unambiguity of factorial base representation

unambiguity of factorial base representation

Theorem 1

.

For all

positive integers

n

, it is the case that

n

!

=

1

+

∑

k

=

1

n

-

1

k

⋅

k

!

Proof.

We first consider two degenerate cases. When

n

=

0

or

n

=

1

, the sum has no terms because the lower limit
is bigger than the upper limit. By the convention that
a sum with no terms equals zero, the

equation

reduces
to

0

!

=

1

or

1

!

=

1

, which is correct.

Next, assume that

n

>

1

. Note that

k

⋅

k

!

=

(

k

+

1

)

⁢

k

!

-

k

!

=

(

k

+

1

)

!

-

k

!

,

hence, we have a telescoping sum:

∑

k

=

1

n

-

1

k

⋅

k

!

=

∑

k

=

1

n

-

1

(

(

k

+

1

)

!

-

k

!

)

=

n

!

-

1

Adding 1, we conclude that

n

!

=

1

+

∑

k

=

1

n

-

1

k

⋅

k

!

∎

Theorem 2

.

If

d

m

⁢

…

⁢

d

2

⁢

d

1

is the

factoradic

representation

of
a positive integer

n

, then

(

d

m

+

1

)

⁢

m

!

>

n

≥

d

m

⋅

m

!

Proof.

By definition,

n

=

d

m

⋅

m

!

+

∑

k

=

1

m

-

1

d

k

⋅

k

!

.

Since each term of the sum is nonnegative, the sum is

positive

,
hence

n

≥

d

m

⋅

m

!

. Using the fact that

0

≤

d

k

<

k

to bound each term in the sum, we have

n

≤

d

m

⋅

m

!

+

∑

k

=

1

m

-

1

k

⋅

k

!

.

By

theorem

1, the right-hand side equals

m

!

-

1

, so we have

n

≤

d

m

⋅

m

!

+

m

!

-

1

, which is the same as saying

(

d

m

+

1

)

⁢

m

!

>

n

.
∎

Theorem 3

.

If

d

m

⁢

…

⁢

d

2

⁢

d

1

and

d

′

m

′

⁢

…

⁢

d

′

2

⁢

d

′

1

are
distinct factoradic representations with

d

m

≠

0

and

d

′

m

′

≠

0

(i.e. not padded with leading zeros), then they denote
distinct

integers

.

Proof.

Let

n

=

∑

k

=

1

m

d

k

⋅

k

!

and let

n

′

=

∑

k

=

1

m

′

d

′

k

⋅

k

!

.

Suppose that

m

and

m

′

are distinct. Without loss of generality,
we may assume that

m

<

m

′

. Then,

(

m

+

1

)

!

≤

m

′

!

. By theorem 2,
we have

n

<

(

d

m

+

1

)

⁢

m

!

and

d

m

′

⋅

m

′

!

≤

n

′

. Because

d

m

≤

m

, we have

(

d

m

+

1

)

⁢

m

!

≤

(

m

+

1

)

⁢

m

!

=

(

m

+

1

)

!

.
Because

d

m

′

≠

0

, we have

n

′

>

m

′

!

. Hence,

n

<

n

′

, so

n

does not equal

n

′

.

To

complete

the proof, we use the method of

infinite descent

. Assume
that factoradic representations are not unique. Then there must exist
a least integer

n

such that

n

has two distinct factoradic
representations

d

m

⁢

…

⁢

d

2

⁢

d

1

and

d

′

m

′

⁢

…

⁢

d

′

2

⁢

d

′

1

with

d

m

≠

0

and

d

′

m

′

≠

0

. By the

conclusion

of the
previous paragraph, we must have

m

=

m

′

. By theorem 2, we must have

d

m

=

d

′

m

. Let

m

1

be the chosen such
that

d

m

1

≠

0

but

d

m

1

+

1

=

⋯

=

d

m

-

1

=

0

and let

m

2

be the chosen such that

d

′

m

2

≠

0

but

d

′

m

2

+

1

=

⋯

=

d

′

m

-

1

=

0

. Then

d

m

1

⁢

…

⁢

d

2

⁢

d

1

and

d

′

m

2

⁢

…

⁢

d

′

2

⁢

d

′

1

would be distinct factoradic representations of

n

-

d

m

⋅

m

!

whose leading digits are not zero but this would
contradict the

assumption

that

n

is the smallest number with this
property.
∎

Title

unambiguity of factorial base representation

Canonical name

UnambiguityOfFactorialBaseRepresentation

Date of creation

2013-03-22 16:38:35

Last modified on

2013-03-22 16:38:35

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 11A63

Related topic

UniquenessOfDigitalRepresentation

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