nyuuzyou/wikis huggingface dataset:
Sequence of Powers of Number less than One/Necessary Condition

== Theorem ==
Let $x \in \R$ be such that $\size{x} < 1$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Then $\sequence {x_n}$ is a null sequence.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also known as ==
This result and Sequence of Powers of Reciprocals is Null Sequence are sometimes referred to as the basic null sequences.
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
, assume that $x \ne 0$.
Observe that :
:$0 < \size x < 1$
Thus by Ordering of Reciprocals:
:$\size x^{-1} > 1$
Define:
:$h = \size x^{-1} - 1 > 0$
Then:
:$x = \dfrac 1 {1 + h}$
By the binomial theorem, we have that:
:$\paren {1 + h}^n = 1 + n h + \cdots + h^n > n h$
because $h > 0$.
By Absolute Value Function is Completely Multiplicative, it follows that:
:$0 \le \size {x^n} = \size x^n = \dfrac 1 {\paren {1 + h}^n} < \dfrac 1 {n h}$
From Sequence of Reciprocals is Null Sequence:
:$\dfrac 1 n \to 0$ as $n \to \infty$
By the Multiple Rule for Real Sequences:
:$\dfrac 1 {n h} \to 0$ as $n \to \infty$
By the :
:$\size {x^n} \to 0$
as $n \to \infty$.
Hence the result, by the definition of a limit.
== Sources ==
* : $3.20 e$
* : $\S 4$: Convergent Sequences: $\S 4.12$: Example
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Necessary Condition/Proof 2

== Theorem ==
== Proof ==
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Suppose that:
:$\exists N \in \N: \size x^N < \epsilon$
Then the result follows by the definition of a limit, because:
:$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$
where Absolute Value Function is Completely Multiplicative is applied.
It remains to consider the case that:
:$\forall n \in \N: \size x^n \ge \epsilon$
By the Axiom of Archimedes, we can choose a natural number $M$ such that:
:$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$
Then, by Sum of Geometric Sequence:
:$\ds M \epsilon \le \sum_{k \mathop = 0}^{M - 1} \size x^k = \frac {1 - \size x^M} {1 - \size x} < M \epsilon$ 
a contradiction.
Hence the result.
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Necessary Condition/Proof 3

== Theorem ==
== Proof ==
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By the Axiom of Archimedes, there exists a natural number $M$ such that:
:$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$
By the Well-Ordering Principle, there exists a smallest natural number $m$ such that:
:$\exists N \in \N: m > M \size x^N$
Note that:
:$m - 1 \le M \size x^{N + 1}$
By elementary algebra, it follows that:
:$1 - \dfrac 1 m = \dfrac {m - 1} m \le \size x < 1 - \dfrac 1 {M \epsilon}$
Hence, $m < M \epsilon$.
Therefore:
:$\size x^N < \epsilon$
By Absolute Value Function is Completely Multiplicative:
:$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$
Hence the result, by the definition of a limit.
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Necessary Condition/Proof 4

== Theorem ==
== Proof ==
Define:
:$\ds L = \inf_{n \mathop \in \N} \size x^n$
By the Continuum Property, such an $L$ exists in $\R$.
Clearly, $L \ge 0$.
 $L > 0$.
Then, by the definition of the infimum, we can choose $n \in \N$ such that $\size x^n < L \size x^{-1}$.
But then $\size x^{n + 1} < L$, which contradicts the definition of $L$.
Therefore, $L = 0$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By the definition of the infimum, there exists an $N \in \N$ such that $\size x^N < \epsilon$.
It follows that:
:$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$
where Absolute Value Function is Completely Multiplicative is applied.
Hence the result, by the definition of a limit.
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring
Let $x \in R$.
Let $\sequence {x_n}$ be the sequence in $R$ defined as $x_n = x^n$.
Then:
:$\norm x < 1$ $\sequence {x_n}$ is a null sequence.
== Proof ==
Let $0_R$ be the zero of $R$.
By the definition of convergence:
:$\ds \lim_{n \mathop \to \infty} x_n = 0_R \iff \lim_{n \mathop \to \infty} \norm {x_n} = 0$
By then for each $n \in \N$:
:$\norm {x_n} = \norm {x^n} = \norm x^n$.
So:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = 0 \iff \lim_{n \mathop \to \infty} \norm x^n = 0$
Since $\norm x \in \R_{\ge 0}$, by Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} \norm x^n = 0 \iff \norm x < 1$
The result follows.
== Sources ==
* : $\S 1.2$: Normed fields: Proposition $1.8$
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One
Category:Normed Division Rings
Sequence of Powers of Number less than One/Rational Numbers

== Theorem ==
Let $x \in \Q$.
Let $\sequence {x_n}$ be the sequence in $\Q$ defined as $x_n = x^n$.
Then:
:$\size x < 1$ $\sequence {x_n}$ is a null sequence.
== Proof ==
By the definition of convergence of a rational sequence:
:$\sequence {x_n}$ is a null sequence in the rational numbers $\sequence {x_n}$ is a null sequence in the real numbers
By Sequence of Powers of Real Number less than One:
:$\sequence {x_n}$ is a null sequence in the real numbers $\size x < 1$
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Sufficient Condition

== Theorem ==
Let $x \in \R$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Let $\sequence {x_n}$ be a null sequence.
Then $\size x < 1$.
== Proof ==
By Reciprocal of Null Sequence:
:$\sequence {x_n}$ converges to $0$ $\sequence {\dfrac 1 {x_n} }$ diverges to $\infty$.
By the definition of divergence to $\infty$:
:$\exists N \in \N: \forall n \ge N: \size {\dfrac 1 {x_n} } > 1$
In particular:
:$\size {\dfrac 1 {x_N} } > 1$
By Ordering of Reciprocals:
:$\size {x_N} < 1$
That is:
:$\size {x_N} = \size {x^N} = \size x^N < 1$
 $\size x \ge 1$.
By Inequality of Product of Unequal Numbers:
:$\size x^N \ge 1^N = 1$
This is a contradiction.
So $\size x < 1$ as required.
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One
Sequence of Powers of Reciprocals is Null Sequence

== Theorem ==
Let $r \in \Q_{>0}$ be a strictly positive rational number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = \dfrac 1 {n^r}$
Then $\sequence {x_n}$ is a null sequence.
=== Real Index ===
If $r \in \R_{>0}$ is a strictly positive real number, the same result applies.
However, the result is specifically stated for a rational index, as this definition is used in the course of derivation of the existence of a power to a real index.
=== Corollary ===
== Proof ==
Let $\epsilon \in \R_{>0}$.
We need to show that:
:$\exists N \in \N: n > N \implies \size {\dfrac 1 {n^r} } < \epsilon$
That is, that $n^r > 1 / \epsilon$.
Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r} }$.
By Reciprocal of Strictly Positive Real Number is Strictly Positive and power of positive real number is positive, it follows that:
:$\paren {\dfrac 1 \epsilon}^{1/r} \gt 0$
Then by Positive Power Function on Non-negative Reals is Strictly Increasing:
:$\forall n > N: n^r > N^r \ge 1 / \epsilon$
== Also see ==
This result and Sequence of Powers of Number less than One‎ are sometimes referred to as the basic null sequences.
* Limit at Infinity of $x^n$
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.6 \ (2)$
Category:Limits of Sequences
Category:Reciprocals
Category:Sequence of Powers of Reciprocals is Null Sequence
Sequence of Powers of Reciprocals is Null Sequence/Corollary

== Corollary to Sequence of Powers of Reciprocals is Null Sequence ==
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = \dfrac 1 n$
Then $\sequence {x_n}$ is a null sequence.
== Proof ==
$n = n^1$ from the definition of power.
As $1 \in \Q_{>0}$ the result follows from Sequence of Powers of Reciprocals is Null Sequence.
Category:Sequence of Powers of Reciprocals is Null Sequence
Sequence of Powers of Reciprocals is Null Sequence/Real Index

== Theorem ==
Let $r \in \R_{>0}$ be a strictly positive real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
: $x_n = \dfrac 1 {n^r}$
Then $\sequence {x_n}$ is a null sequence.
== Proof ==
Let $\epsilon > 0$.
We need to show that:
:$\exists N \in \N: n > N \implies \size {\dfrac 1 {n^r} } < \epsilon$
That is, that $n^r > 1 / \epsilon$.
Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r} }$.
By Reciprocal of Strictly Positive Real Number is Strictly Positive and power of positive real number is positive, it follows that:
:$\paren {\dfrac 1 \epsilon}^{1/r} \gt 0$
Then by Positive Power Function on Non-negative Reals is Strictly Increasing:
:$\forall n > N: n^r > N^r \ge 1 / \epsilon$
== Also see ==
This result and Sequence of Powers of Number less than One‎ are sometimes referred to as the basic null sequences.
* Limit at Infinity of x^n
Category:Sequence of Powers of Reciprocals is Null Sequence
Sequence of Prime Primorial minus 1

== Theorem ==
For prime $p$, let $p \#$ denote the $p$th primorial, defined in the sense that $p \#$ is the product of all primes less than or equal to $p$.
The sequence $\sequence p$ such that $p \# - 1$ is prime begins:
:$3, 5, 11, 13, 41, 89, 317, 337, 991, 1873, 2053, 2377, 4093, 4297, \ldots$
== Also see ==
* Definition:Primorial Prime
* Sequence of Prime Primorial plus 1
== Sources ==
* 
* : $29$
* 
* : $29$
Category:Primorials
Sequence of Prime Primorial minus 1/Mistake

== Source Work ==
:
:The Dictionary
::$29$
:
:The Dictionary
::$29$
== Mistake ==
:Primorial $(n) - 1$ is prime for $3$, $5$, $11$, $13$, $41$, $89$, $317$, $991$, $1873$, $2053$, and no other values below $2377$.
$337$ was omitted from the above list.
However, it needs to be pointed out that the source work that this result was quoted from (the article in by , and , was the source of this inaccuracy:
:The number $N = p \# - 1$ is prime for:
::$3$, $5$, $11$, $13$, $41$, $89$, $317$, $991$, $1873$, $2053$
:and is composite for all other $p < 2377$. The number $2377\# - 1$ is a pseudoprime whose primality has not been verified.
Under the entry for $15,877$ in , interestingly, does include $337$, but this time somehow manages to omit $13$.
== Sources ==
* 
* : $29$
* : $29$
Category:Mistakes/Curious and Interesting Numbers
Sequence of Prime Primorial plus 1

== Theorem ==
For prime $p$, let $p \#$ denote the $p$th primorial, defined in the sense that $p \#$ is the product of all primes less than or equal to $p$.
The sequence $\sequence p$ such that $p \# + 1$ is prime begins:
:$2, 3, 5, 7, 11, 31, 379, 1019, 1021, 2657, 3229, 4547, 4787, 11549, 13649, 18523, 23801, 24029, 42209, 145823, 366439, 392113, \ldots$
== Also see ==
* Sequence of Prime Primorial minus 1
== Sources ==
* 
* : $31$
* : $31$
* : $24,029 \# + 1$
Category:Primorials
Sequence of Record Peaks in Values of Divisor Sum

== Theorem ==
Let $d: \Z_{>0} \to \Z_{>0}$ be the mapping defined as:
:$\map d n = \map {\sigma_1} n - \map {\sigma_1} {n'}$
where:
:$n$ denotes a highly abundant number
:$n'$ denotes the previous highly abundant number.
The following $n \in \Z_{>0}$ have the property that they have a higher value of $\map d n$ than any smaller $n$:
:$\forall m \in \Z_{>0}: m < n \implies \map {\sigma_1} m < \map {\sigma_1} n$
That is, they are the peak divisor sums which exceed the previous peak by a higher number than any previous peak.
That is, they are highly abundant number which have divisor sums whose difference with that of the divisor sum of the previous highly abundant numbers is greater than that with the previous record difference.
:
$n$ 	$n'$ 	$\map {\sigma_1} n$	$\map {\sigma_1} {n'}$	$\map d n$
$2$ 	$1$ 	$3$ 	$1$ 	$2$
$4$ 	$3$ 	$7$ 	$4$ 	$3$
$6$ 	$4$ 	$12$ 	$7$ 	$5$
$12$ 	$10$ 	$28$ 	$18$ 	$10$
$24$ 	$20$ 	$60$ 	$42$ 	$18$
$36$ 	$30$ 	$91$ 	$72$ 	$19$
$48$ 	$42$ 	$124$ 	$96$ 	$28$
$60$ 	$48$ 	$168$ 	$124$ 	$44$
$120$	$108$	$360$ 	$280$ 	$80$
== Sources ==
* : $168$
Category:Highly Abundant Numbers
Sequence of Recursively Defined Terms forms Minimally Inductive Set

== Theorem ==
Let $g$ be a mapping.
Let $\sequence {a_n}$ be a sequence such that:
:$a_0 = \O$
and:
:$\forall n \in \N: a_{n + 1} = \map g {a_n}$
Then the set:
:$\set {a_0, a_1, \ldots, a_n, a_{n + 1}, \ldots}$
is minimally inductive under $g$.
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Exercise $8.6$
Category:Minimally Inductive Classes
Sequence of Row Operations is Row Operation

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\Gamma_1$ be a row operation which transforms $\mathbf A$ to a new matrix $\mathbf B \in \map \MM {m, n}$.
Let $\Gamma_2$ be a row operation which transforms $\mathbf B$ to another new matrix $\mathbf C \in \map \MM {m, n}$.
Then there exists another row operation $\Gamma$ which transforms $\mathbf A$ back to $\mathbf C$ such that $\Gamma$ consists of $\Gamma_1$ followed by $\Gamma_2$.
== Proof ==
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary row operations that compose $\Gamma_1$.
Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary row matrices.
Let $\sequence {f_i}_{1 \mathop \le i \mathop \le l}$ be the finite sequence of elementary row operations that compose $\Gamma_2$.
Let $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$ be the corresponding finite sequence of the elementary row matrices.
From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have:
:$\mathbf R_1 \mathbf A = \mathbf B$
where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
:$\mathbf R_1 = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
Also from Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have:
:$\mathbf R_2 \mathbf B = \mathbf C$
where $\mathbf R_2$ is the product of $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$:
:$\mathbf R_2 = \mathbf F_l \mathbf F_{l - 1} \dotsb \mathbf F_2 \mathbf F_1$
Hence we have:
:$\mathbf R_2 \mathbf R_1 \mathbf A = \mathbf C$
where $\mathbf R := \mathbf R_2 \mathbf R_1$ is the product:
:$\mathbf F_l \mathbf F_{l - 1} \dotsb \mathbf F_2 \mathbf F_1 \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
Let $\Gamma$ be the row operation composed of the finite sequence of elementary row operations $\tuple {e_1, e_2, \ldots, e_{k - 1}, e_k, f_1, f_2, \ldots, f_{l - 1}, f_l}$.
Thus $\Gamma$ is a row operation which transforms $\mathbf A$ into $\mathbf C$.
Hence the result.
== Also see ==
* Sequence of Column Operations is Column Operation
Category:Row Operations
Sequence of Smallest 3 Consecutive Triangular Numbers which are Sphenic

== Theorem ==
The smallest $3$ consecutive triangular numbers which are sphenic is:
:$406$, $435$, $465$
== Proof ==
Let $T_n$ denote the $n$th triangular number.
The smallest sphenic number is $30$.
Hence we need investigate triangular number from where $T_n \ge 30$.
Thus:
== Also see ==
* There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers
Category:Triangular Numbers
Category:Sphenic Numbers
Category:406
Sequence of Smallest Consecutive Composite Numbers longer than 100

== Theorem ==
The $1$st prime gap greater than $100$ is between $370 \, 261$ and $370 \, 373$, of length $112$.
That is, the sequence of the smallest consecutive composite positive integers longer than $100$ is that of $111$ such, from $370 \, 262$ to $370 \, 372$.
== Proof ==
== Sources ==
* 
* : $111$
Category:Prime Gaps
Sequence of Smallest Numbers whose Reciprocal has Period n

== Theorem ==
Let $\sequence {s_n}$ be the sequence defined as:
:$s_n$ is the smallest positive integer the decimal expansion of whose reciprocal has a period of $n$
for $n = 0, 1, 2, \ldots$
Then $\sequence {s_n}$ begins:
:$1, 3, 11, 27, 101, 41, 7, 239, 73, 81, 451, \ldots$
== Proof ==
Demonstrated by inspection and calculation:
== Sources ==
* : $27$
Category:Reciprocals
Sequence of Square Centered Hexagonal Numbers/Mistake

== Source Work ==
:
:The Dictionary
::$169$
== Mistake ==
:The smallest square hexagonal number, apart from $1$ [is $169$]. The next smallest are $32, 761$ and $6,355,441$.
The hexagonal numbers are those of this form:
:
What is intended here is:
:The smallest square centered hexagonal number, apart from $1$.
The centered hexagonal numbers are those of this form:
:
== Sources ==
* : $169$
Category:Mistakes/Curious and Interesting Numbers
Sequence of Square Roots of Natural Numbers is not Cauchy

== Theorem ==
Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:
:$x_n = \sqrt n$
Then, despite the fact that from Difference Between Adjacent Square Roots Converges:
:$\size {\sqrt {n + 1} - \sqrt n} \to 0$ as $n \to \infty$
it is not the case that $\sequence {x_n}$ is a Cauchy sequence.
== Proof ==
 $\sequence {x_n}$ is bounded.
Let $H \in \R$ be an upper bound of $\sequence {x_n}$
By the Axiom of Archimedes:
:$\exists N \in \N: N > H^2$
and so:
:$\exists N \in \N: \sqrt N > H$
But $\sqrt N$ is a term of $\sequence {x_n}$.
So $H$ cannot be an upper bound of $\sequence {x_n}$
Hence by Proof by Contradiction, $\sequence {x_n}$ is not bounded.
From Real Cauchy Sequence is Bounded, it follows by the Rule of Transposition that $\sequence {x_n}$ is not a Cauchy sequence
So, while we have the case that:
:$\size {\sqrt {n + 1} - \sqrt n} \to 0$ as $n \to \infty$
it is not the case that:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \size {\sqrt n - \sqrt m} < \epsilon$
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.21 \ (1)$
Category:Cauchy Sequences
Sequence of Squares Beginning and Ending with n 4s

== Sequence ==
The following sequence:
cannot be continued, as it is not possible for there to be a square number ending in $\ldots 4444$.
== Proof ==
Let $n = 10000 k + 4444$.
We have:
By Square Modulo 4, $\dfrac n 4$ cannot be a square number.
Therefore neither can $n = 4 \times \dfrac n 4$ be a square number.
== Historical Note ==
== Sources ==
* : $444,171,597,444$
Category:Square Numbers
Category:Recreational Mathematics
Sequence of Sum of Squares of Digits

== Theorem ==
For a positive integer $n$, let $\map f n$ be the integer created by adding the squares of digits of $n$.
Let $m \in \Z_{>0}$ be expressed in decimal notation.
Let $\sequence {S_m}_{n \mathop \in \Z_{>0} }$ be the sequence defined as follows:
:$n_k = \begin{cases} m & : n = 1 \\
\map f {n_{k - 1} } & : n > 1 \end{cases}$
Then eventually either $\sequence {S_m}$ sticks at $1$, or goes into the cycle:
:$\ldots, 4, 16, 37, 58, 89, 145, 42, 20, 4, \ldots$
== Proof ==
First note that:
:$1^2 + 9^2 + 9^2 = 163$
:$9^2 + 9^2 + 9^2 = 243$
and it can be seen that for a positive integer $m$ larger than $199$, $\map f m < m$.
Thus it is necessary to investigate numbers only up as far as that.
Starting from the bottom, we have that:
:$\map f 1 = 1^2 = 1$
and so $\sequence {S_1} = 1, 1, 1, \ldots$
We note the sequence:
Hence any $m$ in the set $\set {4, 16, 20, 37, 42, 58, 89, 145}$ stays within that sequence, which we will refer to as $\SS$.
It remains to test the rest.
Let $\TT$ be the set of integers $m$ for which $S_m$ ends up either in $\SS$ or $1$.
Then the elements of $\SS$ are all in $\TT$, as is $1$.
If $m \in \TT$, then any integer $m'$ whose non-zero digits are the same as those of $m$ is also in $\TT$.
Thus:
:$2, 10, 24, 40, 61, 73, 85, 98, 100, 106 \in \TT$
We use the notation:
:$a \to b \to c$
to denote:
:$\map f a = b, \map f b = c$
and work progressively through $\Z_{>0}$.
So:
== Sources ==
* : $4$
* : $89$
* : $4$
* : $89$
Category:Sums of Squares
Category:Recreational Mathematics
Sequence of Triplets of Primitive Pythagorean Triangles with Same Area

== Sequence ==
The sequence of sets of $3$ primitive Pythagorean triangles which all have the same area begins:
:
Generator 	Lengths of sides 	Area
$\left({77, 38}\right)$ 	$\left({4485, 5852, 7373}\right)$ 	$13 \, 123 \, 110$
$\left({78, 55}\right)$ 	$\left({3059, 8580, 9109}\right)$ 	$13 \, 123 \, 110$
$\left({138, 5}\right)$ 	$\left({1380, 19 \, 019, 19 \, 069}\right)$ 	$13 \, 123 \, 110$
$\left({2035, 266}\right)$ 	$\left({4 \, 070 \, 469, 1 \, 082 \, 620, 4 \, 163 \, 285}\right)$ 	$2 \, 203 \, 385 \, 574 \, 390$
$\left({3306, 61}\right)$ 	$\left({10 \, 925 \, 915, 403 \, 332, 7 \, 309 \, 493}\right)$ 	$2 \, 203 \, 385 \, 574 \, 390$
$\left({3422, 55}\right)$ 	$\left({11 \, 707 \, 059, 376 \, 420, 11 \, 713, 109}\right)$ 	$2 \, 203 \, 385 \, 574 \, 390$
$\left({1610, 869}\right)$ 	$\left({1 \, 836 \, 939, 2 \, 798 \, 180, 3 \, 347 \, 261}\right)$ 	$2 \, 570 \, 042 \, 985 \, 510$
$\left({2002, 1817}\right)$ 	$\left({706 \, 515, 7 \, 275 \ , 268, 7 \, 309 \, 493}\right)$ 	$2 \, 570 \, 042 \, 985 \, 510$
$\left({2622, 143}\right)$ 	$\left({6 \, 854 \, 435, 749 \, 892, 6 \, 895 \, 333}\right)$ 	$2 \, 570 \, 042 \, 985 \, 510$
$\left({2201, 1166}\right)$ 	$\left({3 \, 484 \, 845, 5 \, 132 \, 732, 6 \, 203 \, 957}\right)$ 	$8 \, 943 \, 387 \, 723 \, 270$
$\left({2438, 2035}\right)$ 	$\left({1 \, 802 \, 619, 9 \, 922 \, 660, 10 \, 085 \, 069}\right)$ 	$8 \, 943 \, 387 \, 723 \, 270$
$\left({3565, 198}\right)$ 	$\left({12 \, 670 \, 021, 1 \, 411 \, 740, 12 \, 748 \, 429}\right)$ 	$8 \, 943 \, 387 \, 723 \, 270$
$\left({7238, 2465}\right)$ 	$\left({46 \, 312 \, 419, 35 \, 683 \, 340, 6 \, 203 \, 957}\right)$ 	$826 \, 290 \, 896 \, 699 \, 730$
$\left({9077, 1122}\right)$ 	$\left({81 \, 133 \, 045, 20 \, 368 \, 788, 10 \, 085 \, 069}\right)$ 	$826 \, 290 \, 896 \, 699 \, 730$
$\left({10 \, 434, 731}\right)$ 	$\left({108 \, 333 \, 995, 15 \, 254 \, 508, 12 \, 748 \, 429}\right)$ 	$826 \, 290 \, 896 \, 699 \, 730$
$\left({352 \, 538, 2 \, 999 \, 447}\right)$ 	$\left({8 \, 872 \, 399 \, 264 \, 365, 2 \, 114 \, 838 \, 092 \, 972, 9 \, 120 \, 965 \, 347 \, 253}\right)$	$9 \, 381 \, 843 \, 970 \, 167 \, 926 \, 138 \, 271 \, 390$
$\left({1 \, 931 \, 103, 2 \, 398 \, 838}\right)$	$\left({2 \, 025 \, 264 \, 953 \, 635, 9 \, 264 \, 806 \, 516 \, 628, 9 \, 483 \, 582 \, 546 \, 853}\right)$	$9 \, 381 \, 843 \, 970 \, 167 \, 926 \, 138 \, 271 \, 390$
$\left({3 \, 063 \, 347, 3 \, 215 \, 070}\right)$	$\left({952 \, 580 \, 262 \, 491, 19 \, 697 \, 750 \, 078 \, 580, 19 \, 720 \, 769 \, 947 \, 309}\right)$ 	$9 \, 381 \, 843 \, 970 \, 167 \, 926 \, 138 \, 271 \, 390$
These appear to be the only ones known.
== Historical Note ==
== Sources ==
* 
* : $13,123,110$
Category:Pythagorean Triples
Sequence of Triplets of Primitive Pythagorean Triangles with Same Area/Historical Note

== Historical Note on Sequence of Triplets of Primitive Pythagorean Triangles with Same Area ==
The first of these sets of $3$, with area $13 \, 123 \, 110$, was reported by as having been discovered by in $1945$.
The following $3$ sets of $3$ were discovered by in $1986$.
The $5$th set of $3$ was discovered by and in $1990$.
The $6$th set of $3$ was discovered by using an exhaustive computer search over areas, and reported in The Math Forum on the $2$nd of March, $2017$. The search took roughly $35$ GHz-days.
== Sources ==
* 
Category:Historical Notes
Sequence of Triplets of Primitive Pythagorean Triangles with Same Area/Mistake

== Source Work ==
:
:The Dictionary
::$13,123,110$
== Mistake ==
:The area of the next such triplet is $2,570,042,985,510$.
Ordering the sets of $3$ primitive Pythagorean triangles with equal area in order of that area, the $2$nd such area is in fact $2,203,385,574,390$.
== Sources ==
* : $13,123,110$
Category:Mistakes/Curious and Interesting Numbers