nyuuzyou/wikis huggingface dataset:
Sequence of Integers whose Factorial plus 1 is Prime/Mistake

== Source Work ==
:
:The Dictionary
::$1477! + 1$
== Mistake ==
:The largest known prime of the form $n! + 1$, which is also prime for $1$, $2$, $3$, $11$, $27$, $37$, $41$, $73$, $77$, $116$, $154$, $320$, $340$, $399$, $427$, $872$.
$0! + 1 = 1 + 1 = 2$ is mistakenly omitted here.
The same mistake is repeated on the page $1477$.
== Sources ==
* : $1477! + 1$
Category:Mistakes/Curious and Interesting Numbers
Sequence of Mappings Converges Pointwise iff Converges at Each Point

== Theorem ==
Let $D$ be a set.
Let $T$ be a topological space.
For each $n \in \N$, let $f_n : D \to T$ be a mapping.
Let $f : D \to T$ also be a mapping.
Then:
:$\sequence {f_n}$ converges pointwise to $f$
:for every $x \in D$, $\sequence {\map {f_n} x}$ converges to $\map f x$
== Proof ==
Follows from Sequence on Product Space Converges to Point iff Projections Converge to Projections of Point, with:
:$I = D$
:$\forall i \in D: T_i = T$
== Source ==
* : $7$ Complete Metric Spaces and Function Spaces: $\S 46$ Pointwise and Compact Convergence: Theorem $46.1$
Category:Convergence
Sequence of Natural Powers of Right Shift Operator in 2-Sequence Space Converges in Weak Operator Topology

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
Let $R \in \map {CL} {\ell^2}$ be the right shift operator over $\ell^2$.
Let $\sequence {R^n}_{n \mathop \in \N}$ be a sequence.
Let $\mathbf 0 \in \map {CL} {\ell^2}$ be the zero mapping.
Then $\sequence {R^n}_{n \mathop \in \N}$ converges to $\mathbf 0$ in the weak operator topology.
== Proof ==
By Representation Theorem:
:$\ds \forall \phi \in \map {CL} {\ell^2, \C} : \exists \mathbf x_\phi = \sequence {\map {\mathbf x_\phi} k}_{k \mathop \in \N} \in \ell^2 : \forall \mathbf a = \sequence {\map {\mathbf a} k}_{k \mathop \in \N} \in \ell^2 : \map \phi {\mathbf a} = \sum_{k \mathop = 1}^\infty \map {\mathbf a} k \paren{\map {\mathbf x_\phi} k}^*$
where $*$ denotes the complex conjugation.
Furthermore:
Hence, $\sequence {R^n}_{n \mathop \in \N}$ converges to $\mathbf 0 \in \map {CL} {\ell^2}$ in the weak operator topology.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$
Category:Convergence
Category:Operator Theory
Sequence of Natural Powers of Right Shift Operator in 2-Sequence Space does not Converge in Strong Operator Topology

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
Let $R \in \map {CL} {\ell^2}$ be the right shift operator over $\ell^2$.
Let $\sequence {R^n}_{n \mathop \in \N}$ be a sequence.
Let $\mathbf 0 \in \map {CL} {\ell^2}$ be the zero mapping.
Then $\sequence {R^n}_{n \mathop \in \N}$ does not converge to $\mathbf 0$ in the strong operator topology.
== Proof ==
Let $\mathbf e_1 = \tuple {1, 0, \ldots} \in \ell^2$
Then $R^n \mathbf e_1 = \tuple {\underbrace {\ldots, 0}_{n \text{ terms} }, 1, 0, \ldots}$
So:
:$\forall n \in \N : \norm {R^n \mathbf e_1}_2 = 1$
Therefore:
:$\ds \lim_{n \mathop \to \infty} \norm {R^n \mathbf e_1}_2 = 1$
Hence, $\sequence {R^n}_{n \mathop \in \N}$ does not converge to $\mathbf 0$ in the strong operator topology.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$
Category:Convergence
Category:Operator Theory
Sequence of Numbers Divisible by Sequence of Primes

== Theorem ==
The integers in this sequence:
:$788, 789, 790, 791, 792, 793$
are divisible by:
:$2, 3, 5, 7, 11, 13$
respectively.
== Proof ==
== Sources ==
* : $788$
Category:Prime Numbers
Category:Divisors
Category:Recreational Mathematics
Category:788
Category:789
Category:790
Category:791
Category:792
Category:793
Sequence of Numbers with Integer Arithmetic and Harmonic Means of Divisors

== Theorem ==
The following sequence of integers have the property that both the harmonic mean and arithmetic mean of their divisors are integers:
:$1, 6, 140, 270, 672, \ldots$
== Proof ==
Note the integers whose harmonic mean of their divisors are integers are the Ore numbers:
:$1, 6, 28, 140, 270, 496, 672, \ldots$
It remains to calculate the arithmetic mean of their divisors.
Let $\map A n$ denote the arithmetic mean of the divisors of $n$.
Then we have:
:$\map A n = \dfrac {\map {\sigma_1} n} {\map {\sigma_0} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
== Sources ==
* : $140$
Category:Ore Numbers
Sequence of Odd Abundant Numbers without 5 as Factor

== Sequence ==
The sequence of odd abundant numbers which do not have $5$ as a factor begins:
:$81 \, 081, 153 \, 153, 171 \, 171, 189 \, 189, 207 \, 207, 223 \, 839, \ldots$
== Proof ==
== Sources ==
* 
* : $81,081$
Category:Abundant Numbers
Sequence of P-adic Integers has Convergent Subsequence

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Then:
:there exists a convergent subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ of $\sequence{x_n}$
== Proof 1 ==
== Proof 2 ==
Category:P-adic Integers
Category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Lemma 1

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Then there exists a $p$-adic digit $b_0$ such that:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digit $b_0$
== Proof ==
=== Case 1 ===
Let there exist $b \in \set{0, 1, \ldots , p - 2}$:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $y_n$ begins with the $p$-adic digits $b$
Let $b_0 = b$ and the result holds.
=== Case 2 ===
For all $b \in \set{0, 1, \ldots , p - 2}$:
:there exists only a finite number of $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digit $b$
So:
:$\set{n \in \N : \mathop \exists b \in \set{0, 1, \ldots , p - 2} : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digit } b}$ is a finite set.
Let:
:$N = \max \set{n \in \N : \mathop \exists b \in \set{0, 1, \ldots , p - 2} : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b}$
Hence for all $n > N$ and $b \in \set{0, 1, \ldots , p - 2}$:
:$x_n$ does not begin with the $p$-adic digits $b$.
So for all $n > N$:
:the canonical expansion of $y_n$ begins with the $p$-adic digit $p-1$
Let $b_0 = p - 1$ and the result follows.
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Lemma 2

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Let $\sequence{b_0, b_1, \ldots, b_j}$ be a finite sequence of $p$-adic digits such that:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
Then there exists a $p$-adic digit $b_{j + 1}$ such that:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_{j+1}b_j \, \ldots \, b_1 b_0$
== Proof ==
=== Lemma ===
=== Case 1 ===
Let there exist $b \in \set{0, 1, \ldots , p - 2}$:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $y_n$ begins with the $p$-adic digits $b b_j \, \ldots \, b_1 b_0$
Let $b_{j+1} = b$ and the result holds.
=== Case 2 ===
For all $b \in \set{0, 1, \ldots , p - 2}$:
:there exists only a finite number of $n \in \N$ such that the canonical expansion of $y_n$ begins with the $p$-adic digits $b b_j \, \ldots \, b_1 b_0$
So:
:$\set{n \in \N : \mathop \exists b \in \closedint{0}{p-2} : \text{ the canonical expansion of } y_n \text{ begins with the } p \text{-adic digits } b b_j \, \ldots \, b_1 b_0}$ is a finite set.
Let:
:$N = \max \set{n \in \N : \mathop \exists b \in \closedint{0}{p-2} : \text{ the canonical expansion of } y_n \text{ begins with the } p \text{-adic digits } b b_j \, \ldots \, b_1 b_0}$
Hence for all $n > N$ and $b \in \closedint{0}{p-2}$:
:$y_n$ does not begin with the $p$-adic digits $b b_j \, \ldots \, b_1 b_0$.
So for all $n > N$:
:the canonical expansion of $y_n$ begins with the $p$-adic digits $\paren{p-1} b_j \, \ldots \, b_1 b_0$
Let $b_{j+1} = p - 1$ and the result follows.
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Lemma 3

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Then:
:there exists a sequence $\sequence{b_n}$ of $p$-adic digits:
::for all $j \in \N$, there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
The proof proceeds using the second principle of recursive definition.
Let $T$ be the set of $p$-adic digits.
From lemma 1 , there exists $b_0 \in {0, 1, \ldots , p-1}$ such that:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_0$
For each $j \in \N$, let $\map {P_j} {\sequence{b_0, b_1, \ldots, b_j} }$ be the propositional function:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
For any non-empty subset $S$ of $\N$, let $\min S$ denote the smallest element of $S$.
From the Well-Ordering Principle, for any non-empty subset $S$ of $\N$, $\min S$ always exists.
For each $j \in \N$, let $g_j : T^{j+1} \to T$ be defined by:
:$\map {g_j} {\sequence{b_0, b_1, \ldots, b_j} } = \begin{cases}
\min \set{b \in T: \map {P_{j+1}} {\sequence{b_0, b_1, \ldots, b_j, b} } } & : \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } \\
0 & : \lnot \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } 
\end{cases}$
From lemma 2, for each $j \in \N$:
:$g_j$ is well-defined
By definition of $g_j$:
:$\mathop \forall j \in \N: \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } \text{ is true } \implies \map {P_{j+1}} {\sequence{b_0, b_1, \ldots, b_j, \map {g_j} {\sequence{b_0, b_1, \ldots, b_j} } } } \text{ is true }$
From second principle of recursive definition:
:there exists exactly one mapping $b: \N \to T$ such that:
::$\forall j \in \N: b_j = \begin{cases}
\sequence{ b_0 } & : j = 0 \\
\map {g_j} {\sequence{b_0, \ldots, b_n} } & : j = n + 1
\end{cases}$
We have:
:$(1) \quad \map {P_0} {\sequence{b_0} }$ is true
:$(2) \quad \mathop \forall j \in \N: \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } \text{ is true } \implies \map {P_{j+1}} {\sequence{b_0, b_1, \ldots, b_j, b_{j+1} } } \text{ is true }$
From Principle of Mathematical Induction:
:$\forall j \in \N: \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } \text{ is true }$
Hence:
:there exists a sequence $\sequence{b_n}$ of $p$-adic digits:
::for all $j \in \N$, there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Lemma 4

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Let $\sequence{b_n}$ be a sequence of $p$-adic digits such that:
::for all $j \in \N$, there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
Then:
:there exists a subsequence $\sequence{x_{n_j}}_{j \mathop \in \N}$ of $\sequence{x_n}$:
::for all $j \in \N$, the canonical expansion of $x_{n_j}$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
== Proof ==
The proof proceeds using the principle of recursive definition.
For any non-empty subset $S$ of $\N$, let $\min S$ denote the smallest element of $S$.
From the Well-Ordering Principle, for any non-empty subset $S$ of $\N$, $\min S$ always exists.
Let $T = \N \times \N$.
Let $n_0 = \min \set{n \in \N : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b_0}$
By hypothesis and Well-Ordering Principle:
:$n_0$ exists
By definition of smallest element:
:the canonical expansion of $x_{n_0}$ begins with the $p$-adic digit $b_0$
Let $a = \tuple{0, n_0} \in T$.
Let $h: T \to \N$ be the mapping defined by:
:$\map h {j, k} = \min \set{n > k : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b_{j+1}, b_j, \ldots, b_0}$
By hypothesis and Well-Ordering Principle:
:$h$ is well-defined
By definition of smallest element:
:the canonical expansion of $x_{\map h {j, k} }$ begins with the $p$-adic digit $b_{j+1} b_j \ldots b_1 b_0$
Let $g: T \to T$ be the mapping defined by:
:$\map g {j, k} = \tuple {j+1, \map h {j, k} }$
From principle of recursive definition:
:there exists exactly one mapping $f: \N \to T$ such that:
::$\forall j \in \N: \map f j = \begin{cases}
a & : j = 0 \\
\map g {\map f n} & : j = n + 1
\end{cases}$
That is:
:there exists exactly one mapping $f: \N \to T$ such that:
::$\forall j \in \N: \map f j = \begin{cases}
\tuple{0, n_0} & : j = 0 \\
\tuple {j, \map h {n, \map f n} } & : j = n + 1
\end{cases}$
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Lemma 5

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Let $\sequence{b_n}$ be a sequence of $p$-adic digits such that:
:the canonical expansion $\ldots \, b_n \, \ldots \, b_1 b_0$ converges to $x$ in the $p$-adic integers $\Z_p$
Let $\sequence{x_{n_rj}}_{j \mathop \in \N}$ be a subsequence of $\sequence{x_n}$:
:for all $j \in \N$, the canonical expansion of $x_{n_j}$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
Then:
:the subsequence $\sequence{x_{n_j}}_{j \mathop \in \N}$ converges to $x \in \Z_p$
== Proof ==
By definition of the canonical expansion $\ldots \, b_n \, \ldots \, b_1 b_0$ converges to $x$:
:the sequence of partial sums $\ds \sum_{n \mathop = 0}^j b_n p^n$ converges to $x$
Let $\sequence{y_j}$ be the sequence of partial sums:
:$y_j = \ds \sum_{n \mathop = m}^j b_n p^n$
From Null Sequence Test for Convergence, it is sufficient to show that:
:$\sequence{x_{n_j} - y_j}$ is a null sequence
For all $j \in \N$, we have:
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
:$\norm{x_{n_j} - y_j}_p \le \dfrac 1 {p^{j + 1} } < \dfrac 1 {p^j}$
From Sequence of Powers of Number less than One
:$\ds \lim_{j \mathop \to \infty} \dfrac 1 {p^j} = 0$
From Squeeze Theorem for Real Sequences:
:$\ds \lim_{j \mathop \to \infty} \norm{x_{n_j} - y_j}_p = 0$
By definition of convergence:
:$\sequence{x_{n_j} - y_j}$ converges to $0$
It follows that $\sequence{x_{n_j} - y_j}$ is a null sequence by definition.
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Lemma 6

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\sequence{x_n}$ be a sequence of $p$-adic integers.
Let $\sequence{b_0, b_1, \ldots, b_j}$ be a finite sequence of $p$-adic digits, possibly empty, such that:
:there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
Then there exists a subsequence $\sequence{y_n}$ of $\sequence{x_n}$:
:for all $n \in \N$, the canonical expansion of $y_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$
== Proof ==
For any non-empty subset $S$ of $\N$, let $\min S$ denote the smallest element of $S$.
From the Well-Ordering Principle, for any non-empty subset $S$ of $\N$, $\min S$ always exists.
Let $g:\N \to \N$ be the mapping defined by:
:$\map g n = \min \set{n > j : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b_j \, \ldots \, b_1 b_0}$
Let:
:$n_0 = \min \set{n \in \N : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b_j \, \ldots \, b_1 b_0}$
From Well-Ordering Principle:
:$g$ and $n_0$ are well-defined
From Principle of Recursive Definition:
:there exists exactly one mapping $r: \N \to \N$ such that:
::$\forall m \in \N: \map r m = \begin{cases}
a & : m = 0 \\
\map g {\map r n} & : m = n + 1
\end{cases}$
It is noted that:
:
Hence $r$ is a strictly increasing sequence in $\N$.
For all $n \in \N$, let $m_n = \map r n$.
Then $\sequence{x_{m_n}}$ is a subsequence of $\sequence{x_n}$ by definition.
Let $\sequence{y_n} = \sequence{x_{m_n}}$ and the result follows
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Proof 1

== Theorem ==
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
From P-adic Expansion Converges to P-adic Number:
:$\ds \sum_{n \mathop = 0}^\infty b_n p^n$ converges to some $x \in \Q_p$
By definition of $p$-adic integer:
:$x \in \Z_p$
=== Lemma 5 ===
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Theorem $1.34$
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of P-adic Integers has Convergent Subsequence/Proof 2

== Theorem ==
== Proof ==
From P-adic Integers are Compact Subspace:
:$\Z_p$ is a compact subspace in the metric space induced by $\norm{\,\cdot\,}_p$
From Compact Subspace of Metric Space is Sequentially Compact in Itself:
:$\Z_p$ is sequentially compact in itself
By definition of sequentially compact in itself:
:every sequence in $\Z_p$ has a subsequence which converges in the topology to a point in $\Z_p$
From Equivalence of Definitions of Convergence in Normed Division Rings:
:every sequence in $\Z_p$ has a subsequence which converges in the norm to a point in $\Z_p$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Problem $107$
category:Sequence of P-adic Integers has Convergent Subsequence
Sequence of Palindromic Cubes

== Sequence ==
The sequence of positive integers whose cube is palindromic begins:
:$1, 2, 7, 11, 101, 111, 1001, 2201, 10 \, 001, 10 \, 101, \ldots$
Note that $2201$ is the smallest (and only one one known) which is itself non-palindromic.
The corresponding sequence of palindromic cubes begins:
:$1, 8, 343, 1331, 1 \, 030 \, 301, 1 \, 367 \, 631, 1 \, 003 \, 003 \, 001, 10 \, 662 \, 526 \, 601, 1 \, 000 \, 300 \, 030 \, 001, \ldots$
== Also see ==
* Palindromic Cube with Non-Palindromic Root
== Sources ==
* : $2201$
* : $10,662,526,601$
Category:Cube Numbers
Category:Palindromic Numbers
Sequence of Palindromic Lucky Numbers

== Theorem ==
The The sequence of lucky numbers which are also palindromic begins:
:$1, 3, 7, 9, 33, 99, 111, 141, 151, 171, \ldots$
== Proof ==
Of these, the palindromic ones can be picked out by inspection.
== Sources ==
* : $111$
Category:Palindromic Numbers
Category:Lucky Numbers
Sequence of Palindromic Pentagonal Numbers

== Sequence ==
The sequence of pentagonal numbers which are also palindromic begins:
:$1, 5, 22, 1001, 2882, 15 \, 251, 720 \, 027, 7 \, 081 \, 807, 7 \, 451 \, 547, 26 \, 811 \, 862, \ldots$
The corresponding sequence of integers $n$ such that the $n$th pentagonal number is also palindromic begins:
:$1, 2, 4, 26, 44, 101, 693, 2173, 2229, 4228, 6010, \ldots$
== Sources ==
* : $1001$
Category:Pentagonal Numbers
Category:Palindromic Numbers
Sequence of Palindromic Sophie Germain Primes

== Theorem ==
The number $N = 191 \, 918 \, 080 \, 818 \, 091 \, 909 \, 090 \, 909 \, 190 \, 818 \, 080 \, 819 \, 191$ has the property that:
:$N$ is a palindromic Sophie Germain prime
:$2 N + 1$ is also a palindromic Sophie Germain prime
:$2 \left({2 N + 1}\right) + 1$ is also a palindromic prime, but not a Sophie Germain prime.
== Proof ==
By direct calculation:
== Sources ==
* : $191,918,080,818,091,909,090,909,190,818,080,819,191$
Category:Sophie Germain Primes
Category:Palindromic Primes
Sequence of Powers of 10

== Integer Sequence ==
The sequence of powers of $10$ begins:
:$1, 10, 100, 1000, 10 \, 000, 100 \, 000, 1 \, 000 \, 000, 10 \, 000 \, 000, 100 \, 000 \, 000, 1 \, 000 \, 000 \, 000, \ldots$
== Sources ==
* : Chapter $5$: Eternal Triangles: Logarithms
Category:Powers of 10
Sequence of Powers of 11

== Integer Sequence ==
The sequence of powers of $11$ begins:
:$1, 11, 121, 1331, 14 \, 641, 161 \, 051, 1 \, 771 \, 561, 19 \, 487 \, 171, 214 \, 358 \, 881, 2 \, 357 \, 947 \, 691, \ldots$
Category:Powers of 11
Sequence of Powers of 12

== Integer Sequence ==
The sequence of powers of $12$ begins:
:$1, 12, 144, 1728, 20 \, 736, 248 \, 832, 2 \, 985 \, 984, 35 \, 831 \, 808, 429 \, 981 \, 696, \ldots$
Category:Powers of 12
Sequence of Powers of 13

== Integer Sequence ==
The sequence of powers of $13$ begins:
:$1, 13, 169, 2197, 28 \, 561, 371 \, 293, 4 \, 826 \, 809, 62 \, 748 \, 517, 815 \, 730 \, 721, \ldots$
Category:Powers of 13
Sequence of Powers of 2

== Integer Sequence ==
The sequence of powers of $2$ begins:
:$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16 \, 384, 32 \, 768, \ldots$
== Sources ==
* : Chapter $5$: Eternal Triangles: Logarithms
Category:Powers of 2
Sequence of Powers of 4

== Integer Sequence ==
The sequence of powers of $4$ begins:
:$1, 4, 16, 64, 256, 1024, 4096, 16 \, 384, 65 \, 536, 262 \, 144, 1 \, 048 \, 576, 4 \, 194 \, 304, \ldots$
Category:Powers of 4
Sequence of Powers of 5

== Integer Sequence ==
The sequence of powers of $5$ begins:
:$1, 5, 25, 125, 625, 3125, 15 \, 625, 78 \, 125, 390 \, 625, 1 \, 953 \, 125, 9 \, 765 \, 625, \ldots$
Category:Powers of 5
Sequence of Powers of 7

== Integer Sequence ==
The sequence of powers of $7$ begins:
:$1, 7, 49, 343, 2401, 16 \, 807, 117 \, 649, 823 \, 543, 5 \, 764 \, 801, 40 \, 353 \, 607, \ldots$
Category:Powers of 7
Sequence of Powers of 8

== Integer Sequence ==
The sequence of powers of $8$ begins:
:$1, 8, 64, 512, 4096, 32 \, 768, 262 \, 144, 2 \, 097 \, 152, 16 \, 777 \, 216, 134 \, 217 \, 728, \ldots$
Category:Powers of 8
Sequence of Powers of 9

== Integer Sequence ==
The sequence of powers of $9$ begins:
:$1, 9, 81, 729, 6561, 59 \, 049, 531 \, 441, 4 \, 782 \, 969, 43 \, 046 \, 721, 387 \, 420 \, 489, \ldots$
Category:Powers of 9
Sequence of Powers of Number less than One

== Theorem ==
Let $x \in \R$.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.
Then:
:$\size x < 1$ $\sequence {x_n}$ is a null sequence.
=== Rational Numbers ===
=== Complex Numbers ===
=== Normed Division Ring ===
== Proof ==
=== Necessary Condition ===
[For other proofs of the Necessary Condition visit here.]
=== Sufficient Condition ===
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One
Sequence of Powers of Number less than One/Complex Numbers

== Theorem ==
Let $z \in \C$.
Let $\sequence {z_n}$ be the sequence in $\C$ defined as $z_n = z^n$.
Then:
:$\size z < 1$ $\sequence {z_n}$ is a null sequence.
== Proof ==
By the definition of convergence:
:$\ds \lim_{n \mathop \to \infty} z_n = 0 \iff \lim_{n \mathop \to \infty} \size {z_n} = 0$
By Modulus of Product:
:$\forall n \in \N: \size {z_n} = \size {z^n} = \size z^n$
So:
:$\ds \lim_{n \mathop \to \infty} \size {z_n} = 0 \iff \lim_{n \mathop \to \infty} \size z^n = 0$
Since $\size z \in \R_{\ge 0}$, by Sequence of Powers of Real Number less than One:
:$\ds \lim_{n \mathop \to \infty} \size z^n = 0 \iff \size z < 1$
The result follows.
== Also known as ==
This result and Sequence of Powers of Reciprocals is Null Sequence are sometimes referred to as the basic null sequences.
== Sources ==
* : $\S 4.2$. Sequences: Example $\text {(iv)}$
Category:Limits of Sequences
Category:Sequence of Powers of Number less than One