aarjaneiro/planetmath huggingface dataset:
proof of recurrences for derangement numbers

proof of recurrences for derangement numbers

The

derangement numbers

D

n

satisfy

two

recurrence relations

:

D

n

=

(

n

-

1

)

⁢

[

D

n

-

1

+

D

n

-

2

]

(1)

D

n

=

n

⁢

D

n

-

1

+

(

-

1

)

n

(2)

These

formulas

can be derived algebraically (working from the

explicit formula

for the derangement numbers); there is also an enlightening combinatorial proof of (1).

The

exponential generating function

for the derangement numbers is

D

n

=

n

!

⁢

∑

k

=

0

n

(

-

1

)

k

k

!

To derive the formulas algebraically, start with (2):

n

⁢

D

n

-

1

=

n

⁢

(

n

-

1

)

!

⁢

∑

k

=

0

n

-

1

(

-

1

)

k

k

!

=

n

!

⁢

∑

k

=

0

n

-

1

(

-

1

)

k

k

!

=

n

!

⁢

(

-

(

-

1

)

n

n

!

+

∑

k

=

0

n

(

-

1

)

k

k

!

)

=

-

(

-

1

)

n

+

D

n

and (2) follows. To derive (1), use (2) twice:

D

n

=

n

⁢

D

n

-

1

+

(

-

1

)

n

=

(

n

-

1

)

⁢

D

n

-

1

+

D

n

-

1

+

(

-

1

)

n

=

(

n

-

1

)

⁢

D

n

-

1

+

(

(

n

-

1

)

⁢

D

n

-

2

+

(

-

1

)

n

-

1

)

+

(

-

1

)

n

=

(

n

-

1

)

⁢

(

D

n

-

1

+

D

n

-

2

)

Combinatorially, we can see (1) as follows. Write

[

n

]

for

{

1

,

2

,

…

,

n

}

. Let

π

be any

derangement

of

[

n

-

1

]

, i.e. a

permutation

containing no

1

-cycles. Adding

n

before any of the

n

-

1

elements of

π

produces a derangement of

[

n

]

. For fixed

π

, these are clearly all distinct, since

1

has a different

successor

in each case; for distinct

π

, these are equally obviously distinct. Thus each derangement of

[

n

-

1

]

corresponds to exactly

n

-

1

derangements of

[

n

]

. Note also that since

π

had no

1

-cycles that

1

is not a member of a

transposition

(a

2

-cycle).

Now let

π

be a derangement of any

n

-

2

elements

chosen from

[

n

-

1

]

. There are clearly

(

n

-

1

)

⁢

D

n

-

2

such derangements. If the omitted element in

π

is

x

, then adding the transposition

(

1

⁢

x

)

to

π

produces a derangement of

[

n

]

, and all such derangements again are distinct from one another. Finally, since in this case

1

is a member of a transposition, these derangements are distinct from those in the first group. This proves (1).

Title

proof of recurrences for derangement numbers

Canonical name

ProofOfRecurrencesForDerangementNumbers

Date of creation

2013-03-22 19:05:56

Last modified on

2013-03-22 19:05:56

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Result

Classification

msc 05A05

Classification

msc 05A15

Classification

msc 60C05

Generated on Thu Feb 8 20:35:19 2018 by

LaTeXML
rule of product

rule of product

If a process

A

can have altogether

m

different results and another process

B

altogether

n

different results, then the two processes can have altogether

m

⁢

n

different combined results.  Putting it to set-theoretical form,

card

⁢

(

A

×

B

)

=

m

⋅

n

.

The

rule of product

is true also for the

combination

of several processes:  If the processes

A

i

can have

n

i

possible results (

i

=

1

,

2

,

…

,

k

), then their combined process has

n

1

⁢

n

2

⁢

⋯

⁢

n

k

possible results.  I.e.,

card

⁢

(

A

1

×

A

2

×

…

×

A

k

)

=

n

1

⁢

n

2

⁢

⋯

⁢

n

k

.

Example.

Arranging

n

elements

, the first one may be chosen freely from all the

n

elements, the second from the remaining

n

-

1

elements, the third from the remaining

n

-

2

, and so on, the penultimate one from two elements and the last one from the only remaining element; thus by the rule of product, there are in all

n

⁢

(

n

-

1

)

⁢

(

n

-

2

)

⁢

⋯

⁢

2

⋅

1

=

n

!

different

arrangements

, i.e.

permutations

, as the result.

Title

rule of product

Canonical name

RuleOfProduct

Date of creation

2013-03-22 19:13:02

Last modified on

2013-03-22 19:13:02

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Definition

Classification

msc 05A05

Classification

msc 03-00

Synonym

multiplication principle

Related topic

CartesianProduct

Related topic

Combinatorics

Related topic

Cardinality

Related topic

Number

Related topic

Product

Generated on Thu Feb 8 20:35:21 2018 by

LaTeXML
alternating factorial

alternating factorial

The

alternating factorial

a

⁢

f

⁢

(

n

)

of a

positive integer

n

is the sum

a

⁢

f

⁢

(

n

)

=

∑

i

=

1

n

(

-

1

)

n

-

i

⁢

i

!

,

which can also be expressed with the

recurrence relation

a

⁢

f

⁢

(

n

)

=

n

!

-

a

⁢

f

⁢

(

n

-

1

)

with starting condition

a

⁢

f

⁢

(

1

)

=

1

. The notation nÃÂ¡! (alternating an inverted exclamation mark with a

regular

exclamation mark) has been proposed by analogy to that of the

double factorial

, but has not gained much support, in part because of TeX’s lack of support for Spanish

characters

.

The first few alternating factorials, listed in A005165 of Sloane’s OEIS, are 1, 5, 19, 101, 619, 4421.

In 1999, Miodrag Zivković proved that

gcd

⁡

(

n

,

a

⁢

f

⁢

(

n

)

)

=

1

and that the set of alternating factorials that are

prime numbers

is finite.

a

⁢

f

⁢

(

661

)

is the largest such known prime.

Title

alternating factorial

Canonical name

AlternatingFactorial

Date of creation

2013-03-22 16:19:59

Last modified on

2013-03-22 16:19:59

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

7

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 05A10

Generated on Thu Feb 8 20:35:23 2018 by

LaTeXML
birthday problem

birthday problem

The

birthday problem

is:
What is the probability of two or more people out of a of

n

do have the same birthday?

Solution.

Let

A

be the event two or more people out of a group of

n

to have the same birthday.
It is easy to find first the complement of

A

,

A

c

, which is that no two people out of a group of

n

will have

matching

birthdays out of

365

(number of days of one year) equally possible birthdays. We start with an arbitrary person’s birthday. So the second person’s birthday in order to be different from the first one gives us the probability,

1

-

1

365

.
Consecutively, the third person’s birthday is different from the first two and we have

(

1

-

1

365

)

⁢

(

1

-

2

365

)

. Finally for the the

n

th person’s birthday we get the probability

P

⁢

[

A

c

]

=

(

1

-

1

365

)

⁢

(

1

-

2

365

)

⁢

…

⁢

(

1

-

n

-

1

365

)

=

(

365

-

1

)

365

⁢

(

365

-

2

)

365

⁢

…

⁢

(

365

-

(

n

-

1

)

)

365

=

(

365

-

1

)

365

⁢

…

⁢

(

365

-

(

n

-

1

)

)

365

n

=

365

!

(

365

-

n

)

!

⁢

365

n

=

(

365

n

)

⁢

n

!

365

n

By the last expression of the estimated probability we can see that there exists another easier approach to find the probability which is the following:

1.

We set the possible ways for

1

⁢

…

⁢

n

birthdays,

365

n

as the

denominator

.

2.

Using the

binomial coefficient

(

365

n

)

, which represents the number of ways of picking

n

unordered birthdays from

365

days multiplied by the

factorial

n

!

used for the number of orders, in that way we determine the numerator.

Therefore we have

P

⁢

[

A

c

]

=

(

365

n

)

⁢

n

!

365

n

whence

P

⁢

[

A

]

=

1

-

P

⁢

[

A

c

]

=

1

-

(

365

n

)

⁢

n

!

365

n

.

The table below shows the probability

P

⁢

[

A

]

for some values of

n

n

P

⁢

[

A

]

10

0

,

11695

20

0

,

41144

23

0

,

50730

30

0

,

70632

40

0

,

89123

50

0

,

97037

Title

birthday problem

Canonical name

BirthdayProblem

Date of creation

2013-03-22 16:09:38

Last modified on

2013-03-22 16:09:38

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

16

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 05A10

Classification

msc 60C05

Classification

msc 60-00

Synonym

birthday paradox

Generated on Thu Feb 8 20:35:27 2018 by

LaTeXML
Catalan numbers

Catalan numbers

The

Catalan numbers

, or

Catalan sequence

, have many interesting applications in

combinatorics

.

The

n

th Catalan number is given by:

C

n

=

(

2

⁢

n

n

)

n

+

1

,

where

(

n

r

)

represents the

binomial coefficient

. The first several Catalan numbers are

1

,

1

,

2

,

5

,

14

,

42

,

132

,

429

,

1430

,

4862

,…(see OEIS

sequence

http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000108

A000108 for more terms). The Catalan numbers are also generated by the

recurrence relation

C

0

=

1

,

C

n

=

∑

i

=

0

n

-

1

C

i

⁢

C

n

-

1

-

i

.

For example,

C

3

=

1

⋅

2

+

1

⋅

1

+

2

⋅

1

=

5

,

C

4

=

1

⋅

5

+

1

⋅

2

+

2

⋅

1

+

5

⋅

1

=

14

, etc.

The ordinary generating function for the Catalan numbers is

∑

n

=

0

∞

C

n

⁢

z

n

=

1

-

1

-

4

⁢

z

2

⁢

z

.

Interpretations

of the

n

th Catalan number include:

1.

The number of ways to arrange

n

pairs of

matching

parentheses, e.g.:

(

)

(

(

)

)

⁢

⁢

(

)

⁢

(

)

(

(

(

)

)

)

⁢

⁢

(

(

)

⁢

(

)

)

⁢

⁢

(

)

⁢

(

(

)

)

⁢

⁢

(

(

)

)

⁢

(

)

⁢

⁢

(

)

⁢

(

)

⁢

(

)

2.

The number of ways a convex polygon of

n

+

2

sides can be split into

n

triangles

.

3.

The number of rooted binary trees with exactly

n

+

1

leaves.

The Catalan sequence is named for Eugène Charles Catalan, but it was discovered in 1751 by Euler when he was trying to solve the problem of subdividing polygons into triangles.

References

1

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik.

Concrete Mathematics

.

Addison-Wesley, 1998.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0836.00001

Zbl
0836.00001.

Title

Catalan numbers

Canonical

name

CatalanNumbers

Date of creation

2013-03-22 12:29:51

Last modified on

2013-03-22 12:29:51

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

11

Author

bbukh (348)

Entry type

Definition

Classification

msc 05A10

Synonym

Catalan sequence

Related topic

CentralBinomialCoefficient

Related topic

AsymptoticsOfCentralBinomialCoefficient

Generated on Thu Feb 8 20:35:29 2018 by

LaTeXML
central binomial coefficient

central binomial coefficient

The

n

th

central binomial coefficient

is defined to be

(

2

⁢

n

n

)

=

(

2

⁢

n

)

!

(

n

!

)

2

where

(

2

⁢

n

n

)

is a

binomial coefficient

. These numbers have the

generating function

1

1

-

4

⁢

x

=

1

+

2

⁢

x

+

6

⁢

x

2

+

20

⁢

x

3

+

70

⁢

x

4

+

252

⁢

x

5

+

⋯

They are closely related to the

Catalan sequence

, in that

C

n

=

1

n

+

1

⁢

(

2

⁢

n

n

)

Alternate definition

A less frequently-encountered definition for the

n

th central binomial coefficient is

(

n

⌊

n

2

⌋

)

.

Note that the set of these numbers meeting this alternate criterion is a

superset

of those meeting the first criterion, since for

n

=

2

⁢

m

we have

(

n

⌊

n

2

⌋

)

=

(

2

⁢

m

⌊

2

⁢

m

2

⌋

)

=

(

2

⁢

m

m

)

By cancelling terms of one of the

n

!

’s against terms of the

2

⁢

n

!

, one may rewrite the central binomial
coefficient as follows:

(

2

⁢

n

n

)

=

2

⁢

n

⁢

(

2

⁢

n

-

1

)

⁢

⋯

⁢

(

n

+

2

)

⁢

(

n

+

1

)

n

⁢

(

n

-

1

)

⁢

⋯

⁢

3

⋅

2

⋅

1

.

Alternatively, one may cancel each term of the

n

!

against twice itself, leaving

2

’s in the

numerator

:

(

2

⁢

n

n

)

=

2

n

⁢

(

2

⁢

n

-

1

)

⁢

(

2

⁢

n

-

3

)

⁢

⋯

⁢

5

⋅

3

⋅

1

n

⁢

(

n

-

1

)

⁢

⋯

⁢

3

⋅

2

⋅

1

Doubling the terms in the denominator, we obtain an expression for the

central

binomial

coeficient
in terms of a quotient of successive

odd numbers

by successive

even numbers

:

(

2

⁢

n

n

)

=

4

n

⁢

(

2

⁢

n

-

1

)

⁢

(

2

⁢

n

-

3

)

⁢

⋯

⁢

5

⋅

3

⋅

1

2

⁢

n

⁢

(

2

⁢

n

-

2

)

⁢

⋯

⁢

6

⋅

4

⋅

2

By means of these formulae, one may derive some important properties of the central
binomial coeficients. By examining the first two formulae, one may deduce results
about the

prime factors

of central binomial coefficients (for proofs, please see the
attachments to this entry):

Theorem 1

If

n

≥

3

is an

integer

and

p

is a

prime number

such that

n

<

p

<

2

⁢

n

, then

p

divides

(

2

⁢

n

n

)

.

Theorem 2

If

n

≥

3

is an integer and

p

is a prime number such that

2

⁢

n

/

3

<

p

≤

n

, then

p

does not divide

(

2

⁢

n

n

)

.

In

conjunction

with Wallis’

formula

for

π

, the third formula for the central
binomial coefficient may be used to derive an asymptotic expression, as is done in
an attachment to this entry:

(

2

⁢

n

n

)

≈

2

π

⁢

4

n

2

⁢

n

+

1

Title

central binomial coefficient

Canonical name

CentralBinomialCoefficient

Date of creation

2013-03-22 14:25:40

Last modified on

2013-03-22 14:25:40

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 05A10

Classification

msc 11B65

Related topic

BinomialCoefficient

Related topic

CatalanNumbers

Generated on Thu Feb 8 20:35:32 2018 by

LaTeXML
combinations with repeated elements

combinations with repeated elements

Definition 1

.

A

k

-

combination with repeated elements

chosen within the set

X

=

{

x

1

,

x

2

,

…

⁢

x

n

}

is a

multiset

with

cardinality

k

having

X

as the underlying set.

Note 1

.

The

definition

is based on the multiset

concept

and therefore the order of the

elements

within the

combination

is irrelevant.

Note 2

.

The definition generalizes the concept of

combination with distinct elements

.

Lemma 1

.

Given

n

,

k

∈

{

0

,

1

,

2

,

…

}

,

n

≥

k

, the following

formula

holds:

(

n

+

1

k

+

1

)

=

∑

i

=

k

n

(

i

k

)

.

Proof.

The formula is easily demonstrated by repeated application of the Pascal’s Rule for the

binomial coefficient

.
∎

Theorem 1

.

The number

C

n

,

k

′

of the

k

-combinations with repeated elements is given by the formula:

C

n

,

k

′

=

(

n

+

k

-

1

k

)

.

Proof.

The proof is given by finite

induction

(

http://planetmath.org/PrincipleOfFiniteInduction

).

The proof is

trivial

for

k

=

1

, since no repetitions can occur and the number of

1

-combinations is

n

=

(

n

1

)

.

Let’s then prove the formula is true for

k

+

1

, assuming it holds for

k

. The

k

+

1

-combinations can be partitioned in

n

subsets as follows:

•

combinations that include

x

1

at least once;

•

combinations that do not include

x

1

, but include

x

2

at least once;

•

combinations that do not include

x

1

and

x

2

, but include

x

3

at least once;

•

…

•

combinations that do not include

x

1

,

x

2

,…

x

n

-

2

but include

x

n

-

1

at least once;

•

combinations that do not include

x

1

,

x

2

,…

x

n

-

2

,

x

n

-

1

but include

x

n

only.

The number of the subsets is:

C

n

,

k

′

+

C

n

-

1

,

k

′

+

C

n

-

2

,

k

′

+

…

+

C

2

,

k

′

+

C

1

,

k

′

which, by the

inductive hypothesis

and the lemma, equalizes:

(

n

+

k

-

1

k

)

+

(

n

+

k

-

2

k

)

+

(

n

+

k

-

3

k

)

+

…

+

(

k

+

1

k

)

+

(

k

k

)

=

∑

i

=

k

n

+

k

-

1

(

i

k

)

=

(

n

+

k

k

+

1

)

.

∎

Title

combinations with repeated elements

Canonical

name

CombinationsWithRepeatedElements

Date of creation

2013-03-22 17:43:17

Last modified on

2013-03-22 17:43:17

Owner

kfgauss70 (18761)

Last modified by

kfgauss70 (18761)

Numerical id

8

Author

kfgauss70 (18761)

Entry type

Definition

Classification

msc 05A10

Related topic

BinomialCoefficient

Generated on Thu Feb 8 20:35:35 2018 by

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divisibility of central binomial coefficient

divisibility of central binomial coefficient

In this entry, we shall prove two results about the

divisibility of central binomial coefficients

which were stated in the main entry.

Theorem 1

.

If

n

≥

3

is an

integer

and

p

is a

prime number

such that

n

<

p

<

2

⁢

n

, then

p

divides

(

2

⁢

n

n

)

.

Proof.

We will examine the following expression for our

binomial coefficient

:

(

2

⁢

n

n

)

=

2

⁢

n

⁢

(

2

⁢

n

-

1

)

⁢

⋯

⁢

(

n

+

2

)

⁢

(

n

+

1

)

n

⁢

(

n

-

1

)

⁢

⋯

⁢

3

⋅

2

⋅

1

.

Since

n

<

p

<

2

⁢

n

, we find

p

appearing in the

numerator

. However,

p

cannot appear in the denominator because the terms there are all
smaller than

n

. Hence,

p

cannot be cancelled, so it must divide

(

2

⁢

n

n

)

.
∎

Theorem 2

.

If

n

≥

3

is an integer and

p

is a prime number such that

2

⁢

n

/

3

<

p

≤

n

, then

p

does not divide

(

2

⁢

n

n

)

.

Proof.

We will again examine our expression for our binomial coefficient:

(

2

⁢

n

n

)

=

2

⁢

n

⁢

(

2

⁢

n

-

1

)

⁢

⋯

⁢

(

n

+

2

)

⁢

(

n

+

1

)

n

⁢

(

n

-

1

)

⁢

⋯

⁢

3

⋅

2

⋅

1

.

This time, because

2

⁢

n

/

3

<

p

≤

n

, we find

p

appearing in the denominator
and

2

⁢

p

appearing in the numerator. No other

multiples

will appear because,
if

m

>

2

, then

m

⁢

p

>

2

⁢

n

. The two occurrences of

p

noted above cancel, hence

p

is not a

prime factor

of

(

2

⁢

n

n

)

.
∎

Title

divisibility of central binomial coefficient

Canonical name

DivisibilityOfCentralBinomialCoefficient

Date of creation

2013-03-22 17:41:22

Last modified on

2013-03-22 17:41:22

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 05A10

Classification

msc 11B65

Generated on Thu Feb 8 20:35:43 2018 by

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divisibility of prime-power binomial coefficients

divisibility of prime-power binomial coefficients

For

p

a prime,

n

a nonzero

integer

, define

ord

p

⁡

(

n

)

to be the largest integer

r

such that

p

r

∣

n

.

An easy consequence of Kummer’s theorem is:

Theorem 1

.

Let

p

be a prime,

n

≥

1

an integer. If

1

≤

r

⁢

p

s

≤

p

n

where

r

,

s

are

nonnegative integers

with

p

∤

r

, then

ord

p

⁡

(

p

n

r

⁢

p

s

)

=

n

-

s

.

Proof.

The result is clearly true for

r

=

1

,

s

=

n

, so assume that

s

<

n

. By Kummer’s theorem,

ord

p

⁡

(

p

n

r

⁢

p

s

)

is the number of carries when adding

r

⁢

p

s

to

p

n

-

r

⁢

p

s

in base

p

. Consider the base

p

representations

of

r

⁢

p

s

and

p

n

-

r

⁢

p

s

. They each have

n

digits (possibly with leading zeros) when represented in base

p

, and they each have

s

trailing zeros. If the rightmost nonzero digit in

r

⁢

p

s

is

k

, then the rightmost nonzero digit in

p

n

-

r

⁢

p

s

is in the same “decimal” place and has value

p

-

k

. Each pair of corresponding digits (one from

r

⁢

p

s

and one from

p

n

-

r

⁢

p

s

) to the left of that point sum to

p

-

1

(it may help to think about how you subtract a

decimal number

from a power of

10

, and what the result looks like).

It is then clear that adding those two numbers together will result in no carries in the rightmost

s

places, but there will be a carry out of the

s

+

1

st

place and out of each successive place up to and including the

n

th

place, for a total of

n

-

s

carries.
∎

A couple of examples may help to make this proof more transparent. Take

p

=

3

. Then

(

27

4

)

=

17550

=

2

⋅

3

3

⋅

5

2

⋅

13

so that

ord

3

⁡

(

27

4

)

=

3

. Now,

27

10

=

1000

3

and

4

10

=

11

3

, so that

27

-

4

=

23

10

is

212

3

. Adding

212

3

+

11

3

indeed results in carries out of all three places since there are no trailing zeros.

(

27

6

)

=

296010

=

2

⋅

3

2

⋅

5

⋅

11

⋅

13

⋅

23

so that

ord

3

⁡

(

27

6

)

=

2

. Now,

6

10

=

20

3

so that

27

-

6

=

21

10

is

210

3

. When adding

20

3

+

210

3

, there are two carries, out of the

3

’s place and out of the

9

’s place. There is no carry out of the ones place since both numbers have a

0

there.

Title

divisibility of prime-power binomial coefficients

Canonical name

DivisibilityOfPrimepowerBinomialCoefficients

Date of creation

2013-03-22 18:42:29

Last modified on

2013-03-22 18:42:29

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 05A10

Classification

msc 11B65

Related topic

OrderValuation

Generated on Thu Feb 8 20:35:46 2018 by

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double factorial

double factorial

The

double factorial

of a

positive integer

n

is the product

n

!!

of the positive integers less than or equal to

n

that have the same parity as

n

, that is,

n

!!

=

n

⁢

(

n

-

2

)

⁢

(

n

-

4

)

⁢

⋯

⁢

k

n

where

k

n

denotes

1

if

n

is an

odd number

and

2

if

n

is an

even number

.

For example,

7

!!

=

7

⋅

5

⋅

3

⋅

1

=

105

10

!!

=

10

⋅

8

⋅

6

⋅

4

⋅

2

=

3840

Note that

n

!!

is not the same as

(

n

!

)

!

.

Observe that

(

2

⁢

n

)

!!

=

2

n

⁢

n

!

and

(

2

⁢

n

+

1

)

!!

=

(

2

⁢

n

+

1

)

!

2

n

⁢

n

!

.

Title

double factorial

Canonical name

DoubleFactorial

Date of creation

2013-03-22 12:24:54

Last modified on

2013-03-22 12:24:54

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Definition

Classification

msc 05A10

Generated on Thu Feb 8 20:35:48 2018 by

LaTeXML
exponential factorial

exponential factorial

Given a

positive integer

n

, the ”power tower”

n

(

n

-

1

)

(

n

-

2

)

⁢

…

is the

exponential factorial

of

n

. The

recurrence relation

is

a

1

=

1

,

a

n

=

n

a

n

-

1

for

n

>

1

.

So for example,

9

=

3

2

1

,

262144

=

4

3

2

1

. The exponential factorial for 5 has almost two hundred thousand base 10 digits. The ones that are small enough are listed in

sequence

A049384 of Sloane’s OEIS.

The sum of the reciprocals of the exponential factorials is a

Liouville number

.

∑

i

=

1

∞

1

a

i

≈

1.6111149258083767361111111

Title

exponential factorial

Canonical name

ExponentialFactorial

Date of creation

2013-03-22 16:01:38

Last modified on

2013-03-22 16:01:38

Owner

CompositeFan (12809)

Last modified by

CompositeFan (12809)

Numerical id

7

Author

CompositeFan (12809)

Entry type

Definition

Classification

msc 05A10

Related topic

Factorial

Generated on Thu Feb 8 20:35:50 2018 by

LaTeXML
factorial

factorial

For any non-negative integer

n

, the

factorial

of

n

, denoted

n

!

, can be defined by

n

!

=

∏

r

=

1

n

r

where for

n

=

0

the empty product is taken to be

1

.

Alternatively, the factorial can be defined recursively by

0

!

=

1

and

n

!

=

n

⁢

(

n

-

1

)

!

for

n

>

0

.

n

!

is equal to the number of

permutations

of

n

distinct objects.
For example, there are

5

!

ways to arrange the five letters A, B, C, D and E into a word.

For every non-negative integer

n

we have

Γ

⁢

(

n

+

1

)

=

n

!

where

Γ

is Euler’s

gamma function

.
In this way the notion of factorial can be generalized to all

complex

(

http://planetmath.org/Complex

) values except the negative integers.

Title

factorial

Canonical name

Factorial

Date of creation

2013-03-22 11:53:58

Last modified on

2013-03-22 11:53:58

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

22

Author

yark (2760)

Entry type

Definition

Classification

msc 05A10

Classification

msc 11B65

Classification

msc 92-01

Classification

msc 92B05

Synonym

factorial function

Related topic

BinomialCoefficient

Related topic

ExponentialFactorial

Generated on Thu Feb 8 20:35:52 2018 by

LaTeXML