nyuuzyou/wikis huggingface dataset:
Number of Fibonacci Numbers with Same Number of Decimal Digits

== Theorem ==
Let $n$ be an integer such that $n > 1$.
When expressed in decimal notation, there are either $4$ or $5$ Fibonacci numbers with $n$ digits.
== Proof ==
Let $F_m$ be an $n$-digit Fibonacci number.
Then $F_m \ge 10^{n - 1} \ge 10$.
We have:
Thus:
So $F_{m + 5}$ is a number of at least $n + 1$ digits.
Therefore there is at most $5$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}, F_{m + 4}$.
Similarly, $10 \le F_m < 10^n$.
Suppose $F_m$ is the smallest Fibonacci numbers with $n$ digits.
Then $F_{m - 1} < 10^{n - 1}$.
Thus:
So there is at least $4$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}$.
Combining the arguments above, there are either $4$ or $5$ Fibonacci numbers with $n$ digits.
== Historical Note ==
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Number of Fibonacci Numbers with Same Number of Decimal Digits/Mistake

== Source Work ==
:
:The Dictionary
:: $5$
== Mistake ==
:The number of Fibonacci numbers having the same number of digits is either $4$ or $5$.
== Correction ==
There is a notable exception: there are $6$ (or $7$ if you include $F_0 = 0$) Fibonacci numbers with $1$ digit:
:$1, 1, 2, 3, 5, 8$
== Sources ==
* : $5$
* : $5$
Category:Mistakes/Curious and Interesting Numbers
Number of Friday 13ths in a Year

== Theorem ==
In any given year, there are between $1$ and $3$ (inclusive) months in which the $13$th falls on a Friday.
== Proof ==
The day of the week on which the $13$th falls is directy dependent upon the day of the week that the $1$st of the month falls.
From Months that Start on the Same Day of the Week, the months can be grouped into equivalence classes according to which day the month starts:
For a non-leap year, the set of equivalence classes is:
:$\set {\set {\text {January}, \text {October} }, \set {\text {February}, \text {March}, \text {November} }, \set {\text {April}, \text {July} }, \set {\text {May} }, \set {\text {June} }, \set {\text {August} }, \set {\text {September}, \text {December} } }$
For a leap year, the set of equivalence classes is:
:$\set {\set {\text {January}, \text {April}, \text {July} }, \set {\text {February}, \text {August} }, \set {\text {March}, \text {November} }, \set {\text {May} }, \set {\text {June} }, \set {\text {September}, \text {December} }, \set {\text {October} } }$
It can be seen that none of these equivalence classes has fewer than $1$ day, and none has more than $3$.
Hence the result.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $7$
Category:Calendars
Number of Generators of Cyclic Group whose Order is Power of 2

== Theorem ==
Let $G$ be a finite cyclic group.
Let the order of $G$ be $2^k$ for some $k \in \Z_{>0}$.
Then $G$ has $2^{n - 1}$ distinct generators.
== Proof ==
From Finite Cyclic Group has Euler Phi Generators, $G$ has $\map \phi {2^n}$ generators.
The result follows from corollary to Euler Phi Function of Prime Power.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $19$
Category:Cyclic Groups
Category:Generators of Groups
Number of Hamilton Cycles in Complete Graph

== Theorem ==
For all $n \ge 3$, the number of distinct Hamilton cycles in the complete graph $K_n$ is $\dfrac {\paren {n - 1}!} 2$.
== Proof ==
In a complete graph, every vertex is adjacent to every other vertex.
Therefore, if we were to take all the vertices in a complete graph in any order, there will be a path through those vertices in that order.
Joining either end of that path gives us a Hamilton cycle.
From Cardinality of Set of Bijections, there are $n!$ different ways of picking the vertices of $G$ in some order.
Hence there are $n!$ ways of building such a Hamilton cycle.
Not all these are different, though.
On any such cycle, there are:
:$n$ different places you can start
:$2$ different directions you can travel.
So any one of these $n!$ cycles is in a set of $2 n$ cycles which all contain the same sequence of edges.
So there are $\dfrac {n!} {2 n} = \dfrac {\paren {n - 1}!} 2$ distinct Hamilton cycles.
Category:Hamiltonian Graphs
Category:Combinatorics
Number of Injective Restrictions

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $Q$ be the set of all injective restrictions of $f$.
Then the cardinality of $Q$ is:
:$\ds \card {\prod_{i \mathop \in I} \prod_{j \mathop \in J_i} \family {\family {\powerset {S / \RR_f} }_i}_j}$
where:
:$\powerset \cdot$ denotes power set
:$S / \RR_f$ denotes quotient set of the induced equivalence of $f$.
== Proof ==
Category:Injections
Number of Integer Partitions into Sum of Consecutive Primes

== Theorem ==
Let $n$ be a natural number.
Let $\map f n$ denote the number of integer partitions of $n$ where the parts are consecutive prime numbers.
For example:
:$\map f {41} = 3$
because:
:$41 = 11 + 13 + 17 = 2 + 3 + 5 + 7 + 11 + 13$
Then:
:$\ds \lim_{x \mathop \to \infty} \dfrac 1 x \sum_{n \mathop = 1}^x \map f n = \ln 2$
== Proof ==
Let $\mathbb P$ denote the set of prime numbers.
Every set of consecutive primes whose sum is less than $x$ will contribute $1$ to the sum:
:$\map f 1 + \map f 2 + \dotsb + \map f x$
The number of such sets of $r$ primes is clearly at most:
:$\map \pi {x / r}$
and at least:
:$\map \pi {x / r} - r$
where $\pi$ is the prime-counting function.
Hence:
:$\ds \sum_{r \mathop = 1}^k \paren {\map \pi {x / r} - r} \le \map f 1 + \map f 2 + \dotsb + \map f x \le \sum_{r \mathop = 1}^k \map \pi {x / r}$
where $k$ is determined by:
:$(1): \quad p_1 + p_2 + \dotsb + p_k < x < p_1 + p_2 + . . . + p_{k + 1}$
From $(1)$ and the well known $p_r \asymp r \ln r$ we find:
:$(2): \quad k \asymp \sqrt {\dfrac x {\ln x} }$
Here $f \asymp g$ denotes that $\dfrac f g$ is bounded above and below by positive numbers for large arguments.
This implies:
:$\ds \sum_{r \mathop = 1}^k r = \map o x$
where $\map o x$ denotes little-$\oo$ notation.
Hence, if we can show that:
:$\ds \sum_{r \mathop = 1}^k \map \pi {x / r} \approx \ln 2$
it will follow that:
:$\map f 1 + \map f 2 + \dotsb + \map f x \approx x \ln 2$
Using $(2)$ and the Prime Number Theorem, the following estimates are justified:
== Sources ==
* 
* : $0,69314 7805 \ldots$
Category:Partition Theory
Category:Prime Numbers
Number of Lattice Points in Circle

== Conjecture ==
Consider the circle $C$ of radius $\sqrt n$ whose center is at the origin of a cartesian plane.
Let $\map R n$ denote the number of lattice points in $C$.
Let $\map d n$ denote the difference between the area of $C$ and $\map R n$:
:$\map d n = \pi n - \map R n$
It is conjectured that:
:$\map d n = \map \OO {n^{1/4 + \epsilon} } = \map \OO {n^{\theta} }$
== Progress ==
In $1963$, determined that $\theta \le \dfrac {12} {37}$.
Previous to that, the best estimate was $\theta \le \dfrac {17} {53}$, due to .
The lower bound $\dfrac 1 4$ was established by and .
== Sources ==
* 
* : $1/4 \le \theta \le 12/37$
Category:Unsolved Problems
Number of Linearly Independent Eigenvectors of Square Complex Matrix

== Theorem ==
Let $\mathbf A$ be a square matrix of order $n$ over the complex numbers $\C$.
Then $\mathbf A$ has no more than $n$ linearly independent eigenvectors.
== Proof ==
== Sources ==
* 
* 
Category:Eigenvalues
Category:Linear Independence
Category:Square Matrices
Category:Complex Matrices
Number of Matrix Equivalence Classes

== Theorem ==
Let $K$ be a field.
Let $\map {\MM_K} {m, n}$ be the $m \times n$ matrix space over $K$.
Let $\mathbf A$ be an $m \times n$ matrix of rank $r$ over $K$.
Then:
:$\mathbf A \equiv \begin{cases}
\sqbrk {0_K}_{m n} & : r = 0 \\
& \\
\begin{bmatrix}
 \mathbf I_r & \bszero \\
 \bszero & \bszero 
\end{bmatrix} & : 0 < r < \min \set {n, m} \\
& \\
\begin{bmatrix}
 \mathbf I_r & \bszero 
\end{bmatrix} & : r = m < n \\
& \\
\begin{bmatrix}
 \mathbf I_r \\
 \bszero 
\end{bmatrix} & : r = n < m \\
& \\
\mathbf I_r & : r = m = n
\end{cases}$
Thus there are exactly $\min \set {m, n} + 1$ equivalence classes for the relation of equivalence on $\map {\MM_K} {m, n}$, one of which contains only the zero matrix.
== Proof ==
Follows from Equivalent Matrices have Equal Rank.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Equivalence
Number of Maximal Elements is Order Property

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\map M S$ be the number of maximal elements of $\struct {S, \preccurlyeq}$.
Then $\map M S$ is an order property.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be isomorphic ordered sets.
Hence let $\phi: S_1 \to S_2$ be an order isomorphism.
By definition of order property, we need to show that the number of maximal elements of $\struct {S_1, \preccurlyeq_1}$ is equal to the number of maximal elements of $\struct {S_2, \preccurlyeq_2}$.
Let $s \in \struct {S_1, \preccurlyeq_1}$ be a maximal element of $\struct {S_1, \preccurlyeq_1}$.
Then:
:$\forall x \in S_1: s \preccurlyeq_1 x \implies x = s$
Let $\map \phi y, \map \phi s \in S_2$ such that $\map \phi s \preccurlyeq_2 \map \phi y$.
Then as $\phi$ is an order isomorphism:
:$s \preccurlyeq_1 y$
and so as $s$ is a maximal element of $\struct {S_1, \preccurlyeq_1}$:
:$y = s$
From Order Embedding is Injection it follows that $\map \phi s$ is a maximal element of $\struct {S_2, \preccurlyeq_2}$.
Similarly, let $\map \phi s \in \struct {S_2, \preccurlyeq_2}$ be a maximal element of $\struct {S_2, \preccurlyeq_2}$.
Then:
:$\forall \map \phi y \in S_2: \map \phi s \preccurlyeq_2 \map \phi y \implies \map \phi y = \map \phi s$
Let $x \in S_1: x \preccurlyeq_1 s$.
Then as $\phi$ is an order isomorphism:
:$\map \phi s \preccurlyeq_2 \map \phi x$
and so as $\map \phi s$ is a maximal element of $\struct {S_2, \preccurlyeq_2}$:
:$\map \phi x = \map \phi s$
That is:
:$x = s$
and it follows that $s$ is a maximal element of $\struct {S_1, \preccurlyeq_1}$.
Hence:
:all maximal elements of $\struct {S_1, \preccurlyeq_1}$ are also maximal elements of $\struct {S_2, \preccurlyeq_2}$
and:
:all maximal elements of $\struct {S_2, \preccurlyeq_2}$ are also maximal elements of $\struct {S_1, \preccurlyeq_1}$
and the result follows.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Examples of Order Properties
Category:Maximal Elements
Number of Minimal Elements is Order Property

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\map m S$ be the number of minimal elements of $\struct {S, \preccurlyeq}$.
Then $\map m S$ is an order property.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be isomorphic ordered sets.
Hence let $\phi: S_1 \to S_2$ be an order isomorphism.
By definition of order property, we need to show that the number of minimal elements of $\struct {S_1, \preccurlyeq_1}$ is equal to the number of minimal elements of $\struct {S_2, \preccurlyeq_2}$.
Let $s \in \struct {S_1, \preccurlyeq_1}$ be a minimal element of $\struct {S_1, \preccurlyeq_1}$.
Then:
:$\forall x \in S_1: x \preccurlyeq_1 s \implies x = s$
Let $\map \phi y, \map \phi s \in S_2$ such that $\map \phi y \preccurlyeq_2 \map \phi s$.
Then as $\phi$ is an order isomorphism:
:$y \preccurlyeq_1 s$
and so as $s$ is a minimal element of $\struct {S_1, \preccurlyeq_1}$:
:$y = s$
From Order Embedding is Injection it follows that $\map \phi s$ is a minimal element of $\struct {S_2, \preccurlyeq_2}$.
Similarly, let $\map \phi s \in \struct {S_2, \preccurlyeq_2}$ be a minimal element of $\struct {S_2, \preccurlyeq_2}$.
Then:
:$\forall \map \phi y \in S_2: \map \phi y \preccurlyeq_2 \map \phi s \implies \map \phi y = \map \phi s$
Let $x \in S_1: x \preccurlyeq_1 s$.
Then as $\phi$ is an order isomorphism:
:$\map \phi x \preccurlyeq_2 \map \phi s$
and so as $\map \phi s$ is a minimal element of $\struct {S_2, \preccurlyeq_2}$:
:$\map \phi x = \map \phi s$
That is:
:$x = s$
and it follows that $s$ is a minimal element of $\struct {S_1, \preccurlyeq_1}$.
Hence:
:all minimal elements of $\struct {S_1, \preccurlyeq_1}$ are also minimal elements of $\struct {S_2, \preccurlyeq_2}$
and:
:all minimal elements of $\struct {S_2, \preccurlyeq_2}$ are also minimal elements of $\struct {S_1, \preccurlyeq_1}$
and the result follows.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Examples of Order Properties
Category:Minimal Elements
Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order

== Theorem ==
Let $D$ be a deck of $2 m$ cards.
Let $D$ be given a sequence of modified perfect faro shuffles.
Then the cards of $D$ will return to their original order after $n$ such shuffles, where:
:$2^n \equiv 1 \pmod {2 m + 1}$
== Proof ==
From Position of Card after n Modified Perfect Faro Shuffles, after $n$ shuffles a card in position $x$ will be in position $2^n x \pmod {m + 1}$.
So for all $2 m$ cards in $D$, we need to find $n$ such that:
:$2^n x \equiv x \pmod {2 m + 1}$
Because $2 m + 1$ is odd, we have:
:$\gcd \set {2, 2 m + 1}$
and so from Cancellability of Congruences:
:$2^n \equiv 1 \pmod {2 m + 1}$
== Examples ==
== Sources ==
* : $\text {4-3}$ Riffling: $\text {4-3-4}$
Category:Cards
Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples

== Examples of Use of Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order ==
=== Deck of 6 Cards ===
=== Deck of 8 Cards ===
=== Deck of 12 Cards ===
=== Deck of 52 Cards ===
=== Deck of 62 Cards ===
Category:Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order
Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 12 Cards

== Theorem ==
Let $D$ be a deck of $12$ cards.
Let $D$ be given a sequence of modified perfect faro shuffles.
Then after $12$ such shuffles, the cards of $D$ will be in the same order they started in.
== Proof ==
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where:
:$2^n \equiv 1 \pmod {13}$
From Fermat's Little Theorem:
:$2^{12} \equiv 1 \pmod {13}$
so we know that $n$ is at most $12$.
But $n$ may be smaller, so it is worth checking the values:
Inspecting $2^n$ for $n$ from $1$:
Hence the result.
== Sources ==
* : $\text {4-3}$ Riffling: Exercise $1$
Category:Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order
Category:12
Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 62 Cards

== Theorem ==
Let $D$ be a deck of $62$ cards.
Let $D$ be given a sequence of modified perfect faro shuffles.
Then after $6$ such shuffles, the cards of $D$ will be in the same order they started in.
== Proof ==
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where:
:$2^n \equiv 1 \pmod {63}$
We have that:
:$63 = 2^6 - 1$
and so:
:$2^6 \equiv 1 \pmod {63}$
Hence the result.
== Sources ==
* : $\text {4-3}$ Riffling
Category:Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order
Category:62
Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 6 Cards

== Theorem ==
Let $D$ be a deck of $6$ cards.
Let $D$ be given a sequence of modified perfect faro shuffles.
Then after $3$ such shuffles, the cards of $D$ will be in the same order they started in.
== Proof ==
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where:
:$2^n \equiv 1 \pmod 7$
Inspecting $2^n$ for $n$ from $1$:
Hence the result.
== Sources ==
* : $\text {4-3}$ Riffling: Exercise $1$
Category:Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order
Category:6
Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 8 Cards

== Theorem ==
Let $D$ be a deck of $8$ cards.
Let $D$ be given a sequence of modified perfect faro shuffles.
Then after $6$ such shuffles, the cards of $D$ will be in the same order they started in.
== Proof ==
From Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, the cards of $D$ will return to their original order after $n$ such shuffles, where:
:$2^n \equiv 1 \pmod 9$
Inspecting $2^n$ for $n$ from $1$:
Hence the result.
== Sources ==
* : $\text {4-3}$ Riffling: Exercise $1$
Category:Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order
Category:8
Number of Multidimensional Spheres that can touch One Other Sphere

== Theorem ==
Let $S_n$ be a spheres of $n$ dimensions with a given radius $r$.
Let $\map T n$ denote the number of instances of $S_n$ that can touch one other such instance of $S_n$.
The sequence of $\map T n$ begins as follows:
:
$n$	$\map T n$
$0$	$0$
$1$	$2$
$2$	$6$
$3$	$12$
$4$	$24$
$5$	$40$
$6$	$72$
$7$	$126$
$8$	$240$
$9$	$272$
</onlyinclude>
== Proof ==
== Sources ==
* : $12$
* : $12$
Category:Spheres
Number of Multiples less than Given Number

== Theorem ==
Let $m, n \in \N_{\ge 1}$.
The number of multiples of $m$ not greater than $n$ is given by:
:$q = \floor {\dfrac n m}$
where $\floor {\cdot}$ denotes the floor function
== Proof ==
By the Division Theorem:
:$(1): \quad n = q m + r$
where $0 \le r < q$.
As $r < q$, it follows that the greatest multiple of $m$ up to $n$ is $q m$.
So all the multiples of $m$ up to $n$ are:
:$m, 2 m, 3 m, \ldots, q m$
Dividing both sides of $(1)$ by $m$:
:$(2): \quad \dfrac n m = q + \dfrac r m$
Taking the floor of $(2)$:
:$\floor {\dfrac n m} = \floor {q + \dfrac r m}$
But as $0 \le \dfrac r m < 1$:
:$\floor {q + \dfrac r m} = q$
Recall that all the multiples of $m$ up to $n$ are $m, 2 m, 3 m, \ldots, q m$.
It follows that the number of multiples of $m$ up to $n$ is:
:$q = \floor {\dfrac n m}$
== Also see == 
* De Polignac's Formula
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
* : $\S 3.6$
Category:Number Theory
Number of Natural Numbers Less than x which are Squares or Sums of Two Squares

== Theorem ==
Let $x$ be a real number.
The number of natural numbers smaller than $x$ which are either square or the sum of $2$ squares is given by the expression:
:$\map N x \approx \dfrac {k x} {\sqrt {\ln x} }$
where $k$ is given by:
:$k = \sqrt {\dfrac 1 2 \ds \prod_{\substack {r \mathop = 4 n \mathop + 3 \\ \text {$r$ prime} } } \paren {1 - \dfrac 1 {r^2} }^{-1} }$
The number $k$ is known as the Landau-Ramanujan constant:
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $0,764 \ldots$
* 
Category:Analytic Number Theory
Number of Non-Dividing Primes Less than n is Less than Euler Phi Function of n

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\map w n$ denote the number of primes strictly less than $n$ which are not divisors of $n$.
Let $\map \phi n$ denote the Euler $\phi$ function of $n$.
Then:
:$\map w n < \map \phi n$
== Proof ==
Let $P = \set {p < n: p \text { prime}, p \nmid n}$.
Let $Q = \set {0 < q < n: q \perp n}$, where $q \perp n$ denotes that $q$ and $n$ are coprime.
Let $p \in P$.
From Prime not Divisor implies Coprime, $p$ is coprime to $n$.
That is:
:$p \in Q$
So, by definition of subset:
:$P \subseteq Q$
From Integer is Coprime to 1:
:$1 \in Q$
But as One is not Prime:
:$1 \notin P$
Thus $P \subsetneq Q$ and so:
:$\card P < \card Q$
By definition of $\map w n$:
:$\card P = \map w n$
and by definition of Euler $\phi$ function:
:$\card Q = \map \phi n$
Hence the result.
== Sources ==
* : $\text {4-2}$ Residue Systems: Exercise $4$
Category:Euler Phi Function
Category:Prime Numbers
Number of Non-Zero Singular Values equals Rank of Matrix

== Theorem ==
Let $\mathbf A$ be a matrix of order $m \times n$.
Let $\mathbf A = \mathbf U \mathbf D \mathbf V$ be a singular value decomposition of $\mathbf A$.
The number of non-zero singular values of $\mathbf A$ equals the rank of $\mathbf A$.
== Proof ==
== Sources ==
* 
Category:Singular Value Decompositions
Number of Odd Entries in Row of Pascal's Triangle is Power of 2

== Theorem ==
The number of odd entries in a row of Pascal's triangle is a power of $2$.
== Proof ==
Let $n, k \in \Z$.
Let the representations of $n$ and $k$ to the base $p$ be given by:
:$n = 2^r a_r + \cdots + 2 a_1 + a_0$
:$k = 2^r b_r + \cdots + 2 b_1 + b_0$
By Corollary to Lucas' Theorem:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod 2$
By definition, $a_j$ and $b_j$ are either $0$ or $1$.
We have:
:$\dbinom 0 1 = 0$
:$\dbinom 0 0 = \dbinom 1 0 = \dbinom 1 1 = 1$
Thus $\dbinom n k$ is odd none of $\dbinom {a_j} {b_j} = 0$.
We are to determine how many values of $k$ there are such that $\dbinom n k$ is odd.
We construct a $k$ such that $\dbinom n k$ is odd.
For each $a_j = 1$, we can have $b_j = 0$ or $1$.
For each $a_j = 0$, we must have $b_j = 0$.
Hence for each $a_j$, there are $2^{a_j}$ ways for the corresponding $b_j$ to be such that $\dbinom {a_j} {b_j} \ne 0$.
By Product Rule for Counting, the number of choices for $a_0, \dots, a_r$ for which $k$ is such that $\dbinom n k$ is odd is given by: 
:$\ds \prod_{j \mathop = 0}^r 2^{a_j} = 2^{\sum_{j = 0}^r {a_j} }$
which is a power of $2$.
Thus the number of odd entries in a row of Pascal's triangle is a power of $2$.
== Sources ==
* : $35$
Category:Pascal's Triangle
Number of Order p Elements in Group with m Order p Subgroups

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $G$ have $m$ subgroups of order $p$.
The total number of elements of $G$ of order $p$ is $m \paren {p - 1}$.
== Proof ==
Let $H \le G$ be a subgroup of $G$ of order $p$.
From Prime Group is Cyclic, $H$ is a cyclic group.
From Group of Prime Order p has p-1 Elements of Order p, $H$ has $p - 1$ elements of order $p$.
From Intersection of Subgroups of Prime Order, each of the $m$ sets of $p - 1$ non-identity elements of the $m$ subgroups of order $p$ are pairwise disjoint.
Each elements of $G$ order $p$ is the generator of a cyclic group of order $p$.
Therefore there are no elements of order $p$ apart from the $m \paren {p - 1}$ such elements of the $m$ subgroups of order $p$.
Hence the result.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $19$
Category:Order of Group Elements
Number of Parameters of ARMA Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
Let $M$ be an ARMA model on $S$ of order $p$:
:$\tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
Then $M$ has $p + q + 2$ parameters.
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== Proof ==
By definition of the parameters of $M$:
Thus:
:there are $p$ parameters of the form $\phi_i$
:there are $q$ parameters of the form $\theta_j$
:$1$ parameter $\mu$
:$1$ parameter $\sigma_a^2$.
That is: $p + q + 1 + 1 = p + q + 2$ parameters.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Mixed autoregressive -- moving average models
Category:ARMA Models