nyuuzyou/wikis huggingface dataset:
Factor of Mersenne Number

== Theorems ==
Let $p$ and $q$ be prime numbers such that $p$ is a divisor of the Mersenne number $M_q$.
Then the following properties hold:
=== Factor of Mersenne Number $M_p$ is of form $2 k p + 1$===
Thus any factor of a Mersenne number can conveniently be referred to by the value of $k$.
=== Factor of Mersenne Number $M_p$ equivalent to $1 \pmod p$ ===
=== Factor of Mersenne Number equivalent to $\pm 1 \pmod 8$ ===
Category:Mersenne Numbers
Factor of Mersenne Number Mp equivalent to 1 mod p

== Theorem ==
Let $p$ and $q$ be prime numbers such that $q$ is a divisor of the Mersenne number $M_p$.
Then:
:$q \equiv 1 \pmod p$
== Proof ==
Let $q \divides M_p$.
From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$: 
:$q = 2 k p + 1$
and so by definition of congruence modulo an integer:
:$q \equiv 1 \pmod p$
immediately.
== Sources ==
* Proof courtesy of The Prime Pages: Modular restrictions on Mersenne divisors.
Category:Mersenne Numbers
Factor of Mersenne Number Mp is of form 2kp + 1

== Theorem ==
Let $p$ and $q$ be prime numbers such that $q$ is a divisor of the Mersenne number $M_p$.
Then:
:$q = 2 k p + 1$
for some integer $k$.
== Proof ==
Let $q \divides M_p$.
Then:
:$2^p \equiv 1 \pmod q$
From Integer to Power of Multiple of Order, the multiplicative order of $2 \pmod q$ divides $p$.
By Fermat's Little Theorem, the multiplicative order of $2 \pmod q$ also divides $q - 1$.
Hence:
:$q - 1 = 2 k p$
== Historical Note ==
== Sources ==
* : Book $\text{IX}$. Proposition $36$: Footnote
Category:Mersenne Numbers
Factor of Mersenne Number Mp is of form 2kp + 1/Historical Note

== Historical Note on Factor of Mersenne Number $M_p$ is of form $2 k p + 1$ ==
This result was known to by $1640$, and it has been suggested that made use of it when providing his famous list of Mersenne primes in his of $1644$.
== Sources ==
* : $127$
* : $127$
Category:Historical Notes
Factor of Mersenne Number equivalent to +-1 mod 8

== Theorem ==
Let $p$ and $q$ be prime numbers such that $q$ is a divisor of the Mersenne number $M_p$.
Then:
:$q \equiv \pm 1 \pmod 8$
== Proof ==
Suppose $q \divides M_p$, where $\divides$ denotes divisibility.
From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$: 
:$q - 1 = 2 k p$
From above:
:$2^{\paren {q - 1} / 2} \equiv 2 k p \equiv 1 \pmod q$
and so $2$ is a quadratic residue $\pmod q$.
From Second Supplement to Law of Quadratic Reciprocity:
:$q \equiv \pm 1 \pmod 8$
== Sources ==
* Proof courtesy of The Prime Pages: Modular restrictions on Mersenne divisors.
Category:Mersenne Numbers
Factorable Polynomial/Examples/Arbitrary Example 1

== Examples of Factorable Polynomials ==
The polynomial:
:$x^2 + x - 2$
is factorable, as:
:$x^2 + x - 2 = \paren {x + 2} \paren {x - 1}$
== Sources ==
* 
* 
Category:Examples of Factorable Polynomials
Factorial/Examples

== Examples of Factorials ==
The factorials of the first few positive integers are as follows:
$\begin{array}{r|r}
n & n! \\
\hline
0 & 1 \\
1 & 1 \\
2 & 2 \\
3 & 6 \\
4 & 24 \\
5 & 120 \\
6 & 720 \\
7 & 5 \, 040 \\
8 & 40 \, 320 \\
9 & 362 \, 880 \\
10 & 3 \, 628 \, 800 \\
\end{array}$
=== Factorial of $0$ ===
=== Factorial of $1$ ===
=== Factorial of $4$ ===
=== Factorial of $5$ ===
=== Factorial of $6$ ===
=== Factorial of $10$ ===
=== Factorial of $11$ ===
=== Factorial of $12$ ===
=== Factorial of $13$ ===
=== Factorial of $14$ ===
=== Factorial of $15$ ===
=== Factorial of $16$ ===
=== Factorial of $17$ ===
=== Factorial of $18$ ===
=== Factorial of $19$ ===
=== Factorial of $20$ ===
=== Factorial of $21$ ===
=== Factorial of $22$ ===
=== Factorial of $23$ ===
=== Factorial of $24$ ===
=== Factorial of $25$ ===
=== Prime Factors of $39!$ ===
=== Factorial of $52$ ===
=== Factorial of $450$ ===
=== Factorial of $1\,000$ ===
=== Factorial of $1\,000\,000$ ===
== Sources ==
* : Tables: $5$ The Factorials of the Numbers $1$ to $20$
* : $\S 1.2.5$: Permutations and Factorials
* : Tables: $5$ The Factorials of the Numbers $1$ to $20$
Category:Factorials/Examples
Factorial/Examples/0

== Example of Factorial ==
The factorial of $0$ is $1$:
:$0! = 1$
== Proof ==
From the definition of factorial:
:$n! = \ds \prod_{k \mathop = 1}^n k$
where $\prod$ denotes continued product.
When $n = 0$ we have:
:$0! = \ds \prod_{k \mathop = 1}^0 k$
Hence the result, by definition of vacuous product.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Factorial $n$: $3.2$
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $21$
* : $\S 1.2.4$: Factorials and binomial coefficients
* : $\S 1.2.5$: Permutations and Factorials: $(5)$
* 
* 
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Factorial $n$: $3.2.$
Category:Factorials/Examples
Factorial/Examples/1,000,000

== Example of Factorial ==
The factorial of $1 \, 000 \, 000$ has $5 \, 569 \, 709$ digits.
== Proof ==
Let $d$ be the number of digits in $1 \, 000 \, 000!$
From Number of Digits in Factorial:
:$d = 1 + \floor {\paren {n + \dfrac 1 2} \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934}$
from which the result can be calculated by setting $n = 1 \, 000 \, 000$.
== Historical Note ==
== Sources ==
* : $24$
* : $24$
Category:Factorials/Examples
Factorial/Examples/1000

== Example of Factorial ==
The factorial of $1\,000$ starts:
:$402,387,260,077 \ldots$ 
and has $2568$ digits, of which the last $249$ are $0$.
=== Prime Factors of $1000!$ ===
== Proof ==
From Stirling's Formula:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
whence an approximate value for $1 \, 000!$ can be calculated.
Let $d$ be the number of digits in $1\,000!$
From Number of Digits in Factorial:
:$d = 1 + \floor {\paren {n + \dfrac 1 2} \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934}$
from which the result can be calculated by setting $n = 1000$.
From Prime Factors of $1000!$:
:the multiplicity of $5$ in $1000!$ is $249$
:the multiplicity of $2$ in $1000!$ is $994$.
Therefore there is a factor of $10^{249}$ in $1000!$, but not $10^{250}$.
Hence there are $249$ instances of $0$ at the end of $1000!$.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
* : $\S 1.2.5$: Permutations and Factorials: Exercise $4$
Category:Factorials/Examples
Factorial/Examples/11

== Example of Factorial ==
:$11! = 39 \, 916 \, 800$
== Proof ==
From Examples of Factorials:
:$10! = 3 \, 628 \, 800$
Then:
 3 628 800
x 11
----------
 3 628 800
36 288 000
----------
39 916 800
Category:Factorials/Examples
Category:11
Factorial/Examples/12

== Example of Factorial ==
:$12! = 479 \, 001 \, 600$
== Proof ==
From $11$ factorial:
:$11! = 39 \, 916 \, 800$
Then:
 39 916 800
x 12
-----------
 79 833 600
399 168 000
-----------
479 001 600
Category:Factorials/Examples
Category:12
Factorial/Examples/13

== Example of Factorial ==
:$13! = 6 \, 227 \, 020 \, 800$
=== Prime Factors of $13!$ ===
== Proof ==
From $12$ factorial:
:$12! = 479 \, 001 \, 600$
Then:
 479 001 600
x 13
-------------
1 437 004 800
4 790 016 000
-------------
6 227 020 800
Category:Factorials/Examples
Category:13
Factorial/Examples/14

== Example of Factorial ==
:$14! = 87 \, 178 \, 291 \, 200$
== Proof ==
From $13$ factorial:
:$13! = 6 \, 227 \, 020 \, 800$
Then:
 6 227 020 800
x 14
---------------
24 908 083 200
62 270 208 000
--------------
87 178 291 200
Category:Factorials/Examples
Category:14
Factorial/Examples/15

== Example of Factorial ==
:$15! = 1 \, 307 \, 674 \, 368 \, 000$
== Proof ==
From $14$ Factorial:
:$14! = 87 \, 178 \, 291 \, 200$
Then:
 87 178 291 200
x 15
------------------
 435 891 456 000
 871 782 912 000
-----------------
1 307 674 368 000
Category:Factorials/Examples
Category:15
Factorial/Examples/16

== Example of Factorial ==
:$16! = 20 \, 922 \, 789 \, 888 \, 000$
== Proof ==
From $15$ Factorial:
:$15! = 1 \, 307 \, 674 \, 368 \, 000$
Then:
 1 307 674 368 000
x 16
------------------
 7 846 046 208 000
13 076 743 680 000
------------------
20 922 789 888 000
Category:Factorials/Examples
Category:16
Factorial/Examples/17

== Example of Factorial ==
:$17! = 355 \, 687 \, 428 \, 096 \, 000$
== Proof ==
From $16$ Factorial:
:$16! = 20 \, 922 \, 789 \, 888 \, 000$
Then:
 20 922 789 888 000
x 17
-------------------
146 459 529 216 000
209 227 898 880 000
-------------------
355 687 428 096 000
Category:Factorials/Examples
Category:17
Factorial/Examples/18

== Example of Factorial ==
:$18! = 6 \, 402 \, 373 \, 705 \, 728 \, 000$
== Proof ==
From $17$ Factorial:
:$17! = 355 \, 687 \, 428 \, 096 \, 000$
Then:
 355 687 428 096 000
x 18
---------------------
2 845 499 424 768 000
3 556 874 280 960 000
---------------------
6 402 373 705 728 000
Category:Factorials/Examples
Category:18
Factorial/Examples/19

== Example of Factorial ==
:$19! = 121 \, 645 \, 100 \, 408 \, 832 \, 000$
== Proof ==
From $18$ Factorial:
:$18! = 6 \, 402 \, 373 \, 705 \, 728 \, 000$
Then:
 6 402 373 705 728 000
x 19
-----------------------
 57 621 363 351 552 000
 64 023 737 057 280 000
-----------------------
121 645 100 408 832 000
Category:Factorials/Examples
Category:19
Factorial/Examples/20

== Example of Factorial ==
:$20! = 2 \, 432 \, 902 \, 008 \, 176 \, 640 \, 000$
=== Prime Factors of $20!$ ===
== Proof ==
From $19$ Factorial:
:$19! = 121 \, 645 \, 100 \, 408 \, 832 \, 000$
Then:
 121 645 100 408 832 000
x 20
------------------------
2 432 902 008 176 640 000
== Sources ==
* : $24$
* : $24$
Category:Factorials/Examples
Category:20
Factorial/Examples/21

== Example of Factorial ==
:$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$
== Proof ==
From $20$ Factorial:
:$20! = 2 \, 432 \, 902 \, 008 \, 176 \, 640 \, 000$
Then:
 2 432 902 008 176 640 000
x 21
---------------------------
 2 432 902 008 176 640 000
 48 658 040 163 532 800 000
---------------------------
 51 090 942 171 709 440 000 
Category:Factorials/Examples
Category:21
Factorial/Examples/22

== Example of Factorial ==
:$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$
== Proof ==
From $21$ Factorial:
:$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$
Then:
 51 090 942 171 709 440 000
x 22
-----------------------------
 102 181 884 343 418 880 000
1 021 818 843 434 188 800 000
-----------------------------
1 124 000 727 777 607 680 000 
Category:Factorials/Examples
Category:22
Factorial/Examples/23

== Example of Factorial ==
:$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$
== Proof ==
From $22$ Factorial:
:$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$
Then:
 1 124 000 727 777 607 680 000
x 23
------------------------------
 3 372 002 183 332 823 040 000
22 480 014 555 552 153 600 000
------------------------------
25 852 016 738 884 976 640 000 
Category:Factorials/Examples
Category:23
Factorial/Examples/24

== Example of Factorial ==
:$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$
== Proof ==
From $23$ Factorial:
:$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$
Then:
 25 852 016 738 884 976 640 000
x 24
-------------------------------
103 408 066 955 539 906 560 000
517 040 334 777 699 532 800 000
-------------------------------
620 448 401 733 239 439 360 000 
Category:Factorials/Examples
Category:24
Factorial/Examples/25

== Example of Factorial ==
:$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$
== Proof ==
From $24$ Factorial:
:$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$
Then:
 620 448 401 733 239 439 360 000
x 25
----------------------------------
 3 102 242 008 666 197 196 800 000
12 408 968 034 664 788 787 200 000
----------------------------------
15 511 210 043 330 985 984 000 000 
Category:Factorials/Examples
Category:25
Factorial/Examples/450/Historical Note

== Historical Note on $450$ Factorial ==
The factorial of $450$ was calculated by in the $1950$s without the aid of a computer.
As he found it had exactly $1001$ digits, he called it the Arabian Nights factorial:
:The first digit of $450!$ falls in the $1001$th place to the left of the decimal point ... this number may be fancifully dubbed the Arabian Nights factorial.
== Sources ==
* 
* : $24$
* : $450!$
Category:Historical Notes
Factorial/Examples/52

== Example of Factorial ==
The number of ways there are to shuffle a $52$-card deck is given by:
:$52! = 806 \ 58175 \ 17094 \ 38785 \ 71660 \ 63685 \ 64037 \ 66975 \ 28950 \ 54408 \ 83277 \ 82400 \ 00000 \ 00000$
=== Prime Factors of $52!$ ===
== Proof ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $1$
Category:Factorials/Examples
Factorial Divides Product of Successive Numbers

== Theorem ==
Let $m, n \in \N_{\ge 1}$ be natural numbers
Let $m^{\overline n}$ be $m$ to the power of $n$ rising.
Then:
:$m^{\overline n} \equiv 0 \bmod n!$
That is, the factorial of $n$ divides the product of $n$ successive numbers.
== Proof ==
Let $m \in \N_{\ge 1}$.
Consider the set:
:$S = \set{m, m + 1, m + 2, \ldots, m + n - 1}$
Note $S$ has $n$ elements.
By Set of Successive Numbers contains Unique Multiple:
:$\set m$ contains a factor of $1$
:$\set {m, m + 1}$ contains a factor of $2$
and in general:
:$\set {m, m + 1, \ldots, m + j - 1}$ contains a factor of $j$.
It follows that $S$ contains factors of $1, 2, 3, \ldots, n$.
Multiplying all elements of $S$ gives:
Category:Number Theory
Category:Factorials
Factorial Divisible by Binary Root

== Theorem ==
Let $n \in \Z: n \ge 1$.
Let $n$ be expressed in binary notation:
:$n = 2^{e_1} + 2^{e_2} + \cdots + 2^{e_r}$
where $e_1 > e_2 > \cdots > e_r \ge 0$.
Let $n!$ be the factorial of $n$.
Then $n!$ is divisible by $2^{n - r}$, but not by $2^{n - r + 1}$.
== Proof ==
A direct application of Factorial Divisible by Prime Power.
Category:Factorials
Category:Binary Notation
Factorial Divisible by Prime Power

== Theorem ==
Let $n \in \Z: n \ge 1$.
Let $p$ be a prime number.
Let $n$ be expressed in base $p$ notation:
:$\ds n = \sum_{j \mathop = 0}^m r_j p^j$
where $0 \le r_j < p$.
Let $n!$ be the factorial of $n$.
Let $p^\mu$ be the largest power of $p$ which divides $n!$, that is:
:$p^\mu \divides n!$
:$p^{\mu + 1} \nmid n!$
Then:
:$\mu = \dfrac {n - \map {s_p} n} {p - 1}$
where $\map {s_p} n$ is the digit sum of $n$ to base $p$.
== Proof ==
If $p > n$, then $\map {s_p} n = n$ and we have that:
:$\dfrac {n - \map {s_p} n} {p - 1} = 0$
which ties in with the fact that $\floor {\dfrac n p} = 0$.
Hence the result holds for $p > n$.
So, let $p \le n$.
From De Polignac's Formula, we have that:
where $p^s < n < p^{s + 1}$.
From Quotient and Remainder to Number Base: General Result:
:$\ds \floor {\frac n {p^k} } = \sqbrk {r_m r_{m - 1} \ldots r_{k + 1} r_k}_p = \sum_{j \mathop = 0}^{m - k} r_j p^j$
where $0 \le r_j < p$.
Hence:
Hence the result.
== Historical Note ==
Category:Factorials
Category:Prime Numbers
Factorial Experiment/Examples/Chemical Process

== Example of Factorial Experiment ==
A typical factorial experiment is one which measures the yield of a chemical process.
Let factor $A$ be three different temperatures: $100 \cels$, $150 \cels$ and $180 \cels$.
Let factor $B$ be two different pressures: $1$ and $2$ standard atmospheres.
If:
:the yield were to increase as temperature increases at the lower pressure
but:
:the yield were to decrease as temperature increases at the higher pressure
then the factors $A$ and $B$ interact.
== Sources ==
* 
* 
Category:Examples of Factorial Experiments
Factorial Greater than Cube for n Greater than 5

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 5$.
Then $n! > n^3$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $6$
Category:Factorials
Category:Cube Numbers
Category:Factorial Greater than Cube for n Greater than 5
Factorial Greater than Cube for n Greater than 5/Proof 1

== Theorem ==
== Proof ==
We note that:
The proof then proceeds by induction.
For all $n \in \Z_{\ge 6}$, let $\map P n$ be the proposition:
:$n! > n^3$
=== Basis for the Induction ===
$\map P 6$ is the case:
Thus $\map P 6$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 6$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$k! > k^3$
from which it is to be shown that:
:$\paren {k + 1}! > \paren {k + 1}^3$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>5}: n! > n^3$
Category:Factorial Greater than Cube for n Greater than 5
Factorial Greater than Cube for n Greater than 5/Proof 2

== Theorem ==
== Proof ==
For $n > 5$, notice that the following inequalities hold:
:$2 \paren {n - 1} = 2 n - 2 > n + 5 - 2 > n$
:$3 \paren {n - 2} = 3 n - 6 > n + 10 - 6 > n$
And thus:
Category:Factorial Greater than Cube for n Greater than 5
Factorial Greater than Square for n Greater than 3

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 3$.
Then $n! > n^2$.
== Proof ==
We note that:
The proof then proceeds by induction.
For all $n \in \Z_{\ge 4}$, let $\map P n$ be the proposition:
:$n! > n^2$
=== Basis for the Induction ===
$\map P 4$ is the case:
Thus $\map P 4$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 4$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$k! > k^2$
from which it is to be shown that:
:$\paren {k + 1}! > \paren {k + 1}^2$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>3}: n! > n^2$
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $6$
Category:Factorials
Category:Square Numbers
Factorial Number System/Multiplication Table

== Factorial Number System ==
The multiplication table in factoradic can be presented as follows:
$\begin {array} {r|rrrrrrrrrrrrr}
\times & 1 & 10 & 11 & 20 & 21 & 100 & 101 & 110 & 111 & 120 & 121 & 200 & 201 \\ \hline
 1 & 1 & 10 & 11 & 20 & 21 & 100 & 101 & 110 & 111 & 120 & 121 & 200 & 201 \\
 10 & 10 & 20 & 100 & 110 & 120 & 200 & 210 & 220 & 300 & 310 & 320 & 1000 & 1010 \\
 11 & 11 & 100 & 111 & 200 & 211 & 300 & 311 & 1000 & 1011 & 1100 & 1111 & 1200 & 1210 \\
 20 & 20 & 110 & 200 & 220 & 310 & 1000 & 1020 & 1110 & 1200 & 1220 & 1310 & 2000 & 2020 \\
 21 & 21 & 120 & 211 & 310 & 1001 & 1100 & 1121 & 1220 & 1311 & 2010 & 2101 & 2200 & 2221 \\
 100 & 100 & 200 & 300 & 1000 & 1100 & 1200 & 1300 & 2000 & 2100 & 2200 & 2300 & 3000 & 3100 \\
 101 & 101 & 210 & 311 & 1020 & 1121 & 1300 & 2001 & 2110 & 2211 & 2320 & 3021 & 3200 & 3301 \\
 110 & 110 & 220 & 1000 & 1110 & 1220 & 2000 & 2110 & 2220 & 3000 & 3110 & 3220 & 4000 & 4110 \\
 111 & 111 & 300 & 1011 & 1200 & 1311 & 2100 & 2211 & 3000 & 3111 & 3300 & 4011 & 4200 & 4311 \\
 120 & 120 & 310 & 1100 & 1220 & 2010 & 2200 & 2320 & 3110 & 3300 & 4020 & 4210 & 10000 & 10120 \\
 121 & 121 & 320 & 1111 & 1310 & 2101 & 2300 & 3021 & 3220 & 4011 & 4210 & 10001 & 10200 & 10321 \\
 200 & 200 & 1000 & 1200 & 2000 & 2200 & 3000 & 3200 & 4000 & 4200 & 10000 & 10200 & 11000 & 11200 \\
 201 & 201 & 1010 & 1211 & 2020 & 2221 & 3100 & 3301 & 4110 & 4311 & 10120 & 10321 & 11200 & 12001 \\
 210 & 210 & 1020 & 1300 & 2110 & 2320 & 3200 & 4010 & 4220 & 10100 & 10310 & 11120 & 12000 & 12210 \\
 211 & 211 & 1100 & 1311 & 2200 & 3011 & 3300 & 4111 & 10000 & 10211 & 11100 & 11311 & 12200 & 13011 \\
 220 & 220 & 1110 & 2000 & 2220 & 3110 & 4000 & 4220 & 10110 & 11000 & 11220 & 12110 & 13000 & 13220 \\
 221 & 221 & 1120 & 2011 & 2310 & 3201 & 4100 & 4321 & 10220 & 11111 & 12010 & 12301 & 13200 & 14021 \\
 300 & 300 & 1200 & 2100 & 3000 & 3300 & 4200 & 10100 & 11000 & 11300 & 12200 & 13100 & 14000 & 14300 \\
 301 & 301 & 1210 & 2111 & 3020 & 3321 & 4300 & 10201 & 11110 & 12011 & 12320 & 13221 & 14200 & 20101 \\
 310 & 310 & 1220 & 2200 & 3110 & 4020 & 10000 & 10310 & 11220 & 12200 & 13110 & 14020 & 20000 & 20310 \\
 311 & 311 & 1300 & 2211 & 3200 & 4111 & 10100 & 11011 & 12000 & 12311 & 13300 & 14211 & 20200 & 21111 \\
 320 & 320 & 1310 & 2300 & 3220 & 4210 & 10200 & 11120 & 12110 & 13100 & 14020 & 20010 & 21000 & 21320 \\
 321 & 321 & 1320 & 2311 & 3310 & 4301 & 10300 & 11221 & 12220 & 13211 & 14210 & 20201 & 21200 & 22121 \\
1000 & 1000 & 2000 & 3000 & 4000 & 10000 & 11000 & 12000 & 13000 & 14000 & 20000 & 21000 & 22000 & 23000 \\
\end {array}$
Category:Factorial Number System
Factorial as Product of Consecutive Factorials

== Theorem ==
The only factorials which are the product of consecutive factorials are:
== Proof ==
Suppose $m, n \in \N$ and $m > n$.
Write $\map F {n, m} = n! \paren {n + 1}! \cdots m!$.
Suppose we have $\map F {n, m} > r!$ for some $r \in \N$.
Suppose further that there is a prime $p$ where $m < p \le r$.
We claim that $\map F {n, m}$ cannot be a factorial of any number.
 $\map F {n, m} = s!$ for some $s \in \N$.
Since $s! > r!$, we must have $r! \divides s!$.
Since $p \le r$, we must have $p \divides r!$.
Thus we have $p \divides s!$.
However, since $n, n + 1, \dots, m < p$, we must have $p \nmid k!$ for each $n \le k \le m$.
Thus $p \nmid \map F {n, m} = s!$, which is a contradiction.
Hence $\map F {n, m}$ cannot be a factorial of some number.
We have the following lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
We also have:
:$\forall n \in \N: n > 1 \implies \exists p: 2 n < p < 3 n$
=== Case $1$: $m$ is an even number larger than $2$ ===
Write $m = 2 k$, where $k > 1$.
Then:
There is a prime $p$ where $m = 2 k < p \le 3 k - 1$.
Therefore $\map F {n, m}$ is not a factorial of some number.
=== Case $2$: $m$ is an odd number larger than $11$ ===
Write $m = 2 k - 1$, where $k \ge 7$.
Then:
There is a prime $p$ where $m = 2 k - 1 < 2 k < p \le 3 k - 1$.
Therefore $\map F {n, m}$ is not a factorial of some number.
=== Case $3$: Particular values of $m$ ===
The cases above leaves us with $m = 1, 2, 3, 5, 7, 9, 11$ to check.
We have:
Thus there are no more.
== Sources ==
* : $3,628,800$
* : $3,628,800$
Category:Factorials
Category:Factorial as Product of Consecutive Factorials
Factorial as Product of Consecutive Factorials/Lemma 1

== Theorem ==
Let $n \in \N$.
Then $\paren {2 n - 1}! \, \paren {2 n}! > \paren {3 n - 1}!$ for all $n > 1$.
== Proof ==
Let $n, k \in \N_{> 0}$.
Suppose $n > 1$ and $n > k$.
We show that $\paren {k + 1} \paren {2 n - k} > 2 n + k$.
For $k = 1$:
:$2 \paren {2 n - 1} = 4 n - 2 \ge 2 n + 2 > 2 n + 1$
For $k > 1$:
 
 
 
Therefore we have:
 
Category:Factorials
Category:Factorial as Product of Consecutive Factorials
Factorial as Product of Consecutive Factorials/Lemma 2

== Theorem ==
Let $n \in \N$.
Then $\paren {2 n - 2}! \, \paren {2 n - 1}! > \paren {3 n - 1}!$ for all $n \ge 7$.
== Proof ==
We prove the result by induction on $n$.
=== Basis for the Induction ===
The case $n = 7$ is verified by direct calculation:
:$12! \times 13! > 20!$
This is the basis for the induction.
=== Induction Hypothesis ===
We suppose for some $k \ge 7$, we have:
:$\paren {2 k - 2}! \, \paren {2 k - 1}! > \paren {3 k - 1}!$
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
The result follows by the Principle of Mathematical Induction.
Category:Factorials
Category:Factorial as Product of Consecutive Factorials
Factorial as Product of Consecutive Factorials/Mistake

== Source Work ==
:
:The Dictionary
::$3,628,800$
:
:The Dictionary
::$3,628,800$
== Mistake ==
:... the only factorial that is the product of other consecutive factorials apart from the trivial $1! = 0! \times 1!$, $2! = 0! \times 1! \times 2!$ and $1! \times 2! = 2!$.
The author missed the other obvious trivial example: $0! = 0! \times 1!$.
== Sources ==
* : $3,628,800$
* : $3,628,800$
Category:Mistakes/Curious and Interesting Numbers
Factorial as Product of Three Factorials

== Theorem ==
This general pattern can be used to find a factorial which is the product of three factorials:
:$\paren {\paren {n!}!}! = n! \paren {n! - 1}! \paren {\paren {n!}! - 1}!$
while there are instances of factorials which do not fit that pattern.
== Proof ==
== Examples ==
Category:Factorials
Category:Factorial as Product of Three Factorials