nyuuzyou/wikis huggingface dataset:
Prime Gap Size is Unbounded

== Theorem ==
Let $n \in \N$ be a natural number, arbitrarily large.
Then there exist consecutive prime numbers $p_1$ and $p_2$ such that:
:$p_2 - p_1 > n$
That is, the size of prime gaps is unbounded.
That is, there are blocks of consecutive composite numbers whose length exceeds any given $n \in \N$.
== Proof ==
Let $p$ be a prime number greater than $n + 1$.
Consider the primorial:
:$q = 2 \times 3 \times 5 \times \cdots \times p$
All of the $p - 1$ numbers:
:$q + 2, q + 3, q + 4, \ldots, q + p$
are composite.
Hence the result.
== Sources ==
* : $\text I$: The Series of Primes: $1.4$ The sequence of primes: Theorem $5$
Category:Prime Gaps
Prime Gap of 654

== Theorem ==
There exists a prime gap of $654$ between $11\,000\,001\,446\,613\,353$ and $11\,000\,001\,446\,614\,007$.
== Proof ==
$11\,000\,001\,446\,613\,353$ is a prime number.
$11\,000\,001\,446\,614\,007$ is a prime number.
It can be checked that all numbers between these two are composite.
== Historical Note ==
== Sources ==
* 
* : $11,000,001,446,613,353$
Category:Prime Gaps
Prime Gap of 654/Historical Note

== Historical Note on Prime Gap of 654 ==
This prime gap was reported by in his of $1986$ as being a record.
However, in $1993$ a prime gap of 864 was found between the smaller prime numbers $6 \, 505 \, 941 \, 701 \, 960 \, 039$ and $6 \, 505 \, 941 \, 701 \, 960 \, 903$, again by .
== Sources ==
* 
* : $11,000,001,446,613,353$
Category:Prime Gaps
Prime Gaps of 10

== Theorem ==
The following pairs of consecutive prime numbers are those whose difference is $10$:
:$\tuple {139, 149}, \tuple {181, 191}, \tuple {241, 251}, \tuple {283, 293}, \ldots$
== Proof ==
Demonstrated by listing the prime gaps.
== Sources ==
* : $139$
* : $139$
Category:Prime Gaps
Prime Gaps of 18

== Theorem ==
After the prime gap of $14$ between the pairs of consecutive prime numbers:
:$\left({113, 127}\right), \left({293, 307}\right), \left({317, 331}\right), \ldots$
the next prime gap which is greater than $14$ is between the pair of consecutive prime numbers:
:$\left({523, 541}\right)$
for a prime gap of $18$.
== Proof ==
Demonstrated by listing the prime gaps.
== Sources ==
* : $113$
Category:Prime Gaps
Prime Gaps of 8

== Theorem ==
The following pairs of consecutive prime numbers are those whose difference is $8$:
:$\tuple {89, 97}, \tuple {359, 367}, \tuple {389, 397}, \tuple {401, 409}, \ldots$
== Proof ==
Demonstrated by listing the prime gaps.
== Sources ==
* : $89$
* : $89$
Category:Prime Gaps
Prime Group has no Proper Subgroups

== Theorem ==
A nontrivial group $G$ has no proper subgroups except the trivial group $G$ is finite and its order is prime.
== Proof ==
=== Sufficient Condition ===
Suppose $G$ is finite and of prime order $p$.
From Lagrange's Theorem, the order of any subgroup of $G$ must divide the order $p$ of $G$.
From the definition of prime, any subgroups of $p$ can therefore only have order $1$ or $p$.
Hence $G$ can have only itself and the trivial group as subgroups.
=== Necessary Condition ===
Suppose $G$ is not finite and prime.
Let the identity of $G$ be $e$.
Let $h \in G$ be an element of $G$ such that $h \ne e$.
Then $H = \gen h$ is a cyclic subgroup of $G$.
If $H \ne G$ then $H$ is a non-trivial proper subgroup of $G$, and the proof is complete.
Otherwise, $H = G$ is a cyclic group and there are two possibilities:
:$(1): \quad G$ is infinite
:$(2): \quad G$ is finite (and of non-prime order).
First, suppose $G$ is infinite.
From Infinite Cyclic Group is Isomorphic to Integers, $G$ is isomorphic to $\struct {\Z, +}$.
From Subgroups of Additive Group of Integers, $\struct {\Z, +}$ has proper subgroups, for example: $\gen 2$.
Because $G \cong \struct {\Z, +}$, then so does $G$ have proper subgroups, and the proof is complete.
Suppose $G$ is finite, and of order $n$ where $n$ is not prime .
Then:
: $\exists d \in \N: d \divides n, 1 < d < n$
From Subgroup of Finite Cyclic Group is Determined by Order, $G$ has a proper subgroup of order $d$ and again, the proof is complete.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.7 \ \text{(a)}$
* : Chapter $\text{II}$: Groups: Problem $\text{GG}$
* : Chapter $2$: Subgroups and Cosets: $\S 38 \gamma$
* : Chapter $6$: An Introduction to Groups: Exercise $20$
* : $\S 3.4$: Cyclic groups: Exercise $9$
Category:Subgroups
Category:Prime Groups
Prime Group is Cyclic

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group whose order is $p$.
Then $G$ is cyclic.
== Proof ==
Let $a \in G: a \ne e$ where $e$ is the identity of $G$.
From Group of Prime Order p has p-1 Elements of Order p, $a$ has order $p$.
Hence by definition, $a$ generates $G$.
Hence also by definition, $G$ is cyclic.
== Sources ==
* : $\S 6.4$. Lagrange's theorem: Example $118$
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.7$ Corollary
* : $\S 1.9$: Theorem $13$
* : Chapter $2$: Subgroups and Cosets: $\S 43$
* : $0$: Some Conventions and some Basic Facts: Exercise $1$
* : $\S 44.2$ Some consequences of Lagrange's Theorem
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.19$
Category:Prime Groups
Category:Finite Cyclic Groups
Prime Group is Simple

== Theorem ==
Groups of prime order are simple.
== Proof ==
Follows directly from Prime Group has no Proper Subgroups: a group of prime order has only itself and the trivial group as subgroups.
From Trivial Subgroup and Group Itself are Normal, these subgroups are normal.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $123$
* : Chapter $2$: The Sylow Theorems: $\S 59 \epsilon$
Category:Simple Groups
Category:Prime Groups
Prime Groups of Same Order are Isomorphic

== Theorem ==
Two prime groups of the same order are isomorphic to each other.
== Proof ==
Let $G_1$ and $G_2$ be prime groups, both of finite order $p$.
From Prime Group is Cyclic, both $G_1$ and $G_2$ are cyclic.
The result follows directly from Cyclic Groups of Same Order are Isomorphic.
== Sources ==
* : $\S 1.9$
Category:Prime Groups
Category:Group Isomorphisms
Prime Ideal Including Ideal Includes Radical

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal.
Let $\mathfrak a$ be an ideal of $R$ such that:
:$\mathfrak a \subseteq \mathfrak p$
Let $\map \Rad {\mathfrak a}$ be the radical of $\mathfrak a$.
Then:
:$\map \Rad {\mathfrak a} \subseteq \mathfrak p$
== Proof ==
Let $x \in \relcomp R {\mathfrak p}$.
By :
Therefore, by :
:$x \not\in \map \Rad {\mathfrak a}$
Thus:
:$\relcomp R {\mathfrak p} \subseteq \relcomp R {\map \Rad {\mathfrak a} }$
Therefore, by Relative Complement inverts Subsets:
:$\map \Rad {\mathfrak a} \subseteq \mathfrak p$
Category:Radical of Ideals
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
Prime Ideal iff Quotient Ring is Integral Domain

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $J$ be an ideal of $R$.
Then $J$ is a prime ideal of $R$ the quotient ring $R / J$ is an integral domain.
== Proof ==
Since $J \subset R$, it follows from:
:Quotient Ring of Commutative Ring is Commutative
and:
:Quotient Ring of Ring with Unity is Ring with Unity
that $R / J$ is a commutative ring with unity. 
Let $0_{R / J}$ be the zero of $R / J$.
=== Sufficient Condition ===
Let $J$ be a prime ideal.
We need to show that if:
:$x + J, \ y + J \in \struct {R / J, +, \circ}$
such that:
:$\paren {x + J} \circ \paren {y + J} = \paren {x \circ y} + J = 0_{R / J}$
then:
:$x + J = 0_{R/J}$ or $y + J = 0_{R/J}$.
The zero of $R / J$ is $0_{R / J} = J$.
Therefore:
:$\paren {x \circ y} + J = 0_{R / J} \implies x \circ y \in J$
Because $J$ is a prime ideal, it follows that either $x \in J$ or $y \in J$.
 we assume that $x \in J$.
But then:
:$x + J = J = 0_{R / J}$
=== Necessary Condition ===
Let $R / J$ be an integral domain.
Let $x, y \in R$ be such that $x \circ y \in J$.
Then:
:$0_{R / J} = J = x \circ y + J = \paren {x + J} \circ \paren {y + J}$
Because $R / J$ is an integral domain it follows that:
:$x + J = 0_{R / J}$
or:
:$y + J = 0_{R / J}$
If $x + J = 0_{R / J}$, then $x \in J$.
If $y + J = 0_{R / J}$, then $y \in J$.
Thus either $x \in J$ or $y \in J$, and so by definition $J$ is a prime ideal of $R$.
== Sources ==
* : Chapter $9$: Rings: Exercise $13$
Category:Quotient Rings
Category:Prime Ideals of Rings
Category:Integral Domains
Prime Ideal in Lattice

== Theorem ==
Let $\struct {L, \le}$ be a lattice.
Let $I$ be an ideal in $L$.
Then $I$ is a prime ideal :
:$\forall a, b \in L: a \wedge b \in I \implies a \in I \text{ or } b \in I$
where $a \wedge b$ denotes $\min \set {a, b}$, the meet of $a$ and $b$.
== Proof ==
=== Necessary Condition ===
Let $I$ be a prime ideal.
Let $a, b \in L$ such that $a, b \notin I$.
Then $a, b \in L \setminus I$.
By the definition of prime ideal $L \setminus I$ is a filter.
By the definition of a filter:
:$\exists c \in L \setminus I: c \le a, c \le b$
By the definition of meet:
:$c \le a \wedge b$
Since $c \in L \setminus I$, the definition of a filter implies that
:$a \wedge b \in L \setminus I$
Thus $a \wedge b \notin I$.
By contraposition it follows that $a \wedge b \in I$ implies that $a \in I$ or $b \in I$.
=== Sufficient Condition ===
For all $a$ and $b$ in $L$, let $a \wedge b \in I$ imply that $a \in I$ or $b \in I$.
Let $x, y \in L \setminus I$.
Then $x, y \notin I$.
By supposition, $x \wedge y \notin I$, so:
:$x \wedge y \in L \setminus I$
Let $p \in L \setminus I, q \in L$ such that $p \le q$.
Let $q \in I$.
Then by definition of ideal, $p \in I$.
This contradicts $p \in L \setminus I$.
Thus $q \notin I$.
Thus $q \in L \setminus I$.
Therefore, $L \setminus I$ is a filter.
Category:Order Theory
Prime Ideal is Primary Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal of $R$.
Then $\mathfrak p$ is a primary ideal of $R$.
== Proof ==
Let $xy \in \mathfrak p$.
Let $x \not \in \mathfrak p$
By definition of prime ideal:
:$y^1 = y \in \mathfrak p$
Thus, by definition, $\mathfrak p$ is a primary ideal.
Category:Prime Ideals of Rings
Category:Primary Ideals
Prime Ideal is Prime Element

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $I$ be an ideal in $L$.
Then:
:$I$ is a prime ideal
:
:$I$ is a prime element in $\struct {\map {\mathit {Ids} } L, \precsim}$
where:
:$\map {\mathit {Ids} } L$ denotes the set of all ideals in $L$
:$\mathord \precsim := \mathord \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
== Proof ==
=== Sufficient Condition ===
Let $I$ be a prime ideal.
Let $x, y \in \map {\mathit {Ids} } L$ such that
:$x \wedge y \precsim I$
By definition of $\precsim$:
:$x \wedge y \subseteq I$
By Meet is Intersection in Set of Ideals:
:$x \cap y \subseteq I$
:
:$x \not \precsim I$ and $y \not \precsim I$
By definition of $\precsim$:
$x \nsubseteq I$ and $y \nsubseteq I$
By definition of subset:
:$\exists a \in x: a \notin I$
and
:$\exists b \in y: b \notin I$
By Meet Precedes Operands:
:$a \wedge b \preceq a$ and $a \wedge b \preceq b$
By definition lower section:
:$a \wedge b \in x$ and $a \wedge b \in y$
By definitions subset and intersection:
:$a \wedge b \in I$
By Characterization of Prime Ideal:
:$a \in I$ or $b \in I$
This contradicts $a \notin I$ and $b \notin I$
Hence the assumption:
:$x \not \precsim I$ and $y \not \precsim I$
was false.
=== Necessary Condition ===
Let $I$ be prime element in $\struct {\map {\mathit {Ids} } L, \precsim}$
Let $x, y \in S$ such that:
:$x \wedge y \in I$
By Lower Closure of Element is Ideal:
:$X := x^\preceq$ and $Y := y^\preceq$ are ideals in $L$
By Meet is Intersection in Set of Ideals:
:$X \cap Y = X \wedge Y$
We will prove that:
:$X \wedge Y \subseteq I$
Let $a \in X \wedge Y$.
By definition of intersection:
:$a \in X$ and $a \in Y$
By definition of lower closure of element:
:$a \preceq x$ and $a \preceq y$
By definition of infimum:
:$a \preceq x \wedge y$
Thus by definition of lower section:
:$a \in I$
By definition of $\precsim$:
:$X \wedge Y \precsim I$
By definition of prime element:
:$X \precsim I$ or $Y \precsim I$
By definition of $\precsim$:
:$X \subseteq I$ or $Y \subseteq I$
By definition of reflexivity:
:$x \preceq x$ and $y \preceq y$
By definition of lower closure of element:
:$x \in X$ and $y \in Y$
Thus by definition of subset:
:$x \in I$ or $y \in I$
Hence by Characterization of Prime Ideal:
:$I$ is prime ideal.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Category:Prime Elements
Prime Ideal is Prime Filter in Dual Lattice

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $X$ be a subset of $S$.
Then
:$X$ is a prime ideal in $L$
:
:$X$ is a prime filter in $L^{-1}$
where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.
== Proof ==
=== Sufficient Condition ===
Let $X$ be a prime ideal in $L$.
Then
:$X$ is an ideal in $L$.
By Ideal is Filter in Dual Ordered Set:
:$X$ is a filter in $L^{-1}$.
Let $x, y \in S$ such that
:$x \vee' y \in X$
where $\vee'$ denotes the join in $L^{-1}$.
By Join is Dual to Meet:
:$x \wedge y \in X$
Thus by Characterization of Prime Ideal:
:$x \in X$ or $y \in X$
Hence $X$ is a prime filter
=== Necessary Condition ===
This follows by mutatis mutandis.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Category:Dual Orderings
Prime Ideal of Principal Ideal Domain is Maximal

== Theorem ==
Let $D$ be a principal ideal domain whose zero is $0_D$.
Let $J \subseteq D$ be a nonzero prime ideal.
Then $J$ is maximal.
== Proof ==
As $D$ is a principal ideal domain, every ideal of $D$ is a principal ideal $\ideal r$ generated by some $r \in D$.
So, let $\ideal p$ be an arbitrary prime ideal of $D$ generated by $p$, where $p \ne 0_R$.
As $\ideal p$ is prime, $p$ is irreducible.
The result follows from Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal.
== Sources ==
* : Chapter $9$: Rings: Exercise $14$
Category:Principal Ideal Domains
Category:Maximal Ideals of Rings
Category:Prime Ideals of Rings
Prime Ideals of Ring of Integers

== Theorem ==
Let $\struct {\Z, +, \times}$ denote the ring of integers.
Let $J$ be a prime ideal of $\Z$.
Then either:
:$J = \set 0$
or:
:$J = \ideal p$
where:
:$p$ is a prime number
:$\ideal p$ denotes the principal ideal of $\Z$ generated by $p$.
== Proof ==
From Prime Ideal iff Quotient Ring is Integral Domain:
:$J$ is a prime ideal of $\Z$ $\Z / J$ is an integral domain.
From Quotient Ring of Integers and Zero:
:$\Z / \set 0 \cong \Z$
As $\Z$ is an integral domain, it follows that $\set 0$ is a prime ideal of $\Z$.
From Quotient Ring of Integers with Principal Ideal:
:$\struct {\Z, +, \times} / \ideal p$ is isomorphic to $\struct {\Z_p, +_p, \times_p}$, the ring of integers modulo $p$.
From Ring of Integers Modulo Prime is Integral Domain:
:$\struct {\Z_n, +_n, \times_p}$ is an integral domain $p$ is prime.
The result follows.
== Sources ==
* : Chapter $9$: Rings: Exercise $13$
Category:Integers
Category:Prime Ideals of Rings
Prime Magic Square/Examples/Order 12/Smallest with Consecutive Primes from 3/Mistake

== Source Work ==
:
:The Dictionary
::$144$
:
:The Dictionary
::$144$
== Mistake ==
:The smallest magic square composed of consecutive primes comprises the $144$ odd primes from $3$ upwards. The magic constant is $4515$.
The prime magic square being referred to is actually composed of the $143$ odd primes from $3$ upwards, along with $1$.
The magic constant of this square is in fact $4514$, not $4515$.
Elementary arithmetic tells us that the magic constant, being the sum of $12$ (an even number of) odd primes, will in fact be even.
Hence it cannot be $4515$.
It also needs to be pointed out that the smallest magic square composed of consecutive primes is actually of order $3$:
== Sources ==
* : $144$
* : $144$
Category:Mistakes/Curious and Interesting Numbers
Prime Magic Square/Examples/Order 3

== Examples of Order $3$ Prime Magic Squares ==
There are many order $3$ prime magic squares.
==== Smallest Order $3$ Prime Magic Square ====
The order $3$ prime magic square which has the smallest elements is as follows:
==== Smallest Order $3$ Prime Magic Square with Consecutive Primes ====
The smallest prime magic square which has the smallest elements which are consecutive primes is the following one of order $3$:
Category:Prime Magic Squares
Prime Magic Square/Examples/Order 3/Smallest with Consecutive Primes

== Example of Order $3$ Prime Magic Square ==
This order $3$ prime magic square is the smallest whose elements are consecutive odd primes:
:$\begin{array}{|c|c|c|}
\hline 1 \, 480 \, 028 \, 159 & 1 \, 480 \, 028 \, 153 & 1 \, 480 \, 028 \, 201 \\
\hline 1 \, 480 \, 028 \, 213 & 1 \, 480 \, 028 \, 171 & 1 \, 480 \, 028 \, 129 \\
\hline 1 \, 480 \, 028 \, 141 & 1 \, 480 \, 028 \, 189 & 1 \, 480 \, 028 \, 183 \\
\hline
\end{array}$
== Proof ==
== Also see ==
* Magic Constant of Smallest Prime Magic Square with Consecutive Primes
== Historical Note ==
== Sources ==
* 
* 
* : $1,480,028,171$
* 
Category:Prime Magic Squares
Prime Number/Sequence

== Sequence of Prime Numbers ==
The sequence of prime numbers starts:
:$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, \ldots$
=== Table of Prime Numbers ===
== Sources ==
* : $\text {2-2}$ Divisibility
* : Chapter $1$: Properties of the Natural Numbers: $\S 22$
* : $\text I$: The Series of Primes: $1.4$ The sequence of primes
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Theorem $5$
* : Chapter $\text {A}.31$: Chebyshev ()
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : Chapter $\text {B}.16$: The Sequence of Primes
* 
* : $\text A$. Prime Numbers
* 
* : Chapter $7$: Patterns in Numbers: Primes
Category:Prime Numbers
Prime Number/Table

== Table of Prime Numbers ==
The prime numbers smaller than $1000$ are tabulated as follows.
In the following, $\map p n$ denotes the $n$th prime number.
$\quad \begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
1 & 2 \\
2 & 3 \\
3 & 5 \\
4 & 7 \\
5 & 11 \\
6 & 13 \\
7 & 17 \\
8 & 19 \\
9 & 23 \\
10 & 29 \\
11 & 31 \\
12 & 37 \\
13 & 41 \\
14 & 43 \\
15 & 47 \\
16 & 53 \\
17 & 59 \\
18 & 61 \\
19 & 67 \\
20 & 71 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
21 & 73 \\
22 & 79 \\
23 & 83 \\
24 & 89 \\
25 & 97 \\
26 & 101 \\
27 & 103 \\
28 & 107 \\
29 & 109 \\
30 & 113 \\
31 & 127 \\
32 & 131 \\
33 & 137 \\
34 & 139 \\
35 & 149 \\
36 & 151 \\
37 & 157 \\
38 & 163 \\
39 & 167 \\
40 & 173 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
41 & 179 \\
42 & 181 \\
43 & 191 \\
44 & 193 \\
45 & 197 \\
46 & 199 \\
47 & 211 \\
48 & 223 \\
49 & 227 \\
50 & 229 \\
51 & 233 \\
52 & 239 \\
53 & 241 \\
54 & 251 \\
55 & 257 \\
56 & 263 \\
57 & 269 \\
58 & 271 \\
59 & 277 \\
60 & 281 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
61 & 283 \\
62 & 293 \\
63 & 307 \\
64 & 311 \\
65 & 313 \\
66 & 317 \\
67 & 331 \\
68 & 337 \\
69 & 347 \\
70 & 349 \\
71 & 353 \\
72 & 359 \\
73 & 367 \\
74 & 373 \\
75 & 379 \\
76 & 383 \\
77 & 389 \\
78 & 397 \\
79 & 401 \\
80 & 409 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
81 & 419 \\
82 & 421 \\
83 & 431 \\
84 & 433 \\
85 & 439 \\
86 & 443 \\
87 & 449 \\
88 & 457 \\
89 & 461 \\
90 & 463 \\
91 & 467 \\
92 & 479 \\
93 & 487 \\
94 & 491 \\
95 & 499 \\
96 & 503 \\
97 & 509 \\
98 & 521 \\
99 & 523 \\
100 & 541 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
101 & 547 \\
102 & 557 \\
103 & 563 \\
104 & 569 \\
105 & 571 \\
106 & 577 \\
107 & 587 \\
108 & 593 \\
109 & 599 \\
110 & 601 \\
111 & 607 \\
112 & 613 \\
113 & 617 \\
114 & 619 \\
115 & 631 \\
116 & 641 \\
117 & 643 \\
118 & 647 \\
119 & 653 \\
120 & 659 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
121 & 661 \\
122 & 673 \\
123 & 677 \\
124 & 683 \\
125 & 691 \\
126 & 701 \\
127 & 709 \\
128 & 719 \\
129 & 727 \\
130 & 733 \\
131 & 739 \\
132 & 743 \\
133 & 751 \\
134 & 757 \\
135 & 761 \\
136 & 769 \\
137 & 773 \\
138 & 787 \\
139 & 797 \\
140 & 809 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
141 & 811 \\
142 & 821 \\
143 & 823 \\
144 & 827 \\
145 & 829 \\
146 & 839 \\
147 & 853 \\
148 & 857 \\
149 & 859 \\
150 & 863 \\
151 & 877 \\
152 & 881 \\
153 & 883 \\
154 & 887 \\
155 & 907 \\
156 & 911 \\
157 & 919 \\
158 & 929 \\
159 & 937 \\
160 & 941 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|}
\hline
n & \map p n \\
\hline
161 & 947 \\
162 & 953 \\
163 & 967 \\
164 & 971 \\
165 & 977 \\
166 & 983 \\
167 & 991 \\
168 & 997 \\
\hline
\end{array}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $2$
* : Tables: $4$ The Prime Numbers less than $1000$
* : Tables: $4$ The Prime Numbers less than $1000$
* : Appendix $18$: Prime numbers less than $1000$
Category:Prime Numbers
Prime Number Formed by Concatenating Consecutive Integers down to 1

== Theorem ==
Let $N$ be an integer whose decimal representation consists of the concatenation of all the integers from a given $n$ in descending order down to $1$.
Let the $N$ that is so formed be prime.
The only $n$ less than $100$ for which this is true is $82$.
That is:
:$82 \, 818 \, 079 \, 787 \, 776 \ldots 121 \, 110 \, 987 \, 654 \, 321$
is the only prime number formed this way starting at $100$ or less.
== Proof ==
Can be determined by checking all numbers formed in such a way for primality.
== Historical Note ==
== Sources ==
 
* : $82,818,079,787,776,757,473,727,170 \, \ldots \, 10,987,654,321$
* 
Category:Prime Numbers
Prime Number Formed by Concatenating Consecutive Integers down to 1/Historical Note

== Historical Note on Prime Number Formed by Concatenating Consecutive Integers down to 1 ==
 reports in his of $2004$ that this result was determined by and , but he provides no citation and it has not been corroborated.
== Sources ==
 
* : $82,818,079,787,776,757,473,727,170 \, \ldots \, 10,987,654,321$
* 
Category:Historical Notes
Prime Number Race/Examples/3n+1 vs. 3n-1

== Example of Prime Number Race ==
In the prime number race between prime numbers of the form $3 n - 1$ and $3 n + 1$, the prime numbers of the form $3 n + 1$ overtake those of the form $3 n - 1$ for the first time at $608 \, 981 \, 813 \, 029$.
== Historical Note ==
== Sources ==
* 
* : $608,981,813,029$
* : $608,981,813,029$
Category:Prime Number Races
Prime Number Race/Examples/3n+1 vs. 3n-1/Historical Note

== Historical Note on Prime Number Race: $3 n + 1$ vs. $3 n - 1$ ==
 and discovered this fact on Christmas day $1976$.
It has been suggested that this may have been the only opportunity they had to obtain the appropriate computer time.
== Sources ==
* : $608,981,813,029$
* : $608,981,813,029$
Category:Historical Notes
Prime Number Race/Examples/4n+1 vs. 4n-1

== Example of Prime Number Race ==
The sequence of prime numbers at which the prime number race between prime numbers of the form $4 n - 1$ and $4 n + 1$ are tied begins:
:$2, 5, 17, 41, 461, 26 \, 833, 26 \, 849, 26 \, 863, 26 \, 881, 26 \, 893, 26 \, 921, 616 \, 769, \ldots$
The details of this prime number race is as follows:
== Proof ==
== Sources ==
* 
* : $26,861$
* : $461$
* : $26,861$
Category:Prime Number Races