nyuuzyou/wikis huggingface dataset:
Sequence Lemma

== Lemma ==
Let $A$ be a subset of a topological space $X$.
If there is a sequence of points of $A$ converging to $x$, then $x \in \bar A$.
The converse holds if $X$ is first-countable.
== Proof ==
Assume the sequence of points of $A$ that converges to $x$ is $\sequence {x_i}$.
Then for any open set $U$ of $x$, there exists a positive natural number $N$ such that when $i > N$, $x_i \in U$.
Thus $U \cap A$ is nonempty, $x \in \bar A$.
Let the topological space $X$ be first-countable.
Then there is a countable collection of open neighbourhood $\family {U_i}_{i \mathop \in \Bbb Z_+}$ of $x$ such that any open neighbourhood $U$ of $x$ contains at least one of the sets $U_i$.
Because $x \in \bar A$, $U_1 \cap A$ is nonempty, we can select a point $x_1$ in it.
In a similar manner, $U_1 \cap U_2 \cap A$ is nonempty.
Hence we can select a point $x_2$ in it.
The point $x_i$ is selected from:
:$U_1 \cap U_2 \cap \cdots \cap U_i \cap A$
We then obtain a sequence $\sequence {x_i}$.
For any open neighbourhood $U$ of $x$, it contains at least one of the set $U_N$, $N \in \Bbb Z_+$ of $\family {U_i}_{i \mathop \in \Bbb Z_+}$.
Thus it contains the set:
:$U_1 \cap U_2 \cap \cdots \cap U_N \cap A$
:$U_1 \cap U_2 \cap \cdots \cap U_N \cap U_{N + 1}\cap A$
:$U_1 \cap U_2 \cap \cdots \cap U_N \cap U_{N + 1} \cap U_{N + 2} \cap A$
:$\ldots$
or the set $U_1 \cap U_2 \cap \cdots \cap U_i \cap A$ with $i > N$, hence the points $x_i$ with $i > N$.
The sequence $\sequence {x_i}$ converges to $x$.
== Sources ==
* Chapter $2$ Topological Spaces and Continuous Functions: $\S 21$ The Metric Topology (contiuned): Lemma $21.2$
Category:Topology
Category:Countability Axioms
Category:First-Countable Spaces
Sequence in Indiscrete Space converges to Every Point

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\sequence {s_n}$ be a sequence in $T$.
Then $\sequence {s_n}$ converges to every point of $S$.
== Proof ==
Let $\alpha \in S$.
By definition, $\sequence {s_n}$ converges to $\alpha$ if every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {s_n}$.
But as $T$ has only one open set containing any points at all, every point of $\sequence {s_n}$ is contained in every open set in $T$ containing $\alpha$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $4$
Category:Indiscrete Topology
Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ :
:$\norm {x_n - x} \to 0$
== Proof ==
From the definition of a convergent sequence in a normed vector space, we have that: 
:$x_n$ converges to $x$
:
:for each $\epsilon > 0$ there exists $N \in \N$ such that $\norm {x_n - x} < \epsilon$.
From the definition of a convergent real sequence, we have that: 
:$\norm {x_n - x} \to 0$
:
:for each $\epsilon > 0$ there exists $N \in \N$ such that $\size {\norm {x_n - x} - 0} < \epsilon$.
Since the norm is non-negative, we have that:
:$\norm {x_n - x} \to 0$
:
:for each $\epsilon > 0$ there exists $N \in \N$ such that $\norm {x_n - x} < \epsilon$.
We can therefore immediately deduce the result.
Category:Convergent Sequences (Normed Vector Spaces)
Sequence is Bounded in Norm iff Bounded in Metric

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} } $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Then:
:$\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,} }$ $\sequence {x_n} $ is a bounded sequence in the metric space $\struct {R, d}$.
== Proof ==
=== Necessary Condition ===
=== Sufficent Condition ===
== Also see ==
* Definition:Bounded Sequence in Normed Division Ring
* Definition:Metric Induced by Norm on Division Ring
* Definition:Bounded Sequence in Metric Space
Category:Normed Division Rings
Category:Convergence
Category:Metric Spaces
Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n} $ be a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,}}$ 
Then:
:$\sequence {x_n} $ is a bounded sequence in the metric space $\struct {R, d}$
== Proof ==
Let $\sequence {x_n} $ be a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.
Then:
:$\exists K \in \R_{\gt 0} : \forall n : \norm {x_n} \le K$
Then $\forall n, m \in \N$:
Hence the sequence $\sequence {x_n} $ is bounded by $2 K$ in the metric space $\struct {R, d}$.
== Also see ==
* Definition:Bounded Sequence in Normed Division Ring
* Definition:Metric Induced by Norm on Division Ring
* Definition:Bounded Sequence in Metric Space
Category:Normed Division Rings
Category:Boundedness
Category:Metric Spaces
Sequence is Bounded in Norm iff Bounded in Metric/Sufficient Condition

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$
Then:
:$\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,} }$
== Proof ==
Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$.
Then:
:$\exists K \in \R_{> 0} : \forall n, m : \map d {x_n , x_m} \le K$
By the definition of the metric induced by a norm this is equivalent to:
:$\exists K \in \R_{> 0} : \forall n, m : \norm {x_n - x_m} \le K$
Then $\forall n \in \N$:
Hence the sequence $\sequence {x_n}$ is bounded by $K + \norm {x_1}$ in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
== Also see ==
* Definition:Bounded Sequence in Normed Division Ring
* Definition:Metric Induced by Norm on Division Ring
* Definition:Bounded Sequence in Metric Space
Category:Normed Division Rings
Category:Boundedness
Category:Metric Spaces
Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Q$ denote the rational numbers identified as a dense subfield of $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
Let $\norm {\,\cdot\,} ^\Q _p$ denote the $p$-adic norm on the rational numbers.
Let $\sequence{x_n}$ be a sequence in the rational numbers $\Q$.
Then $\sequence {x_n}$ is a Cauchy sequence in $\struct{\Q, \norm{\,\cdot\,} ^\Q _p}$ $\sequence {x_n}$ is a Cauchy sequence in $\struct{\Q_p, \norm{\,\cdot\,}_p}$
== Proof ==
From Rational Numbers are Dense Subfield of P-adic Numbers:
:the $p$-adic norm $\norm {\,\cdot\,}_p$ on the $p$-adic numbers is an extension of the $p$-adic norm $\norm {\,\cdot\,} ^\Q _p$ on the rational numbers.
That is,
:$\norm {\,\cdot\,}_p \restriction_\Q \mathop = \norm {\,\cdot\,} ^\Q _p$
The result follows from Cauchy Sequence of Subring iff Cauchy Sequence of Normed Division Ring
Category:P-adic Number Theory
Sequence of 11 Primes by Trebling and Adding 16

== Theorem ==
The process of multiplication by $3$ and then adding $16$ produces a sequence of $11$ primes when starting from $587$:
:$587, 1777, 5347, 16 \, 057, 48 \, 187, 144 \, 577, 433 \, 747, 1 \, 301 \, 257, 3 \, 903 \, 787, 11 \, 711 \, 377, 35 \, 134 \, 147$
== Proof ==
== Sources ==
* : $587$
* : $587$
Category:Prime Numbers
Category:Polynomial Expressions for Primes
Sequence of 4 Consecutive Integers with Equal Number of Divisors

== Theorem ==
The following sequence of integers are sets of $4$ consecutive integers which all have the same number of divisors:
:$\map {\sigma_0} m = \map {\sigma_0} {m + 1} = \map {\sigma_0} {m + 2} = \map {\sigma_0} {m + 3}$
where $\map {\sigma_0} n$ denotes the divisor count function.
:$242, 243, 244, 245, 3655, 3656, 3657, 3658, 4503, 4504, 4505, 4506, \ldots$
== Proof ==
== Sources ==
* : $242$
* : $242$
Category:Divisor Count Function
Sequence of 4 Consecutive Square-Free Triplets

== Theorem ==
The following sets of $4$ consecutive triplets of integers, with one integer between each triplet, are square-free:
:$29, 30, 31; 33, 34, 35; 37, 38, 39; 41, 42, 43$
:$101, 102, 103; 105, 106, 107; 109, 110, 111; 113, 114, 115$
== Proof ==
Note that $32, 36, 40$ and $104, 108, 112$ are all divisible by $4 = 2^2$, so are by definition not square-free.
Then inspecting each number in turn:
== Also see ==
* Sequence of Square-Free Triplets
== Sources ==
* : $29$
Category:Square-Free Integers
Sequence of 5 Consecutive Non-Primable Numbers by Changing 1 Digit

== Theorem ==
The following sequence of $5$ consecutive positive integers cannot be made into prime numbers by changing just one digit:
:$872\,894, 872\,895, 872\,896, 872\,897, 872\,898$
== Proof ==
Numbers ending in $0$, $2$, $4$, $6$ and $8$ are not prime because by Divisibility by 2 they are divisible by $2$.
Numbers ending in $0$ and $5$ are not prime because by Divisibility by 5 they are divisible by $5$.
Hence each of $872\,894$, $872\,895$, $872\,896$ and $872\,898$ remain composite when you change any of their digits except the last one.
So we inspect the prime factors of the following, only bothering to check the numbers ending in $1$, $3$, $7$ and $9$:
All we need to do now is to inspect $872\,897$.
The result has been proven.
== Sources ==
* 
* : $872,894$
Category:Prime Numbers
Category:Recreational Mathematics
Category:Numbers that cannot be made Prime by changing 1 Digit
Sequence of 8 Consecutive Primes with Same Pattern of Differences as from 11

== Theorem ==
The $8$ prime numbers starting at $15 \, 760 \, 091$ have the same prime gaps as the $8$ primes starting at $11$.
Those $8$ primes are:
:$15 \, 760 \, 091, 15 \, 760 \, 093, 15 \, 760 \, 097, 15 \, 760 \, 099, 15 \, 760 \, 103, 15 \, 760 \, 109, 15 \, 760 \, 111, 15 \, 760 \, 117$
Their prime gaps are:
:$2, 4, 2, 4, 6, 2, 6$
The $8$ primes starting at $11$:
:$11, 13, 17, 19, 23, 29, 31, 37$
Their prime gaps are:
:$2, 4, 2, 4, 6, 2, 6$
== Sources ==
* 
* : $15,760,091$
Category:Prime Gaps
Sequence of 9 Consecutive Integers each with 48 Divisors

== Theorem ==
The $9$ integers beginning $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044$ each has $48$ divisors.
== Proof ==
In the below, $\sigma_0$ denotes the divisor count function.
but then:
== Sources ==
* 
* : $17,796,126,877,482,329,126,044$
Category:Divisor Count Function
Sequence of 9 Consecutive Integers each with 48 Divisors/Mistake

== Source Work ==
:
:The Dictionary
::$17,796,126,877,482,329,126,044$
== Mistake ==
:The first in a sequence of $9$ consecutive integers, each with $9$ divisors.
== Correction ==
That should read:
:The first in a sequence of $9$ consecutive integers, each with $48$ divisors.
== Sources ==
* : $17,796,126,877,482,329,126,044$
Category:Mistakes/Curious and Interesting Numbers
Sequence of 9 Primes of form 4n+1

== Theorem ==
The following sequence of $9$ consecutive prime numbers are all of the form $4 n + 1$:
:$11 \, 593, 11 \, 597, 11 \, 617, 11 \, 621, 11 \, 633, 11 \, 657, 11 \, 677, 11 \, 681, 11 \, 689$
== Proof ==
It remains to be noted that:
:the prime number before $11 \, 593$ is $11 \, 587$ which is $4 \times 2897 - 1$
:the prime number after $11 \, 689$ is $11 \, 699$ which is $4 \times 2925 - 1$
confirming that they are not of the form $4 n + 1$.
== Historical Note ==
== Sources ==
* : $11,593$
* : $11,593$
Category:Prime Numbers
Sequence of 9 Primes of form 4n+1/Historical Note

== Historical Note on Sequence of $9$ Primes of form $4 n + 1$ ==
In his of $1986$, reports that discusses this in his of $1981$.
In of $1997$, he then attributes this result to a certain Den Haan, but gives no context.
It remains to identify who this is.
== Sources ==
* : $11,593$
* : $11,593$
Category:Historical Notes
Sequence of Best Rational Approximations to Square Root of 2

== Theorem ==
A sequence of best rational approximations to the square root of $2$ starts:
:$\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
where:
:the numerators are half of the Pell-Lucas numbers, $\dfrac 1 2 Q_n$
:the denominators are the Pell numbers $P_n$
starting from $\dfrac {\tfrac12 Q_1} {P_1}$.
== Proof ==
Let $\tuple {a_0, a_1, \ldots}$ be the continued fraction expansion of $\sqrt 2$.
By Continued Fraction Expansion of Root 2:
:$\sqrt 2 = \sqbrk {1, \sequence 2} = \sqbrk {1, 2, 2, 2, \ldots}$
From Convergents are Best Approximations, the convergents of $\sqbrk {1, \sequence 2}$ are the best rational approximations of $\sqrt 2$.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of the continued fraction expansion of $\sqrt 2$.
Then $\dfrac {p_n} {q_n}$ is the $n$th convergent of $\sqbrk {1, \sequence 2}$.
By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $p_n$ and $q_n$ are coprime for all $n \ge 0$.
It remains to show that for all $n \ge 1$:
:$Q_n = 2 p_{n - 1}$
:$P_n = q_{n - 1}$
It is sufficient to prove that they satisfy the same recurrence relation.
By definition:
so that:
:$\tuple {Q_1, P_1} = \tuple {2, 1} = \tuple {2 p_0, q_0}$
so that:
:$\tuple {Q_2, P_2} = \tuple {6, 2} = \tuple {2 p_1, q_1}$
The result follows by definition of Pell numbers and Pell-Lucas numbers.
== Historical Note ==
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $13$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $13$
Category:Square Root of 2
Sequence of Best Rational Approximations to Square Root of 2/Historical Note

== Historical Note on Sequence of Best Rational Approximations to Square Root of 2 ==
The sequence of best rational approximations to the square root of $2$ was known to in the $2$nd century C.E.
More precisely, he knew that if $\dfrac p q$ is an approximation to $\sqrt 2$, then $\dfrac {p + 2 q} {p + q}$ is a better one, which result is demonstrated in Relation between Adjacent Best Rational Approximations to Root 2.
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Historical Notes
Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index

== Corollary to Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\dbinom n k$ be the binomial coefficient of $n$ over $k$ for a positive integer $k \in \Z_{\ge 0}$.
Let $S_n = \left\langle{x_k}\right\rangle$ be the sequence defined as:
:$x_k = \dbinom n k$
Then $S_n$ is strictly decreasing exactly where $\dfrac n 2 < k \le n$.
== Proof ==
If $k > \dfrac n 2$ then it follows that $n - k < \dfrac n 2$.
Thus:
Hence the result.
Category:Binomial Coefficients
Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\dbinom n k$ be the binomial coefficient of $n$ over $k$ for a positive integer $k \in \Z_{\ge 0}$.
Let $S_n = \sequence {x_k}$ be the sequence defined as:
:$x_k = \dbinom n k$
Then $S_n$ is strictly increasing exactly where $0 \le k < \dfrac n 2$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
When $k \ge 0$, we have:
In order for $S_n$ to be strictly increasing, it is necessary for $\dfrac {n - k} {k + 1} > 1$.
So:
Thus $\dbinom n {k + 1} > \dbinom n k$ $k + 1$ is less than half of $n$.
Hence the result.
Category:Binomial Coefficients
Sequence of Column Operations is Column 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 column operation which transforms $\mathbf A$ to a new matrix $\mathbf B \in \map \MM {m, n}$.
Let $\Gamma_2$ be a column operation which transforms $\mathbf B$ to another new matrix $\mathbf C \in \map \MM {m, n}$.
Then there exists another column 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 column 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 column matrices.
Let $\sequence {f_i}_{1 \mathop \le i \mathop \le l}$ be the finite sequence of elementary column 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 column matrices.
From Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices, we have:
:$\mathbf A \mathbf R_1 = \mathbf B$
where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
:$\mathbf R_1 = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k$
Also from Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices, we have:
:$\mathbf B \mathbf R_2 = \mathbf C$
where $\mathbf R_2$ is the product of $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$:
:$\mathbf R_2 = \mathbf F_1 \mathbf F_2 \dotsb \mathbf F_{l - 1} \mathbf F_l$
Hence we have:
:$\mathbf A \mathbf R_1 \mathbf R_2 = \mathbf C$
where $\mathbf R := \mathbf R_1 \mathbf R_2$ is the product:
:$\mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k \mathbf F_1 \mathbf F_2 \dotsb \mathbf F_{l - 1} \mathbf F_l$
Let $\Gamma$ be the column operation composed of the finite sequence of elementary column 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 column operation which transforms $\mathbf A$ into $\mathbf C$.
Hence the result.
== Also see ==
* Sequence of Row Operations is Row Operation
Category:Column Operations
Sequence of Composite Mersenne Numbers

== Theorem ==
The sequence of Mersenne numbers which are composite begins:
:$2047, 8 \, 388 \, 607, 536 \, 870 \, 911, 137 \, 438 \, 953 \, 471, 2 \, 199 \, 023 \, 255 \, 551,\ldots$
The sequence of corresponding indices $p$ such that $2^p - 1$ is composite begins:
:$11, 23, 29, 37, 41, 43, 47, 53, 59, 67, 71, 73, 79, 83, \ldots$
The sequence of corresponding integers $n$ such that the $n$th prime number $p \left({n}\right)$ is such that $2^{p \left({n}\right)} - 1$ is composite begins:
:$5, 9, 10, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, \ldots$
== Examples of Composite Mersenne Numbers ==
== Proof ==
Established by inspecting the sequence of Mersenne numbers:
:$3, 7, 31, 127, 2047, 8191, 131 \, 071, 524 \, 287, 8 \, 388 \, 607, 536 \, 870 \, 911, 2 \, 147 \, 483 \, 647, \ldots$
and removing from it the sequence of Mersenne primes:
:$3, 7, 31, 127, 8191, 131 \, 071, 524 \, 287, 2 \, 147 \, 483 \, 647, \ldots$
== Sources ==
* : $2047$
* : $2047$
Category:Mersenne Numbers
Sequence of Consecutive Integers with Same Number of Divisors

== Theorem ==
The following sequence of consecutive integers all have the same number of divisors, that is, $8$:
:$40 \, 311, 40 \, 312, 40 \, 313, 40 \, 314, 40 \, 315$
This is the longest such sequence known.
== Proof ==
In the below, $\sigma_0$ denotes the divisor count function.
Then we have:
== Sources ==
* 
* : $40,311$
* : $40,311$
Category:Divisor Count Function
Sequence of Differences on Generalized Pentagonal Numbers

== Theorem ==
Recall the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$
Consider the sequence defined as $\Delta_n = GP_{n + 1} - GP_n$:
:$1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, \ldots$
Then:
:The values of $\Delta_n$ for odd $n$ consist of the odd numbers
:The values of $\Delta_n$ for even $n$ consist of the natural numbers.
== Proof ==
Recall the definition of the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$
:$GP_n = \begin{cases} \dfrac {m \paren {3 m + 1} } 2 & : n = 2 m \\
\dfrac {m \paren {3 m - 1} } 2 & : n = 2 m - 1 \end{cases}$
for $n = 0, 1, 2, \ldots$
Hence:
which defines the sequence of natural numbers.
Then:
which defines the sequence of odd numbers.
== Sources ==
* : $22$
* : $22$
Category:Generalized Pentagonal Numbers
Sequence of Dudeney Numbers

== Theorem ==
The only Dudeney numbers are:
:$0, 1, 8, 17, 18, 26, 27$
two of which are themselves cubes, and one of which is prime.
== Proof ==
We have trivially that:
Then:
A quick empirical test shows that when $n = 46$, it is already too large to be the sum of the digits of its cube.
For $46 < n \le 54$, $n^3 \le 54^3 < 200 \, 000$.
Hence the sum of the digits of $n^3$ is less than:
:$1 + 5 \times 9 = 46 < n$
For $54 < n < 100$, $n^3 < 10^6$.
Hence the sum of the digits of $n^3$ is less than:
:$6 \times 9 = 54 < n$
For $n \ge 100$, let $n$ be a $d$-digit number, where $d \ge 3$.
Then $10^{d - 1} \le n < 10^d$ and $n^3 < 10^{3 d}$.
Hence the sum of the digits of $n^3$ is less than:
so no numbers greater than $46$ can have this property.
== Also reported as ==
Some sources (either deliberately or by oversight) do not include $0$ in this list.
== Also see ==
* Definition:Armstrong Number, with which the numbers in this entry appear frequently to be conflated
== Historical Note ==
== Sources ==
* : Solutions: $69$. -- Root Extraction
* : Answers: $120$. Root Extraction
* 
* : $27$
* : $17$
* : $18$
* : $26$
* : $27$
* : $4913$
* : $27$
* 
* : $17$
* : $18$
* : $26$
* : $27$
* : $4913$
* 
Category:Dudeney Numbers
Category:8
Category:17
Category:18
Category:26
Category:27
Sequence of Dudeney Numbers/Historical Note

== Historical Note on Sequence of Dudeney Numbers ==
The earliest known appearance of this result is from in $1879$, although exactly where this was published is still to be identified.
 subsequently published it in one of his own collections.
As a result, a number which equals the sum of the digits of its cube is now called a Dudeney number.
It continues to crop up occasionally in publications devoted to recreational mathematics.
Category:Historical Notes
Sequence of Even Poulet Numbers

== Sequence ==
The sequence of even Poulet numbers begins:
:$161 \, 038, 215 \, 326, 2 \, 568 \, 226, 3 \, 020 \, 626, 7 \, 866 \, 046, 9 \, 115 \, 426, 49 \, 699 \, 666, \ldots$
== Also defined as ==
Some sources include $2$ in this list, but as a Poulet number is usually presupposed on to be composite, this has not been declared so here.
== Historical Note ==
== Sources ==
* : $161,038$
* : $161,038$
Category:Poulet Numbers
Sequence of Fibonacci Numbers ending in Index

== Theorem ==
Let $F_k$ denote the $k$th Fibonacci number.
For all $k \in \Z$, let $F_k$ be expressed in decimal notation.
The sequence of integers $\sequence n$ such that $F_n$ ends in $n$ starts:
:$0, 1, 5, 25, 29, 41, 49, 61, 65, 85, 89, 101, 125, 145, 149, 245, 265, 365, 385, 485, 505, 601, \ldots$
== Sources ==
* 
* : $5$
Category:Fibonacci Numbers
Sequence of Fibonacci Numbers ending in Index/Mistake

== Source Work ==
:
:The Dictionary
:: $5$
== Mistake ==
:The sequence of numbers, $n$, such that $F_n$ ends in $n$, starts: $1 \ 5 \ 25 \ 29 \ 41 \ 49 \ldots$
== Correction ==
The sequence starts with $0$:
:$0, 1, 5, 25, 29, 41, 49, \ldots$
== Sources ==
* : $5$
Category:Mistakes/Curious and Interesting Numbers
Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent

== Theorem ==
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$. 
Then $\sequence {f_n}$ is uniformly Cauchy on $S$ $\sequence {f_n}$ converges uniformly on $S$.
== Proof ==
=== Sufficient Condition ===
Let $\sequence {f_n}$ be uniformly Cauchy on $S$.
=== Necessary Condition ===
Let $\sequence {f_n}$ converge uniformly on $S$ to $f$.
== Sources ==
* : $\S 9.5$: The Cauchy Condition for Uniform Convergence: Theorem $9.3$
Category:Uniform Convergence
Category:Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent