nyuuzyou/wikis huggingface dataset:
Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then the non-zero elements of the $n$th row of Pascal's triangle are all odd :
:$n = 2^m - 1$
for some $m \in \Z_{\ge 0}$.
As can be seen, the entries in rows $0, 1, 3, 7$ are all odd.
== Proof ==
The statement:
:$\dbinom n k$ is odd
is equivalent to:
:$\dbinom n k \equiv 1 \pmod p$
The corollary to Lucas' Theorem gives:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where:
:$n, k \in \Z_{\ge 0}$ and $p$ is prime
:the representations of $n$ and $k$ to the base $p$ are given by:
::$n = a_r p^r + \cdots + a_1 p + a_0$
::$k = b_r p^r + \cdots + b_1 p + b_0$
When $p = 2$, the digits $a_j, b_j$ are either $0$ or $1$.
In order for $\dbinom n k \equiv 1 \pmod p$, it is necessary and sufficient that $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $j \in \set {0, 1, 2, \ldots, r}$.
In order for this to happen, either:
:$\dbinom {a_j} {b_j} = \dbinom 0 0$
or:
:$\dbinom {a_j} {b_j} = \dbinom 1 0$
or:
:$\dbinom {a_j} {b_j} = \dbinom 1 1$
Suppose $a_i = 0$ for some $i \in \set {0, 1, 2, \ldots, r}$.
Then if $b_i = 1$:
:$\dbinom {a_j} {b_j} = \dbinom 0 1 = 0$
and so:
:$\dbinom n k \equiv 0 \pmod p$
for whichever $k$ (and there will be at least one) has digit $b_i = 1$.
So the only way it can be assured that all $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $k \in \set {0, 1, 2, \ldots, n}$ is for $a_j = 0$ for all $j \in \set {0, 1, 2, \ldots, r}$.
That is, for $n = 2^m - 1$ for some $m \in \Z_{\ge 0}$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $12$
Category:Pascal's Triangle
Numbers n whose Euler Phi value Divides n + 1

== Theorem ==
The following integers $n$ satisfy the equation:
:$\exists k \in \Z: k \, \map \phi n = n + 1$
where $\phi$ denotes the Euler $\phi$ function:
:$83 \, 623 \, 935, 83 \, 623 \, 935 \times 83 \, 623 \, 937$
== Proof ==
From :
:$\map \phi {83 \, 623 \, 935} = 41 \, 811 \, 968$
and then:
Then we have that $83 \, 623 \, 937$ is the $868 \, 421$st prime number.
From Euler Phi Function of Prime:
:$\map \phi {83 \, 623 \, 937} = 83 \, 623 \, 936$
Then we have that:
and so:
== Historical Note ==
== Sources ==
* 
* : $\mathbf B$: Divisibility: $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
* : $83,623,935$
* : $\mathbf B$: Divisibility: $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
Category:Euler Phi Function
Numbers n whose Euler Phi value Divides n + 1/Historical Note

== Historical Note on Numbers $n$ whose Euler Phi value Divides $n + 1$ ==
This result appears in an article from $1932$ by , where he performs a complete analysis of the situation where $k \map \phi n = n + 1$ where $n$ has fewer than $7$ distinct prime factors.
It appears that may have made the same observation, as attributes it to him (without providing a citation) in a note in his of $1994$.
Interestingly, while citing 's $1932$ article in the context of a conjecture about $k \map \phi n = n - 1$, appears to completely fail to notice his analysis of $k \map \phi n = n + 1$.
In his of $2004$, he does now report on 's $1932$ article, but continues to credit with the observation that $83 \, 623 \, 935 \times 83 \, 623 \, 937 \times \paren {83 \, 623 \, 935 \times 83 \, 623 \, 937 + 2}$ would also be a solution if $83 \, 623 \, 935 \times 83 \, 623 \, 937 + 2$ were prime.
However, again, the latter result also appears in 's $1932$ article.
In , then reports that established that $83 \, 623 \, 935 \times 83 \, 623 \, 937 + 2 = 6 \, 992 \, 962 \, 672 \, 132 \, 097$ is not prime, as it has $73$ as a prime factor.
== Sources ==
* 
* : $\mathbf B$: Divisibility: $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
* : $83,623,935$
* : $\mathbf B$: Divisibility: $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
Category:Historical Notes
Numbers n whose Euler Phi value Divides n + 1/Mistake

== Source Work ==
:
:$\mathbf B$: Divisibility
::$\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
:
:$\mathbf B$: Divisibility
::$\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
=== Second Edition ===
=== Third Edition ===
Category:Mistakes/Unsolved Problems in Number Theory
Numbers n whose Euler Phi value Divides n + 1/Mistake/Second Edition

== Source Work ==
:
:$\mathbf B$: Divisibility
::$\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
== Mistake ==
: notes that $\map \phi n$ sometimes divides $n + 1$, e.g. for $n = n_1 = 3 \cdot 5 \cdot 17 \cdot 353 \cdot 929$ and $n = n_1 \cdot 83623937$. [Note that $353 = 11 \cdot 2^5 + 1, 929 = 29 \cdot 2^5 + 1, 83623937 = 11 \cdot 29 \cdot 2^{18} + 1$ and $\paren {353 - 2^8} \paren {929 - 2^8} = 2^{16} - 2^8 + 1$.]
== Correction ==
This was in fact published by in an article in $1932$.
As this article was itself cited by in the references of this actual chapter, it is an oversight of his not to have noticed it.
In his of $2004$, correctly attributes the result to , although still misses his final observation in an endnote of the above article:
:In the same way if $6992962672132097$ is a prime
::$3 \cdot 5 \cdot 17 \cdot 353 \cdot 929 \cdot 83623937 \cdot 6992962672132097 = 48901526933832864378258473353215$
:is a solution of $(2)$.
 continues to attribute the above to .
== Also see ==
* Numbers $n$ whose Euler Phi value Divides $n + 1$
== Sources ==
* 
* : $\mathbf B$: Divisibility: $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
Category:Mistakes/Unsolved Problems in Number Theory
Numbers n whose Euler Phi value Divides n + 1/Mistake/Third Edition

== Source Work ==
:
:$\mathbf B$: Divisibility
::$\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
== Mistake ==
: gives $8$ solutions to $\map \phi n \mid n + 1$, namely $n = 2$, $n = 2^{2^k} - 1$ for $1 \le k \le 5$, $n = n_1 = 3 \cdot 5 \cdot 17 \cdot 353 \cdot 929$ and $n = n_1 \cdot 83623937$. [Note that $353 = 11 \cdot 2^5 + 1, 929 = 29 \cdot 2^5 + 1, 83623937 = 11 \cdot 29 \cdot 2^{18} + 1$ and $\paren {353 - 2^8} \paren {929 - 2^8} = 2^{16} - 2^8 + 1$.] This exhausts the solutions with less than seven factors. notes that $n = n_1 \cdot 83623937 \cdot 699296672132097$ would be a solution were the largest factor a prime, put notes that this is divisible by $73$.
== Correction ==
While attributes that final observation to , it does in fact appear in 's original $1932$ article as a footnote:
:In the same way if $6992962672132097$ is a prime
::$3 \cdot 5 \cdot 17 \cdot 353 \cdot 929 \cdot 83623937 \cdot 6992962672132097 = 48901526933832864378258473353215$
:is a solution of $(2)$.
Also note that $699296672132097$ is a misprint.
The correct number is $6992962672132097$.
== Also see ==
* Numbers $n$ whose Euler Phi value Divides $n + 1$
== Sources ==
* 
* : $\mathbf B$: Divisibility: $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
Category:Mistakes/Unsolved Problems in Number Theory
Numbers not Expressible as Sum of Distinct Pentagonal Numbers

== Theorem ==
The positive integer which cannot be expressed as the sum of distinct pentagonal numbers are:
:$2, 3, 4, 7, 8, 9, 10, 11, 14, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30,$
:$31, 32, 33, 37, 38, 42, 43, 44, 45, 46, 49, 50, 54, 55, 59, 60, 61, 65,$
:$66, 67, 72, 77, 80, 81, 84, 89, 94, 95, 96, 100, 101, 102, 107, 112, 116,$
:$124, 136, 137, 141, 142, 147, 159$
== Proof ==
It will be proved that the largest integer which cannot be expressed as the sum of distinct pentagonal numbers is $159$.
The remaining integers in the sequence can be identified by inspection.
We prove this using a variant of Second Principle of Mathematical Induction.
Let $\map P n$ be the proposition:
:$n$ can be expressed as the sum of distinct pentagonal numbers.
=== Basis for the induction ===
We verify the result up to $P_{19} = 532$.
Then we conclude all $n$ with $160 \le n \le 532$ can be expressed as the sum of distinct pentagonal numbers.
So $\map P n$ is true for all $160 \le n \le 532$.
This is the basis for the induction.
=== Induction Hypothesis ===
Suppose for some $k > 532$, $\map P j$ is true for all $160 \le j < k$.
That is, every integer between $160$ and $k - 1$ can be expressed as the sum of distinct pentagonal numbers.
This is the induction hypothesis.
=== Induction Step ===
This is the induction step:
We find the largest integer $i$ such that $P_i < k \le P_{i + 1}$.
We show that $k - P_{i - 4}$ can be expressed as the sum of distinct pentagonal numbers, and the sum does not involve $P_{i - 4}$.
Then $k$ can be expressed as the sum of distinct pentagonal numbers.
Since $k > 160 = P_{19}$, we must have $i \ge 19$.
Hence $k > k - P_{i - 4} \ge P_i - P_{i - 4} = \paren {3 i - 2} + \paren {3 i - 5} + \paren {3 i - 8} + \paren {3 i - 11} \ge 55 + 52 + 49 + 47 = 202 > 160$.
Thus $k - P_{i - 4}$ can be expressed as the sum of distinct pentagonal numbers by Induction Hypothesis.
A sufficient condition such that the sum does not involve $P_{i - 4}$ is $P_{i - 4} > k - P_{i - 4}$.
We have:
Therefore we have $P_{i - 4} > k - P_{i - 4}$.
Hence $\map P k$ is true.
By the Second Principle of Mathematical Induction, $\map P n$ is true for all $n \ge 160$.
Thus every integer greater than $159$ can be expressed as the sum of distinct pentagonal numbers.
== Sources ==
* : $159$
Category:Pentagonal Numbers
Numbers not Expressible as Sum of Fewer than 19 Fourth Powers

== Theorem ==
The following positive integer are the only ones which cannot be expressed as the sum of fewer than $19$ fourth powers:
:$79, 159, 239, 319, 399, 479, 559$
== Proof ==
On a case-by-case basis:
From Smallest Number not Expressible as Sum of Fewer than 19 Fourth Powers:
From 159 is not Expressible as Sum of Fewer than 19 Fourth Powers:
From 239 is not Expressible as Sum of Fewer than 19 Fourth Powers:
From 319 is not Expressible as Sum of Fewer than 19 Fourth Powers
From 399 is not Expressible as Sum of Fewer than 19 Fourth Powers
From 479 is not Expressible as Sum of Fewer than 19 Fourth Powers
From 559 is not Expressible as Sum of Fewer than 19 Fourth Powers
== Also see ==
* Hilbert-Waring Theorem
== Historical Note ==
== Sources ==
* : $559$
* : $559$
Category:Fourth Powers
Category:Hilbert-Waring Theorem
Numbers not Expressible as Sum of Fewer than 19 Fourth Powers/Historical Note

== Historical Note on Numbers not Expressible as Sum of Fewer than 19 Fourth Powers ==
It was noted by that there are no other positive integers less than $4100$ needing $19$ fourth powers to express them.
This limit was reported by in his of $1986$.
, and extended this to $10^{245}$ in $2000$.
In $2005$, , and showed that the sequence is complete beyond $10^{220}$.
== Sources ==
* 
* : $559$
* : $559$
* 
* 
Category:Historical Notes
Numbers not Expressible as Sum of Less than 9 Positive Cubes

== Theorem ==
The following are the only positive integers cannot be expressed as the sum of less than $9$ positive cubes:
== Proof ==
== Sources ==
* : $23$
* : $239$
* : $23$
* : $239$
Category:Hilbert-Waring Theorem
Category:Cube Numbers
Category:23
Category:239
Numbers not Sum of Distinct Squares

== Theorem ==
The positive integers which are not the sum of $1$ or more distinct squares are:
:$2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128$
== Proof ==
It will be proved that the largest integer which cannot be expressed as the sum of distinct squares is $128$.
The remaining integers in the sequence can be identified by inspection.
We prove this using a variant of Second Principle of Mathematical Induction.
Let $\map P n$ be the proposition:
:$n$ can be expressed as the sum of distinct squares.
=== Basis for the induction ===
We verify the result up to $18^2 = 324$.
Then we conclude all $n$ with $129 \le n \le 324$ can be expressed as the sum of distinct squares.
So $\map P n$ is true for all $129 \le n \le 324$.
This is the basis for the induction.
=== Induction Hypothesis ===
Suppose for some $k > 324$, $\map P j$ is true for all $129 \le j < k$.
That is, every integer between $129$ and $k - 1$ can be expressed as the sum of distinct squares.
This is the induction hypothesis.
=== Induction Step ===
This is the induction step:
We find the largest integer $i$ such that $i^2 < k \le \paren {i + 1}^2$.
We show that $k - \paren {i - 4}^2$ can be expressed as the sum of distinct squares, and the sum does not involve $\paren {i - 4}^2$.
Then $k$ can be expressed as the sum of distinct squares.
Since $k > 324 = 18^2$, we must have $i \ge 18$.
Hence $k > k - \paren {i - 4}^2 > i^2 - \paren {i - 4}^2 = 4 \paren {2 i - 4} \ge 4 \times 32 = 128$.
Thus $k - \paren {i - 4}^2$ can be expressed as the sum of distinct squares by Induction Hypothesis.
A sufficient condition such that the sum does not involve $\paren {i - 4}^2$ is $\paren {i - 4}^2 > k - \paren {i - 4}^2$.
We have:
Therefore we have $\paren {i - 4}^2 > k - \paren {i - 4}^2$.
Hence $\map P k$ is true.
By the Second Principle of Mathematical Induction, $\map P n$ is true for all $n \ge 128$.
Thus every integer greater than $128$ can be expressed as the sum of distinct square numbers.
== Sources ==
* 
* 
* : $128$
* : $128$
Category:Sums of Squares
Numbers not Sum of Square and Prime

== Sequence ==
The sequence of (strictly) positive integers which is not the sum of a square and a prime begins:
:$1, 10, 25, 34, 58, 64, 85, 91, 121, 130, 169, 196, \ldots$
== Proof ==
$1$ is trivial.
No prime number is in this sequence, as trivially:
:$p = p + 0^2$
and so is the sum of a prime (itself), and $0^2$, which is square.
From Square of n such that 2n-1 is Composite is not Sum of Square and Prime, the sequence contains all $n^2$ such that $2 n - 1$ is composite.
The sequence of such $n$ is infinite, and begins::
:$1, 5, 8, 11, 13, 14, 17, 18, 20, 23, 25, 26, 28, 29, 32, 33, 35, 38, 39, 41, 43, 44, 46, 47, 48, 50, 53, 56, 58, 59, 60, 61, 62, 63, 65, \ldots$
The sequence of remaining integers is:
:$10, 34, 58, 85, 91, 130, 214, 226, 370, 526, 706, 730, 771, 1255, 1351, 1414, 1906, 2986, 3676, 9634, 21679$
From Non-Square Positive Integers not Sum of Square and Prime, this sequence is conjectured to be complete.
== Sources ==
* : $10$
Category:Number Theory
Category:Square Numbers
Category:Prime Numbers
Category:Numbers not Sum of Square and Prime
Numbers of Primes with at most n Digits

== Theorem ==
Let $p: \Z_{>0} \to \Z_{>0}$ be the mapping defined as:
:$\forall n \in \Z_{>0}: \map p n = $ the number of prime numbers with no more than $n$ digits
Then the value of $p$ for the first few numbers is given below:
:
$n$ 	$\map p n$
$1$ 	$4$
$2$ 	$25$
$3$ 	$168$
$4$ 	$1229$
$5$ 	$9592$
$6$ 	$78 \, 498$
$7$ 	$664 \, 579$
$8$ 	$5 \, 761 \, 455$
$9$ 	$50 \, 847 \, 534$
$10$	$455 \, 052 \, 511$
== Historical Note ==
== Sources ==
* : $455,052,511$
* 
* : $168$
* : $50,847,534$
* : $455,052,511$
Category:Prime Numbers
Numbers of Zeroes that Factorial does not end with

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $n!$ denote the factorial of $n$.
Let $n!$ be expressed in decimal notation.
Then $n!$ cannot end in the following numbers of zeroes:
:$5, 11, 17, 23, 29, 30, 36, 42, \ldots$
== Proof ==
== Sources ==
* 
* : $5$
Category:Factorials
Numbers of form 1 + 2m over 1 + 2n form Infinite Abelian Group under Multiplication

== Theorem ==
Let $S$ be the set of integers defined as:
:$S = \set {\dfrac {1 + 2 m} {1 + 2 n}: m, n \in \Z}$
Then $\struct {S, \times}$ is an infinite abelian group.
== Proof ==
Let $k \in \Z$.
Then $1 + 2 k \ne 0$.
Thus:
:$\forall x \in S: x \in \Q_{\ne 0}$
Thus by definition of subset:
:$S \subseteq \Q_{\ne 0}$
From Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group:
:$\struct {\Q_{\ne 0}, \times}$ is an infinite abelian group.
It is noted that $S$ is an infinite set and so trivially $S \ne \O$.
Consider $a = \dfrac {1 + 2 m} {1 + 2 n} \in S$.
We have that:
:$\dfrac {1 + 2 m} {1 + 2 n} \times \dfrac {1 + 2 n} {1 + 2 m} = 1$
and so $\dfrac {1 + 2 n} {1 + 2 m}$ is the inverse of $b \in \struct {\Q_{>0}, \times}$.
By inspection it can be seen that $b^{-1} \in S$.
Let $a, b \in S$.
Then:
:$\exists m_1, n_1 \in \Z: a = \dfrac {1 + 2 m_1} {1 + 2 n_1}$
:$\exists m_2, n_2 \in \Z: b = \dfrac {1 + 2 m_2} {1 + 2 n_2}$
Then we have:
Hence by the Two-Step Subgroup Test, $\struct {S, \times}$ is a subgroup of $\struct {\Q_{>0}, \times}$.
It has been established that $S$ is an infinite set.
Hence by definition $\struct {S, \times}$ is an infinite group.
Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(1) \ \text {(b)}$
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Numbers of form 31 x 16^n are sum of 16 4th Powers

== Theorem ==
Let $m \in \Z$ be an integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$.
Then in order to express $m$ as the sum of fourth powers, you need $16$ of them.
== Proof ==
We have:
:$31 \times 16^n = \paren {2^{n + 1} }^4 + 15 \times \paren {2^n}^4$
so every integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$ can be expressed as the sum of 16 fourth powers.
Now we show that we cannot use less than $16$ fourth powers.
Observe that for an even number $2 k$:
For an odd number $2 k + 1$:
It is obvious that for $n = 0$, $31$ requires $16$ fourth powers to express:
:$31 = 2^4 + 15 \times 1^4$
 for some $n > 0$, $31 \times 16^n$ requires less than $16$ fourth powers to express.
Let $m$ be the smallest of those $n$.
Suppose $x$ of the summands are odd, where $x < 16$.
By the above, we must have:
:$31 \times 16^m \equiv x \pmod {16}$
Since $31 \times 16^m$ is divisible by $16$, $x = 0$.
Hence each summand is even.
Dividing each summand by $2$ gives a representation of $31 \times 16^{m - 1}$ as a sum of less than $16$ fourth powers.
This contradicts the minimality condition on $m$.
Hence the result by Proof by Contradiction.
Category:Hilbert-Waring Theorem
Category:16
Numbers such that Divisor Count divides Phi divides Divisor Sum

== Theorem ==
The sequence of integers $n$ with the property that:
:$\map {\sigma_0} n \divides \map \phi n \divides \map {\sigma_1} n$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count of $n$
:$\phi$ denotes the Euler $\phi$ (phi) function: the count of smaller integers coprime to $n$
:$\sigma_1$ denotes the divisor sum of $n$
begins:
:$1, 3, 15, 30, 35, 56, 70, 78, 105, 140, 168, 190, 210, 248, 264, \ldots$
== Proof ==
By inspection and investigation.
== Examples ==
== Sources ==
* : $210$
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:Divisor Count Function
Category:Euler Phi Function
Category:Divisor Sum Function
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
=== 1 ===
=== 3 ===
=== 15 ===
=== 30 ===
=== 35 ===
=== 56 ===
=== 70 ===
=== 78 ===
=== 105 ===
=== 140 ===
=== 168 ===
=== 190 ===
=== 210 ===
=== 248 ===
=== 264 ===
=== 357 ===
=== 420 ===
=== 570 ===
=== 616 ===
=== 630 ===
== Sources ==
* : $210$
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/1

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $1$ has the property that:
:$\map {\sigma_0} 1 \divides \map \phi 1 \divides \map {\sigma_1} 1$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/105

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $105$ has the property that:
:$\map {\sigma_0} {105} \divides \map \phi {105} \divides \map {\sigma_1} {105}$
where:
:$\divides $ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:105
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/140

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $140$ has the property that:
:$\map {\sigma_0} {140} \divides \map \phi {140} \divides \map {\sigma_1} {140}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:140
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/15

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $15$ has the property that:
:$\map {\sigma_0} {15} \divides \map \phi {15} \divides \map {\sigma_1} {15}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:15
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/168

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $168$ has the property that:
:$\map {\sigma_0} {168} \divides \map \phi {168} \divides \map {\sigma_1} {168}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:168
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/190

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $190$ has the property that:
:$\map {\sigma_0} {190} \divides \map \phi {190} \divides \map {\sigma_1} {190}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:190
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/210

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $210$ has the property that:
:$\map {\sigma_0} {210} \divides \map \phi {210} \divides \map {\sigma_1} {210}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
== Sources ==
* : $210$
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:210
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/248

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $248$ has the property that:
:$\map {\sigma_0} {248} \divides \map \phi {248} \divides \map {\sigma_1} {248}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:248
Numbers such that Divisor Count divides Phi divides Divisor Sum/Examples/264

== Examples of Numbers such that Divisor Count divides Phi divides Divisor Sum ==
The number $264$ has the property that:
:$\map {\sigma_0} {264} \divides \map \phi {264} \divides \map {\sigma_1} {264}$
where:
:$\divides$ denotes divisibility
:$\sigma_0$ denotes the divisor count function
:$\phi$ denotes the Euler $\phi$ (phi) function
:$\sigma_1$ denotes the divisor sum function.
== Proof ==
Category:Numbers such that Divisor Count divides Phi divides Divisor Sum
Category:264