Bob Proctor

Injections
$$T(n, k) = n*T(n-1, k-1) = k*T(n-1, k-1)+T(n-1, k)$$

all functions

$$k^n$$ $$OGef = lin(k.exp(n)) $$ $$eegf = exp(k.exp(n)) $$

B3: Onto functions
A019538 onto functions

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

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

$$e.g.f. = A(x,y) = exp(y.e^x-y)$$

for m-th column: $$e.g.f. = {(e^x-1)^m \over m!}$$

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

Row sums: Bell number A000110 $$e.g.f. = exp(e^x-1))$$

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.

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

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

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

Something new, partial surjections
https://oeis.org/A028246 - the triangle

https://oeis.org/A000629 - the sum of rows, $$ E(X) \cdot L(E^*(X)) $$