Combinatorial Numbers

**By: Tao Steven Zheng (郑涛)**

A problem I created in Grade 11.

## **Problem**

Let there be a recurrence relation

P

a

(
n
)

=
n
⋅

P

a

(

n
−
1

)

a

n

{\displaystyle {P}\_{a}\left(n\right)=n\cdot {P}\_{a}\left(n-1\right)+{a}^{n}}

, where

P

a

(
0
)

=
1

{\displaystyle {P}\_{a}\left(0\right)=1}

. Show that

P

a

(
n
)

∼

e

a

n
!

{\displaystyle {P}\_{a}\left(n\right)\sim {e}^{a}n!}

 for large

n

{\displaystyle n}

.

## **Solution**

We first show that the above recurrence relation constructs the sum

P

a

(
n
# )
n
!

(

∑

# k
0

n

a

k

k
!

)

.

{\displaystyle {P}\_{a}(n)=n!\left(\sum \_{k=0}^{n}{\frac {{a}^{k}}{k!}}\right).}

Now we proceed to prove by induction.

When

# k
1

{\displaystyle k = 1}

1
!

(

a

0

0
!

a

1

1
!

)

=
1
(
1
)
a

{\displaystyle 1!\left({\frac {{a}^{0}}{0!}}+{\frac {{a}^{1}}{1!}}\right)=1(1)+a}

1
!

(

∑

# k
0

1

a

k

k
!

)

=
1
⋅

P

a

(
0
)

a

1

=

P

a

(
1
)

{\displaystyle 1!\left(\sum \_{k=0}^{1}{\frac {{a}^{k}}{k!}}\right)=1\cdot {P}\_{a}(0)+{a}^{1}={P}\_{a}(1)}

When

# k
n

{\displaystyle k=n}

n
!

(

a

0

0
!

a

1

1
!

.
.
.

a

n

n
!

)

=
n
(
n
−
1
)
!

(

a

0

0
!

a

1

1
!

.
.
.

a

n
−
1

(
n
−
1
)
!

)

a

n

{\displaystyle n!\left({\frac {{a}^{0}}{0!}}+{\frac {{a}^{1}}{1!}}+...+{\frac {{a}^{n}}{n!}}\right)=n(n-1)!\left({\frac {{a}^{0}}{0!}}+{\frac {{a}^{1}}{1!}}+...+{\frac {{a}^{n-1}}{(n-1)!}}\right)+{a}^{n}}

n
!

(

∑

# k
0

n

a

k

k
!

)

=
n
⋅

P

a

(
n
−
1
)

a

n
−
1

=

P

a

(
n
)
.

{\displaystyle n!\left(\sum \_{k=0}^{n}{\frac {{a}^{k}}{k!}}\right)=n\cdot {P}\_{a}(n-1)+{a}^{n-1}={P}\_{a}(n).}

When

# k
n
1

{\displaystyle k=n+1}

(
n
1
)
!

(

a

0

0
!

a

1

1
!

.
.
.

a

n
1

(
n
1
)
!

)

=
(
n
1
)
(
n
)
!

(

a

0

0
!

a

1

1
!

.
.
.

a

n

n
!

)

a

n
1

{\displaystyle (n+1)!\left({\frac {{a}^{0}}{0!}}+{\frac {{a}^{1}}{1!}}+...+{\frac {{a}^{n+1}}{(n+1)!}}\right)=(n+1)(n)!\left({\frac {{a}^{0}}{0!}}+{\frac {{a}^{1}}{1!}}+...+{\frac {{a}^{n}}{n!}}\right)+{a}^{n+1}}

(
n
1
)
!

(

∑

# k
0

n
1

a

k

k
!

)

=
(
n
1
)
⋅

P

a

(
n
)

a

n
1

=

P

a

(
n
1
)
.

{\displaystyle (n+1)!\left(\sum \_{k=0}^{n+1}{\frac {{a}^{k}}{k!}}\right)=(n+1)\cdot {P}\_{a}(n)+{a}^{n+1}={P}\_{a}(n+1).}

Notice that the sum

∑

# k
0

n

a

k

k
!

{\displaystyle \sum \_{k=0}^{n}{\frac {{a}^{k}}{k!}}}

 approaches

e

a

{\displaystyle {e}^{a}}

 for large

n

{\displaystyle n}

. Hence,

P

a

(
n
)

∼

e

a

n
!

{\displaystyle {P}\_{a}\left(n\right)\sim {e}^{a}n!}

 for large

n

{\displaystyle n}

.

**Note:** The way I found this recurrence relation is by investigating the integral

P

a

(
n
# )

∫

a

∞

e

a
−
x

x

n

d
x
.

{\displaystyle {P}\_{a}(n)=\int \_{a}^{\infty }{{e}^{a-x}}{x}^{n}dx.}

As an exercise, prove that this integral is equivalent to the defined recurrence equation.

When

# a
−
1

{\displaystyle a = -1}

, the integral solves the derangement problem.

When

# a
0

{\displaystyle a = 0}

, the integral becomes the Gamma Function, and solves the permutation problem.

When

# a
1

{\displaystyle a = 1}

, the integral solves the combinatorial problem of counting the number of paths between two nodes in a network graph where all nodes are interconnected (pictured above).

I created the problem Nine Islands, which uses the case when

# a
1

{\displaystyle a = 1}

.