nyuuzyou/wikis huggingface dataset:
Numbers with Square-Free Binomial Coefficients/Lemma

== Theorem ==
Let $n$ be a (strictly) positive integer. 
Let $p$ be a prime number.
By Basis Representation Theorem, there is a unique sequence $\sequence {a_j}_{0 \mathop \le j \mathop \le r}$ such that:
:$(1): \quad \ds n = \sum_{k \mathop = 0}^r a_k p^k$
:$(2): \quad \ds \forall k \in \closedint 0 r: a_k \in \N_b$
:$(3): \quad r_t \ne 0$
Suppose $r \ge 2$ and $p^2 \nmid \dbinom n m$ for all $0 \le m \le n$. 
Then:
:$p^{r - 1} \divides \paren {n + 1}$
that is:
:$p^{r - 1}$ divides $\paren {n + 1}$.
== Proof ==
Suppose $\forall i: 0 \le i \le r - 2: a_i = p - 1$. 
Then:
Since $p a_r + a_{r - 1} + 1$ is an integer,
we have $p^{r - 1} \divides \paren {n + 1}$.
Now we prove the contrapositive of our theorem.
Suppose $r > 2$ and $p^{r - 1} \nmid \paren {n + 1}$.
By the contrapositive of the result above:
:$\exists i: 0 \le i \le r - 2, a_i \neq p - 1$
Let $I$ be the smallest integer for which $a_I \neq p - 1$.
Then for every $0 \le i < I$, we have:
:$a_i = p - 1$
Let $m = p^r - 1$.
Then $\ds m = \paren {p - 1} \sum_{i \mathop = 0}^r p^i$ is a representation of $m$ to the base $p$.
We also have:
Notice that:
:$0 \le a_r - 1 < p - 1$
:$0 \le a_i < p$ for $I + 1 \le i \le r - 1$
:$1 \le a_I + 1 < p - 1 + 1 = p$
So $\ds \paren {a_r - 1} p^r + \sum_{i = I + 1}^{r - 1} a_i p^i + \paren {a_I + 1} p^I$ is a representation of $n - m$ to the base $p$ (possibly with leading zeroes).
Now we add $n - m$ and $m$ in base $p$:
 place value p^r p^{r-1} p^{I+1} p^I p^{I-1} p^{I-2} 1 
 n - m = (a_r-1) a_{r-1} ... a_{I+1} (a_I+1) 0 0 ... 0 
 +) m = (p-1) (p-1) (p-1) (p-1) (p-1) (p-1) 
 -------------------------------------------------------------------------
 n = * * a_{I+1} a_I (p-1) (p-1) (p-1) 
 carry carry 
So there are at least two carries in this addition, at place values $p^I$ and $p^{I + 1}$.
By Kummer's Theorem, we have:
:$p^2 \divides \dbinom {n - m + m} m = \dbinom n m$
Thus the contrapositive is proved.
== Source ==
* 
Category:Square-Free Integers
Category:Binomial Coefficients
Numbers with Square-Free Binomial Coefficients/Mistake

== Source Work ==
:
:The Dictionary
::$23$
== Mistake ==
:For every $n$ greater than $23$, none of the binomial coefficients $\dbinom n k$ are square-free.
== Correction ==
Take $\dbinom {26} 1 26$ for example, which is indeed square-free.
What appears to be meant is:
:For every $n$ greater than $23$, there exists a binomial coefficient $\dbinom n k$ that is not square-free.
As is seen in Numbers with Square-Free Binomial Coefficients, the list of numbers $n$ such that $\dbinom n k$ are squarefree for all $k = 0, \dots, n$ is given by:
:$1, 2, 3, 5, 7, 11, 23$
== Sources ==
* : $23$
Category:Mistakes/Curious and Interesting Numbers
Numerator of p-1th Harmonic Number is Divisible by Prime p

== Theorem ==
Let $p$ be an odd prime.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p$.
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $17$
Category:Harmonic Numbers
Category:Numerator of p-1th Harmonic Number is Divisible by Prime p
Numerator of p-1th Harmonic Number is Divisible by Prime p/Historical Note

== Historical Note on Numerator of p-1th Harmonic Number is Divisible by Prime p ==
 reports in his ($1997$) that this result was established by in $1782$, but he gives no further details.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $17$
Category:Historical Notes
Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 1

== Theorem ==
== Proof ==
Add the terms of $H_{p - 1}$ using the definition of rational addition to obtain $\dfrac m n$.
Do not cancel common prime factors from $m$ and $n$.
It is seen that $n = \paren {p - 1}!$
Hence $p$ is not a divisor of $n$.
The numerator $m$ is seen to be:
:$m = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}$
Thus it is sufficient to show that $m$ is a multiple of $p$.
Each term in this sum is an integer of the form $\dfrac {\paren {p - 1}!} k$.
For each $k \in \set {1, 2, \ldots, p - 1}$, define $k'= - \dfrac {\paren {p - 1}!} k \bmod p$.
By Wilson's Theorem
:$k k' \equiv -\paren {p - 1}! \equiv 1 \pmod p$
Therefore
:$k' \equiv k^{-1} \pmod p$
From the corollary to Reduced Residue System under Multiplication forms Abelian Group:
:$\struct {\Z'_p, \times}$ is an abelian group.
Since Inverse in Group is Unique, the set:
:$\set {1', 2', \ldots, \paren {p - 1}'}$
is merely the set:
:$\set {1, 2, \ldots, p - 1}$
in a different order.
Thus 
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $17$
Category:Numerator of p-1th Harmonic Number is Divisible by Prime p
Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 2

== Theorem ==
== Proof ==
Note that for any integer $x$:
By comparing coefficients:
:$\ds {p \brack p} \equiv 1 \pmod p$
:$\ds {p \brack 1} \equiv -1 \pmod p$
:$\ds {p \brack k} \equiv 0 \pmod p$ for $k \ne 1, p$
or in a more compact form:
:$\ds {p \brack k} \equiv \delta_{k p} - \delta _{k 1} \pmod p$
where:
:$\ds {p \brack k}$ denotes an unsigned Stirling number of the first kind
:$\delta$ is the Kronecker delta.
From Harmonic Number as Unsigned Stirling Number of First Kind over Factorial:
:$\ds H_{p - 1} = \frac {p \brack 2} {\paren {p - 1}!}$
From the above we have:
:$\ds p \divides {p \brack 2}$
By Prime iff Coprime to all Smaller Positive Integers we also have:
:$p \nmid \paren {p - 1}!$
Hence the numerator of $H_{p - 1}$, when expressed in canonical form, is divisible by $p$.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $17$
Category:Numerator of p-1th Harmonic Number is Divisible by Prime p
Numerator of p-1th Harmonic Number is Divisible by p^2 for Prime Greater than 3

== Theorem ==
Let $p$ be a prime number such that $p > 3$.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p^2$.
== Proof ==
== Sources ==
* : $7.8$: A theorem of Wolstenholme: Theorem $116$
* : $\S 1.2.7$: Harmonic Numbers: Exercise $17$ (Solution)
Category:Harmonic Numbers
ODE/(D^4 - 1) y = sin x

== Theorem ==
The second order ODE:
:$(1): \quad \paren {D^4 - 1} y' = \sin x$
has a general solution:
:$y = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x + \dfrac {x \cos x} 4$
== Proof ==
First we solve the reduced equation of $(1)$:
:$(2): \quad \paren {D^4 - 1} y' = 0$
The auxiliary equation of $(1)$ is:
:$(3): \quad: m^4 - 1 = 0$
From Complex $4$th Roots of Unity, the roots of $(2)$ are:
:$m_1 = 1$
:$m_2 = i$
:$m_3 = -1$
:$m_4 = -i$
So from Solution of Constant Coefficient Linear nth Order ODE, the general solution of $(2)$ is:
:$y_g = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x$
Because $\sin x$ is already a solution of $(2)$, we try:
:$y = A x e^{i x}$
with a view to taking the real part in due course.
It follows that
:$-4 i A - 1$
and so:
:$A = -\dfrac 1 4$
Thus a particular solution to $(1)$ can be given as:
and the general solution is:
:$y = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x + \dfrac {x \cos x} 4$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.1$ The $n$th order equation: Example $1$
Category:Examples of Linear ODE
OTTFFSSE

== Puzzle ==
What letter follows:
:$\text {OTTFFSSE} \ldots$
?
== Solution ==
:$\text N$
They are the initial letters of the first few natural numbers:
:$\text {one, two, three, four, five, six, seven, eight, nine} \ldots$
== Sources ==
* : Preface to the Dover Edition
Category:Classic Problems
Obliquity of Ecliptic

== Physical Law ==
The plane of the ecliptic is at an inclination of $23 \degrees \, 26 \minutes$ to the plane of the celestial equator.
This inclination is known as the obliquity of the ecliptic.
== Sources ==
* 
* 
* 
* 
Category:Ecliptic
Category:Celestial Mechanics
Category:Astronomy
Category:Oblique
Occam's Razor

== Philosophical Position ==
Occam's Razor is a philosophical position which can be expressed in several ways, including the following:
* The simplest explanation is probably the correct one.
* When seeking to explain a phenomenon, choose the hypothesis which requires the fewest postulates.
* Demonstration by fewer postulates or hypotheses (in short from fewer premises) is ceteris paribus superior; for, given that all of these are equally well known, where they are fewer, knowledge will be more speedily acquired, and that is a desideratum. (, from Analytica Posteriora).
* We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances. Therefore, to the same natural effects we must, so far as possible, assign the same causes (attributed to ).
* (used as a design principle).
It has profound philosophical implications, and considerable scope for misuse.
== Also known as ==
More accurately, the name should strictly be given as Ockham's Razor, but this usage is rare.
Occam's Razor is also known in Latin as the lex parsimoniae (which translates as the law of economy or law of succinctness). 
== Historical Note ==
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $1$: The Search for Knowledge: The Greek View of Motion
Category:Philosophical Positions
Occam's Razor/Historical Note

== Historical Note on Occam's Razor ==
In the words of himself, Occam's Razor was expressed as:
 
:entia non sunt multiplicanda praeter necessitatem (entities must not be multiplied beyond necessity)
or:
:pluralitas non est ponenda sine necessitate (plurality should not be posited without necessity).
He was not the first to voice this philosophy. For example, had expressed a similar philosophical position.
Category:Historical Notes
Occurrence in Prefix Notation has Unique Scope

== Theorem ==
Let $\FF$ be a formal language in prefix notation.
Let $\mathbf A$ be a well-formed formula of $\FF$.
Let $a$ be an occurrence in $\mathbf A$.
Then $a$ has a unique scope.
== Proof ==
From the formal definition of prefix notation, it follows that $a$ must be introduced by the rule of formation:
:$a \mathbf A_1 \cdots \mathbf A_n$
for some well-formed formulas $\mathbf A_1, \ldots, \mathbf A_n$.
By Unique Readability for Prefix Notation, the $\mathbf A_i$ are uniquely determined.
Then $\mathbf A' = a \mathbf A_1 \cdots \mathbf A_n$ is a well-formed part of $\mathbf A$.
Moreover, any well-formed part of $\mathbf A$ containing $a$ must contain $\mathbf A'$.
Hence $\mathbf A'$ is the scope of $a$.
== Sources ==
* : $\mathrm{II}.4$ Polish Notation: Lemma $\mathrm{II}.4.5$
Category:Prefix Notation
Category:Formal Languages
Occurrence of Event/Examples/Electric Circuit 1

== Examples of Occurrences of Events ==
Consider the electric circuit:
:
Let event $A$ be that switch $A$ is open.
Let event $B_n$ for $n = 1, 2, 3$ be that switch $B_n$ is open.
Let $C$ be the event that no current flows from $M$ to $N$.
Then:
:$C = A \cup \paren {B_1 \cap B_2 \cap B_3}$
:$\overline C = \overline A \cap \paren {\overline {B_1} \cup \overline {B_2} \cup \overline {B_3} }$
== Proof ==
The circuit is open between $M$ and $N$ :
:switch $A$ is open
or:
:all of $B_1$, $B_2$ and $B_3$ are open.
The corresponding events are:
:$A$
and
:$B_1 \cap B_2 \cap B_3$
from which $C = A \cup \paren {B_1 \cap B_2 \cap B_3}$ follows.
Then:
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Example $1.4$
Category:Examples of Occurrences of Events
Octagonal Number/Sequence

== Definition ==
The sequence of octagonal numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 1, 8, 21, 40, 65, 96, 133, 176, 225, 280, 341, \ldots$
== Sources ==
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
* : $8$
* : Tables: $2$ The First $20$ Pentagonal, Hexagonal, Heptagonal and Octagonal Numbers
Category:Octagonal Numbers
Octal Notation/Examples/371.24/Mistake

== Source Work ==
:
:octal
== Mistake ==
:... for example, $371.24_8$ represents the number
::$\paren {3 \times 8^2} + \paren {7 \times 8^1} + \paren {1 \times 8^0} + \paren {2 \times 8^{-1} } + \paren {4 \times 8^{-2} } = 249.25_{10}$.
== Correction ==
The actual number in decimal is $249.3125_{10}$.
It appears that the last octal digit was omitted from the calculation.
== Sources ==
* 
Category:Mistakes/Dictionary of Mathematics
Octonion Conjugation is Involution

== Theorem ==
Let $x = \tuple {a, b}: a, b \in \mathbb H$ be a octonion.
Let $\overline x$ be the conjugate of $x$.
Then:
:$\overline \cdot: x \mapsto \overline x$
is an involution.
That is:
:$\overline {\paren {\overline x} } = x$
== Proof ==
Category:Octonions
Category:Involutions
Octonions form Algebra

== Theorem ==
The set of octonions $\Bbb O$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An alternative algebra, but not an associative algebra.
:$(2): \quad$ A normed division algebra.
:$(3): \quad$ A nicely normed $*$-algebra.
== Proof ==
The octonions $\Bbb O$ are formed by the Cayley-Dickson construction from the quaternions $\Bbb H$.
From Quaternions form Algebra, we have that $\Bbb H$ forms:
:$(1): \quad$ An associative algebra
:$(2): \quad$ A normed division algebra
:$(3): \quad$ A nicely normed $*$-algebra.
From Cayley-Dickson Construction forms Star-Algebra, $\Bbb O$ is a $*$-algebra.
From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\Bbb O$ is a nicely normed $*$-algebra.
From Nicely Normed Cayley-Dickson Construction from Associative Algebra is Alternative, $\Bbb O$ is an alternative algebra.
Now suppose $\Bbb O$ formed an associative algebra.
Then from Cayley-Dickson Construction from Commutative Associative Algebra is Associative, that would mean $\Bbb H$ is a commutative algebra.
But from Quaternions form Algebra it is explicitly demonstrated that $\Bbb H$ is not a commutative algebra.
So $\Bbb O$ cannot be a associative algebra.
=== Proof of Normed Division Algebra ===
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
* 
Category:Octonions
Category:Algebras
Odd-Even Topology is First-Countable

== Theorem ==
Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is first-countable.
== Proof ==
From Odd-Even Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is First-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $3$
Category:First-Countable Spaces
Category:Odd-Even Topology
Odd-Even Topology is Lindelöf

== Theorem ==
Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is Lindelöf.
== Proof ==
From Odd-Even Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is Lindelöf.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $3$
Category:Lindelöf Spaces
Category:Odd-Even Topology
Odd-Even Topology is Second-Countable

== Theorem ==
Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is second-countable.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
From Basis for Partition Topology, the set:
:$\BB := \set {\set {2 k - 1, 2 k}: k \in \Z, k > 0}$
is a basis for $T$.
There is an obvious one-to-one correspondence $\phi: \Z_{>0} \leftrightarrow \BB$ between $\Z_{>0}$ and $\BB$:
:$\forall x \in \Z_{>0}: \map \phi x = \set {2 x - 1, 2 x}$
But $\Z_{>0} \subseteq \Z$, and Integers are Countably Infinite.
So from Subset of Countably Infinite Set is Countable, $\Z_{>0}$ is countable.
Thus $\BB$ is also countable by definition of countability.
So we have that $T$ has a countable basis, and so is second-countable by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $3$
Category:Second-Countable Spaces
Category:Odd-Even Topology
Odd-Even Topology is Separable

== Theorem ==
Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is separable.
== Proof ==
From Odd-Even Topology is Second-Countable, $T$ is second-countable.
The result follows from Second-Countable Space is Separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $3$
Category:Separable Spaces
Category:Odd-Even Topology
Odd-Even Topology is Weakly Countably Compact

== Theorem ==
Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is weakly countably compact.
== Proof ==
Let $H \subseteq \Z_{>0}$ such that $H$ is infinite.
Let $x \in H$.
By definition, the odd-even topology is a partition topology.
So $U$ is a union of sets of the form $\set {2 k - 1, 2 k}$.
Now if $x \in U$, it will be of the form $2 k - 1$ or $2 k$.
So there will exist $y \in U$ of the form $2 k$ or $2 k - 1$.
So, by definition, $x$ is a limit point of $H$.
So, by definition, $T$ is weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $3$
Category:Weakly Countably Compact Spaces
Category:Odd-Even Topology
Odd-Even Topology is not Countably Compact

== Theorem ==
Let $T = \struct {\Z_{>0}, \tau}$ be a topological space where $\tau$ is the odd-even topology on the strictly positive integers $\Z_{>0}$.
Then $T$ is not countably compact.
== Proof ==
By definition, the odd-even topology is a partition topology.
Let $\PP$ be the partition which is the basis for $T$:
:$\PP = \set {\set {2 k - 1, 2 k}: k \in \Z_{>0} }$
Then $\PP$ is a countable open cover of $S$ which has no finite subcover.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $3$
Category:Countably Compact Spaces
Category:Odd-Even Topology
Odd Amicable Pair/Examples

== Examples of Odd Amicable Pairs ==
=== $12 \, 285$ and $14 \, 595$ ===
=== $67 \, 095$ and $71 \, 145$ ===
=== $69 \, 615$ and $87 \, 633$ ===
=== $100 \, 485$ and $124 \, 155$ ===
=== $1 \, 175 \, 265$ and $1 \, 438 \, 983$ ===
Category:Amicable Pairs
Odd Amicable Pair/Examples/1,175,265-1,438,983

== Example of Odd Amicable Pair ==
$1 \, 175 \, 265$ and $1 \, 438 \, 983$ are the $9$th odd amicable pair:
:$\map {\sigma_1} {1 \, 175 \, 265} = \map {\sigma_1} {1 \, 438 \, 983} = 2 \, 614 \, 240 = 1 \, 175 \, 265 + 1 \, 438 \, 983$
== Proof ==
By definition, $m$ and $n$ form an amicable pair :
:$\map {\sigma_1} m = \map {\sigma_1} n = m + n$
where $\map {\sigma_1} n$ denotes the divisor sum function.
Thus:
== Historical Note ==
== Sources ==
* : $1,175,265$
* : $1,175,265$
Category:Amicable Pairs
Odd Amicable Pair/Examples/1,175,265-1,438,983/Historical Note

== Historical Note on the Amicable Pair $1 \, 175 \, 265$ and $1 \, 438 \, 983$ ==
The odd amicable pair $1 \, 175 \, 265$ and $1 \, 438 \, 983$ was discovered by in the $17$th century.
It was the $1$st odd amicable pair to be discovered.
== Sources ==
* : $1,175,265$
* : $1,175,265$
Category:Historical Notes
Odd Amicable Pair/Examples/12,285-14,595

== Example of Odd Amicable Pair ==
$12 \, 285$ and $14 \, 595$ are the $7$th amicable pair and the smallest odd amicable pair:
:$\map {\sigma_1} {12 \, 285} = \map {\sigma_1} {14 \, 595} = 26 \, 880 = 12 \, 285 + 14 \, 595$
== Proof ==
By definition, $m$ and $n$ form an amicable pair :
:$\map {\sigma_1} m = \map {\sigma_1} n = m + n$
where $\map {\sigma_1} n$ denotes the divisor sum function.
Thus:
It can be determined by inspection of the aliquot sums of all smaller odd integers that there is no smaller odd amicable pair.
== Historical Note ==
== Sources ==
* 
* : $12,285$
* : $12,285$
Category:Amicable Pairs
Odd Amicable Pair/Examples/12,285-14,595/Historical Note

== Historical Note on the Amicable Pair $12 \, 285$ and $14 \, 595$ ==
The odd amicable pair $12 \, 285$ and $14 \, 595$ was discovered by , who reported on them in the in $1939$.
== Sources ==
* 
* : $12,285$
* : $12,285$
Category:Historical Notes
Odd Amicable Pair/Examples/29,912,035,725-34,883,817,075

== Example of Odd Amicable Pair ==
$29 \, 912 \, 035 \, 725$ and $34 \, 883 \, 817 \, 075$ are an odd amicable pair:
:$\map {\sigma_1} {29 \, 912 \, 035 \, 725} = \map {\sigma_1} {34 \, 883 \, 817 \, 075} = 64 \, 795 \, 852 \, 800 = 29 \, 912 \, 035 \, 725 + 34 \, 883 \, 817 \, 075$
== Proof ==
By definition, $m$ and $n$ form an amicable pair :
:$\map {\sigma_1} m = \map {\sigma_1} n = m + n$
where $\sigma_1$ denotes the divisor sum function.
Thus:
== Sources ==
* : $64,795,852,800$
Category:Amicable Pairs
Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525

== Example of Odd Amicable Pair ==
$31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are an odd amicable pair:
:$\map {\sigma_1} {31 \, 695 \, 652 \, 275} = \map {\sigma_1} {33 \, 100 \, 200 \, 525} = 64 \, 795 \, 852 \, 800 = 31 \, 695 \, 652 \, 275 + 33 \, 100 \, 200 \, 525$
== Proof ==
By definition, $m$ and $n$ form an amicable pair :
:$\map {\sigma_1} m = \map {\sigma_1} n = m + n$
where $\sigma_1$ denotes the divisor sum function.
Thus:
== Sources ==
* : $64,795,852,800$
Category:Amicable Pairs
Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275

== Example of Odd Amicable Pair ==
$32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are an odd amicable pair:
:$\map {\sigma_1} {32 \, 129 \, 958 \, 525} = \map {\sigma_1} {32 \, 665 \, 894 \, 275} = 64 \, 795 \, 852 \, 800 = 32 \, 129 \, 958 \, 525 + 32 \, 665 \, 894 \, 275$
== Proof ==
By definition, $m$ and $n$ form an amicable pair :
:$\map {\sigma_1} m = \map {\sigma_1} n = m + n$
where $\map {\sigma_1} n$ denotes the divisor sum function.
Thus:
== Sources ==
* : $64,795,852,800$
Category:Amicable Pairs
Odd Bernoulli Numbers Vanish

== Theorem ==
Let $B_n$ denote the $n$th Bernoulli Number.
Then:
:$B_{2n + 1} = 0$
for $n \ge 1$.
== Proof ==
By definition, the Bernoulli numbers are given by:
:$\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
We have:
Take $\map f x := \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$, and note that:
and so $\map f x := \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$ is an even function.
Rewriting the definition of Bernoulli numbers
We now have:
If we set $n$ to be an odd integer where $n > 1$, we have:
Therefore: $\forall n \in \N: n \ge 1$:
:$B_{2 n + 1} = 0$
== Sources ==
* : Chapter $\text {B}.20$: The Bernoulli Numbers and some Wonderful Discoveries of Euler
Category:Bernoulli Numbers
Category:Odd Bernoulli Numbers Vanish
Odd Convergents of Simple Continued Fraction are Strictly Decreasing

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, \ldots}$ be a simple continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Let $\sequence {C_0, C_1, \ldots}$ be its sequence of convergents.
The odd convergents satisfy $C_1 > C_3 > C_5 > \cdots$
== Proof ==
Let $ k \ge 3$ be an odd integer.
From Difference between Adjacent Convergents But One of Simple Continued Fraction:
:$C_k - C_{k - 2} = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} } = \dfrac {-a_k} {q_k q_{k - 2} }$
By definition of simple continued fraction, $a_k > 0$.
By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $q_k > 0$ and $q_{k - 2} > 0$.
Thus $C_k < C_{k - 2}$.
== Also see ==
* Even Convergents of Simple Continued Fraction are Strictly Increasing
* Properties of Convergents of Continued Fractions
Category:Simple Continued Fractions