaystyle e.g.f. = A(x,y) = exp(y.e^x-y)}

for m-th column:

e
.
g
.
f
# .

(

e
x

−
1

)
m

m
!

{\displaystyle e.g.f. = {(e^x-1)^m \over m!}}

(

n
k

)

=
k

(

n
−
1

k

)

(

n
−
1

k
−
1

)

{\displaystyle {n \choose k} = k{n-1 \choose k} + {n-1 \choose k-1}}

Row sums: Bell number A000110

e
.
g
.
f
# .
e
x
p
(

e
x

−
1
)
)

{\displaystyle e.g.f. = exp(e^x-1))}

### **F3: Stirling II types**

A008284 Triangle of partition numbers: T(n,k) = number of partitions of n in which the greatest part is k, 1<=k<=n. Also number of partitions of n into k positive parts, 1<=k<=n.

(

n
k

)

=

(

n
−
1

k
−
1

)

(

n
−
k

k

)

{\displaystyle {n \choose k} = {n-1 \choose k-1} + {n-k \choose k}}

Row sums: A000041 number of partitions of n (the partition numbers)
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101,..

5+1+1...21 ways to label
4+2+1...105 ways to label
3+3+1...70 ways to label
3+2+2...105 ways to label total 301

(

(

n
k

)

)

{\displaystyle \left(\!\!{n\choose k}\!\!\right) }

## Something new, partial surjections

https://oeis.org/A028246 - the triangle

https://oeis.org/A000629 - the sum of rows,

E
(
X
)
⋅
L
(

E
∗

(
X
)
)

{\displaystyle E(X) \cdot L(E^\*(X)) }