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g
[3] =
x
1
f
[3] + 6
x
2
f
[2]
f
[1] + 6
x
3
f
[1]
3
g
[4] =
x
1
f
[4] + 8
x
2
f
[3]
f
[1] + 6
x
2
f
[2]
2
+36
x
3
f
[2]
f
[1]
2
+24
x
4
f
[1]
4
That concludes the illustrative example. In Table 2, which follows, we list a few
of many possible variations on the theme. For the record, entry 1 is Leibniz' rule,
entry 5 is Faa di Bruno, and entries 6 and 7 are essentialy due to Lagrange.
Table 2.
Some (
H
(
z
)
↔ h
[
n
]) pairs
H
(
z
)
h
A
[
n
]
h
B
[
n
]
1
.
1
(
z
)
···
F
k
(
z
)
f
1
[
α
1
]
···
f
k
[
α
k
]
A
(
c
)
f
1
[
c
1
]
···
f
k
[
c
k
]
α
∈
C
n,k
c
∈
C
n,k
x
|
f
[
α
]
x
|
A
(
λ
)
B
(
λ
)
f
[
λ
]
2
.
a,d
F
(
z
)
x
α
|
λ
|
α
∈
C
n
λ
∈
P
n
3
.
b,c
G
(
F
(
z
))
f
g
[
π
]
C
(
λ
)
f
g
[
λ
]
π
∈
P
n
λ
∈
P
n
H
(
z
)=
G
(
F
(
z
))
F
(
z
)
(
−
1)
|
π
|
f
[
π
]
(
−
1)
|λ|
C
(
λ
)
f
[
λ
]
4
.
b
F
−
1
(
z
)
R
(
π
)
∈
DP
n−
1
λ
∈
DP
n−
1
F
(
z
)=
1
G
(
F
(
z
))
1)
|
π
|
f
1)
|λ|
C
(
λ
)
f
5
.
b,c
G
(
F
−
1
(
z
))
(
−
g
[
π
]
(
−
g
[
λ
]
λ
∈
DV
n−
1
R
(
π
)
∈
DV
n−
1
6
.
(
H, z
)=0 (See[4])
7
.
b
H
=
G
(
w
)
,z
=
F
(
w
) (see no. 5)
8
.
(
z
)=
G
(
F
1
(
z
)
,...F
q
(
z
)) (requires
q
-chromatic partitions)
........................................................................................
(
a
)
F
(0) = 1,
(
b
)
F
(0) = 0
,F
(0) = 1,
(
c
)
f g
[
α
] =
f
[
α
]
g
[
|α|
],
(
d
)
x ∈
R
c
an integer composition
λ
an integer partition
C
n
a set of integer compositions
P
n
a set of integer partitions
α
a set composition
π
a set partition
C
n
a set of set compositions
P
n
a set of set partitions
n
sum of parts
k
number of parts
∗
j
number of null parts
null parts ok
7
The Lagrange Inversion Formula
The following is required for some of our proofs.
Lagrange Inversion Formula [2] [6].
Suppose
F
(0) = 0, and
F
0
(
z
)=
F
(
z
)
/z
.
Then
z
n
n
!
F
−
1
(
z
)=
z
n−
1
(
n
F
0
(
z
)
−n
.
−
1)!