nyuuzyou/wikis huggingface dataset:
194

== Number ==
$194$ (one hundred and ninety-four) is:
:$2 \times 97$
:The $6$th positive integer solution after $1, 3, 15, 104, 164$ to $\map \phi n = \map \phi {n + 1}$:
::$\map \phi {194} = 96 = \map \phi {195}$
:The $28$th nontotient:
::$\nexists m \in \Z_{>0}: \map \phi m = 194$
:where $\map \phi m$ denotes the Euler $\phi$ function
:The number of conjugacy classes of the Fischer-Griess monster $\mathrm M$
== Also see ==
* Number of Conjugacy Classes of Fischer-Griess Monster
* 
* 
Category:Specific Numbers
Category:194
195

== Number ==
$195$ (one hundred and ninety-five) is:
:$3 \times 5 \times 13$
:With $140$, an element of the $2$nd quasiamicable pair:
::$\map {\sigma_1} {140} = \map {\sigma_1} {195} = 336 = 140 + 195 + 1$
:The $9$th inconsummate number after $62$, $63$, $65$, $75$, $84$, $95$, $161$, $173$:
::$\nexists n \in \Z_{>0}: n = 195 \times \map {s_{10} } n$
:The $19$th sphenic number after $30$, $42$, $66$, $70$, $78$, $102$, $105$, $110$, $114$, $130$, $138$, $154$, $165$, $170$, $174$, $182$, $186$, $190$:
::$195 = 3 \times 5 \times 13$
:The $38$th lucky number:
::$1$, $3$, $7$, $9$, $13$, $15$, $21$, $\ldots$, $133$, $135$, $141$, $151$, $159$, $163$, $169$, $171$, $189$, $193$, $195$, $\ldots$
== Also see ==
* 
* 
* 
* 
Category:Specific Numbers
Category:195
1951

== Number ==
$1951$ (one thousand, nine hundred and fifty-one) is:
:The $297$th prime number
:The $26$th centered hexagonal number after $1$, $7$, $19$, $37$, $61$, $91$, $127$, $\ldots$, $817$, $919$, $1027$, $1141$, $1261$, $1387$, $1519$, $1657$, $1801$:
::$1951 = \ds 1 + \sum_{k \mathop = 1}^{26 - 1} 6 k = 26^3 - 25^3$
:The $2$nd of the $60$th pair of twin primes:
::$1951 - 1949 = 2$
== Also see ==
* 
* 
* 
Category:Specific Numbers
Category:1951
1953

== Number ==
$1953$ (one thousand, nine hundred and fifty-three) is:
:$3^2 \times 7 \times 31$
:The $62$nd triangular number after $1$, $3$, $6$, $10$, $15$, $\ldots$, $1326$, $1378$, $1431$, $1485$, $1540$, $1596$, $1653$, $1711$, $1770$, $1770$, $1830$, $1891$:
::$1953 = \ds \sum_{k \mathop = 1}^{62} k = \dfrac {62 \times \left({62 + 1}\right)} 2$
== Also see ==
* 
Category:Specific Numbers
Category:1953
196

== Number ==
$196$ (one hundred and ninety-six) is:
:$2^2 \times 7^2$
:The $1$st candidate Lychrel number.
:The $2$nd power of $14$ after $(1)$, $14$:
::$196 = 14^2$
:The $6$th heptagonal pyramidal number after $1$, $8$, $26$, $60$, $115$:
::$196 = 1 + 7 + 18 + 34 + 55 + 81 = \dfrac {6 \paren {6 + 1} \paren {5 \times 6 - 2} } 6$
:and the $2$nd after $1$ which is square 
:The $7$th pentagonal pyramidal number after $1$, $6$, $12$, $40$, $75$, $126$:
::$196 = 1 + 5 + 12 + 22 + 35 + 51 + 70 = \dfrac {7^2 \paren {7 + 1} } 2$
:and the $2$nd after $1$ which is square
:The $12$th positive integer which cannot be expressed as the sum of a square and a prime:
::$1$, $10$, $25$, $34$, $58$, $64$, $85$, $91$, $121$, $130$, $169$, $196$, $\ldots$
:The $14$th square number after $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$, $121$, $144$, $169$:
::$196 = 14 \times 14$
:The $22$nd powerful number after $1$, $4$, $8$, $9$, $16$, $25$, $27$, $32$, $36$, $49$, $64$, $72$, $81$, $100$, $108$, $121$, $125$, $128$, $144$, $169$
== Also see ==
* Heptagonal Pyramidal Numbers which are Square
* 
* 
* 
* 
* 
* 
* 
== Sources ==
* : $196$
* : $196$
Category:Specific Numbers
Category:196
196,560

== Number ==
$196 \, 560$ (one hundred and ninety-six thousand, five hundred and sixty) is:
:$2^4 \times 3^3 \times 5 \times 7 \times 13$
:The number of spheres touching any one spheres in a $24$-dimensional Leech lattice.
== Also see ==
* Spheres in 24 Dimensions in Leech Lattice
== Sources ==
* : $196,560$
* : $196,560$
Category:Specific Numbers
Category:196,560
1962

== Number ==
$1962$ (one thousand, nine hundred and sixty-two) is:
:$2 \times 3^2 \times 109$
:The $36$th second pentagonal number after $2$, $7$, $15$, $26$, $\ldots$, $1107$, $1190$, $1276$, $1365$, $1457$, $1552$, $1650$, $1751$, $1855$:
::$1962 = \dfrac {36 \paren {3 \times 36 + 1} } 2$
:The $72$nd generalized pentagonal number after $1$, $2$, $5$, $7$, $12$, $15$, $\ldots$, $1335$, $1365$, $1426$, $1457$, $1520$, $1552$, $1617$, $1650$, $1717$, $1751$, $1820$, $1855$, $1926$:
::$1962 = \dfrac {36 \paren {3 \times 36 + 1} } 2$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1962
1963

== Number ==
$1963$ (one thousand, nine hundred and sixty-three) is:
:$13 \times 151$
:The $18$th integer $m$ such that $m! - 1$ (its factorial minus $1$) is prime:
::$3$, $4$, $6$, $7$, $12$, $14$, $30$, $32$, $33$, $38$, $94$, $166$, $324$, $379$, $469$, $546$, $974$, $1963$
== Also see ==
* 
Category:Specific Numbers
Category:1963
197

== Number ==
$197$ (one hundred and ninety-seven) is:
:The $45$th prime number
:The $6$th prime number of the form $n^2 + 1$ after $2$, $5$, $17$, $37$, $101$:
::$197 = 14^2 + 1$
:The $7$th Keith number after $14$, $19$, $28$, $47$, $61$, $75$:
::$1$, $9$, $7$, $17$, $33$, $57$, $107$, $197$, $\ldots$
:The smaller of the $15$th pair of twin primes, with $199$
:The $20$th left-truncatable prime after $2$, $3$, $5$, $7$, $13$, $17$, $23$, $37$, $43$, $47$, $53$, $67$, $73$, $83$, $97$, $113$, $137$, $167$, $173$
== Also see ==
* 
* 
* 
* 
* 
Category:Specific Numbers
Category:197
1976

== Number ==
$1976$ (one thousand, nine hundred and seventy-six) is:
:$2^3 \times 13 \times 19$
:The $26$th octagonal number, after $1$, $8$, $21$, $40$, $65$, $\ldots$, $645$, $736$, $833$, $936$, $1045$, $1160$, $1281$, $1408$, $1541$, $1680$, $1825$:
::$1976 = \ds \sum_{k \mathop = 1}^{26} \paren {6 k - 5} = 26 \paren {3 \times 26 - 2}$
== Also see ==
* 
Category:Specific Numbers
Category:1976
1980

== Number ==
$1980$ (one thousand, nine hundred and eighty) is:
:$2^2 \times 3^2 \times 5 \times 11$
:The smallest term of the amicable triplet $\tuple {1980, 2016, 2556}$:
::$\map {\sigma_1} {1980} = \map {\sigma_1} {2016} = \map {\sigma_1} {2556} = 6552 = 1980 + 2016 + 2556$
:The $2$nd integer after $954$ whose digits are unchanged when subtracting its reversal:
::$1980 - 0891 = 1089$
== Also see ==
* 
* 
== Sources ==
* : $\S 13$: The fundamental theorem of arithmetic
* : $1980$
* : $1980$
Category:Specific Numbers
Category:1980
199

== Number ==
$199$ (one hundred and ninety-nine) is:
:The $46$th prime number
:The $1$st of the smallest $10$ primes in arithmetic sequence:
::$199 = 0 \times 210 + 199$
:The $7$th Lucas prime after $2$, $3$, $7$, $11$, $29$, $47$
:The $7$th near-repdigit prime after $101$, $113$, $131$, $151$, $181$, $191$
:The $11$th Lucas number after $(2)$, $1$, $3$, $4$, $7$, $11$, $18$, $29$, $47$, $76$, $123$:
::$199 = 76 + 123$
:The $15$th emirp after $13$, $17$, $31$, $37$, $71$, $73$, $79$, $97$, $107$, $113$, $149$, $157$, $167$, $179$
:The larger of the $15$th pair of twin primes, with $197$
:The $16$th permutable prime after $2$, $3$, $5$, $7$, $11$, $13$, $17$, $31$, $37$, $71$, $73$, $79$, $97$, $113$, $131$
== Also see ==
* 
* 
* 
* 
* 
* 
* 
* 
== Sources ==
* : $199$
* : $199$
Category:Specific Numbers
Category:199
1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways

== Theorem ==
The number $1$ can be expressed as the sum of $4$ distinct unit fractions in $6$ different ways:
== Proof 1 ==
== Proof 2 ==
== Also see ==
* 2520 equals Sum of 4 Divisors in 6 Ways
Category:1
Category:Unit Fractions
Category:Recreational Mathematics
Category:1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways
1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways/Proof 1

== Theorem ==
== Proof ==
Let:
:$1 = \dfrac 1 v + \dfrac 1 w + \dfrac 1 x + \dfrac 1 y$
where $ 1 < v < w < x < y$
Suppose $v = 3$ and take the largest potential solution that can be generated:
:$1 \stackrel {?} {=} \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6$
But we find:
:$1 > \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6$
Therefore, there can be no solutions where $v \ge 3$, as that solution was the largest possible.
Hence, $v = 2$ if there are any solutions.
Repeating the above anaylsis on $w$:
Potential solutions are located where $w = 3, 4, 5$.
Now that $v$ and $w$ are known, the variable $y$ can be written in terms of $x$:
:$y = \dfrac 1 {\dfrac {w - 2} {2 w} - \dfrac 1 x}$
Solutions are only positive when:
As $x < y$:
Therefore solutions exist only in the domain:
:$\dfrac {2 w} {w - 2} < x < \dfrac {4 w} {w - 2}$
and:
:$w < x$
Case $w = 3$:
Integer solutions in the above domain can then be found by inspection:
:$\tuple {7, 42}, \tuple {8, 24}, \tuple {9, 18}, \tuple {10, 15}$
Case $w = 4$:
Integer solutions in the above domain can again be found by inspection:
:$\tuple {5, 20}, \tuple {6, 12}$
Case $w = 5$:
and it is immediately seen that there are no integer solutions in this domain.
All solutions have therefore been found:
:$\tuple {2, 3, 7, 42}, \tuple {2, 3, 8, 24}, \tuple {2, 3, 9, 18}, \tuple {2, 3, 10, 15}, \tuple {2, 4, 5, 20}, \tuple {2, 4, 6, 12}$
Hence:
Category:1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways
1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways/Proof 2

== Theorem ==
== Proof ==
From Sum of 4 Unit Fractions that equals 1:
This includes repeated unit fractions.
The full list is:
The result follows from inspection: solutions $(1)$, $(2)$, $(3)$, $(4)$, $(6)$ and $(7)$ are the only solutions of the above such that the denominators of the summands are distinct.
Category:1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways
1 is Limit Point of Sequence in Sierpiński Space

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
The sequence in $T$:
:$\sigma = \sequence {0, 1, 0, 1, \ldots}$
has $1$ as a limit.
== Proof ==
By definition, $\alpha$ is a limit of $\sigma$ :
:$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \notin U}$ is finite.
The only open set of $T$ containing $1$ is $\set {0, 1}$.
It contains all but a finite number (that is: $0$) elements of $\sigma$.
Hence, by definition, $1$ is a limit of $\sigma$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:Examples of Limit Points
Category:Sierpiński Space
1 plus Perfect Power is not Power of 2

== Theorem ==
The equation:
:$1 + a^n = 2^m$
has no solutions in the integers for $n, m > 1$.
This is an elementary special case of Catalan's Conjecture.
== Proof ==
 there is a solution.
Then:
$a$ is immediately seen to be odd.
By Square Modulo 4, $n$ must also be odd.
Now:
The latter sum has $n$ powers of $a$, which sums to an odd number.
The only odd divisor of $2^m$ is $1$.
However, if the sum is $1$, we have:
:$a^n + 1 = a + 1$
giving $n = 1$, contradicting our constraint $n > 1$.
Hence the result by Proof by Contradiction.
Category:Number Theory
1 plus Perfect Power is not Prime Power except for 9

== Theorem ==
The only solution to:
:$x^m = y^n + 1$
is:
:$\tuple {x, m, y, n} = \tuple {3, 2, 2, 3}$
for positive integers $x, y, m, n > 1$, and $x$ is a prime number.
This is a special case of Catalan's Conjecture.
== Proof ==
It suffices to show the result for prime values of $n$.
The case $n = 2$ is covered in 1 plus Square is not Perfect Power.
So we consider the cases where $n$ is an odd prime.
where $Q$ is a polynomial in one unknown and $R$ is a degree zero polynomial, so $R$ is a constant.
We have:
Hence we have $x^m = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}$.
Since $x$ is a prime, we have:
:$x \divides y + 1$
:$x \divides \map Q y \paren {y + 1} + n$
Hence $x \divides n$.
Since $x > 1$ and $n$ is a prime, we must have $x = n$.
Now we write $y + 1 = x^\alpha$.
Then we have:
For $\alpha > 1$, $x \nmid \map Q y x^{\alpha - 1} + 1$.
Hence $\alpha = 1$.
This gives $y + 1 = x = n$.
The equation now simplifies to:
:$\paren {y + 1}^m = y^n + 1$
Expanding:
hence we must have $y \divides m$.
By Absolute Value of Integer is not less than Divisors, $y \le m$.
Moreover, from $\ds \sum_{j \mathop = 1}^m \binom m j y^{j - 1} = y^{n - 1}$ we also have:
:$y^{n - 1} > \dbinom m m y^{m - 1} = y^{m - 1}$
Therefore we also have $n > m$.
This gives $y = n - 1 \ge m$.
The two inequalities forces $y = m$.
Now our original equation is further simplified to:
From Real Sequence (1 + x over n)^n is Convergent:
:$\paren {1 + \dfrac 1 y}^y$ is increasing and has limit $e$.
Then we have for all $y \in \N$:
:$y + \dfrac 1 {y^y} < e < 3$
Since $\dfrac 1 {y^y} > 0$ and $y > 1$, we can only have $y = 2$.
This gives the solution $3^2 = 2^3 + 1$, and there are no others.
== Also see ==
* Cube which is One Less than a Square
== Sources ==
* 
Category:Number Theory
1 plus Power of 2 is not Perfect Power except 9

== Theorem ==
The only solution to:
:$1 + 2^n = a^b$
is:
:$\tuple {n, a, b} = \tuple {3, 3, 2}$
for positive integers $n, a, b$ with $b > 1$.
== Proof ==
It suffices to prove the result for prime values of $b$.
For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a perfect power.
For $n > 0$, $1 + 2^n$ is odd.
Hence for the equation to hold $a$ must be odd as well.
Writing $a = 2 m + 1$ we have:
Since all factors of $2^n$ are powers of $2$:
:$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is a power of $2$.
But since each summand is non-negative:
:$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1} \ge 2$
we must have $\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is even.
We have:
Therefore we must have $b = 2$, the only even prime.
In that case:
So $m$ and $m + 1$ are powers of $2$.
The only $m$ satisfying this is $1$, giving the solution:
== Also see ==
* Catalan's Conjecture, of which this is a special case.
Category:Number Theory
1 plus Square is not Perfect Power

== Theorem ==
The equation:
:$x^p = y^2 + 1$
has no solution in the integers for $x, y, p > 1$.
== Proof ==
Suppose $p$ is even. 
Write $p = 2 k$.
Then:
Since both $y - x^k$ and $y + x^k$ are integers, they must be equal to $\pm 1$.
Summing them up, we have $2 y$ is one of $-2, 0, 2$.
Thus $y$ is one of $-1, 0, 1$, and we ignore these solutions due to our condition $y > 1$.
Now suppose $p$ is odd.
Suppose $y$ is odd.
Then $x^p = y^2 + 1$ is even.
Hence $x$ is even.
Then:
which is a contradiction.
Hence $y$ must be even, and $x$ must be odd.
From Gaussian Integers form Euclidean Domain, we can define greatest common divisors on $\Z \sqbrk i$, and it admits unique factorization.
We factorize $y^2 + 1$:
:$x^p = y^2 + 1 = \paren {1 + i y} \paren {1 - i y}$
The greatest common divisors of $1 + i y$ and $1 - i y$ must divide their sum and product.
Their sum is $2$ while their product is $y^2 + 1$, which is odd.
Therefore we see that $1 + i y$ and $1 - i y$ are coprime.
From unique factorization we must have that both $1 + i y$ and $1 - i y$ is a product of a unit and a $p$th power.
By Units of Gaussian Integers, the units are $\pm 1$ and $\pm i$.
Hence
:$\exists u \in \set {\pm 1, \pm i}: \exists \alpha \in \Z \sqbrk i: 1 + i y = u \alpha^p, 1 - i y = \bar u \bar \alpha^p$
Since $p$ is odd:
:$1^p = 1$
:$\paren {-1}^p = -1$
:$i^p = \pm i$
:$\paren {-i}^p = -i^p = \mp i$
therefore there is some unit $u' \in \set {\pm 1, \pm i}$ such that $u'^p = u$.
By writing $\beta = u' \alpha$:
:$1 + i y = u'^p \alpha^p = \beta^p, 1 - i y = \bar \beta^p$
Write $\beta = a + i b$, where $a, b \in \Z$.
By Sum of Two Odd Powers:
:$2 a = \beta + \bar \beta \divides \beta^p + \bar \beta^p = 2$
this gives $a = \pm 1$.
We also have:
since $1 + y^2$ is odd, $b$ must be even.
Hence:
In particular, comparing real parts gives $1 \equiv a^p \pmod 4$.
Since $p$ is odd, we have $a = 1$.
Now we have:
The summands on the may not be an integer, but if we can show:
:In canonical form, the numerator of each summand is even
then the equation is never satisfied.
This is because the sum of all the terms will be a rational number with even numerator and odd denominator, which cannot equal to $1$.
Since $2 k + 1$ is always odd and $\paren {-1}^k \dbinom {p - 2} {2 k - 2}$ is always an integer, we only need to check $\dfrac {b^{2 k - 2} } k$.
Since $b$ is even:
:$2^{2 k - 2} \divides b^{2 k - 2}$
But we have:
Hence the largest power of $2$ that divides $k$ is less than $2^{2 k - 2}$.
Therefore the numerator of $\dfrac {b^{2 k - 2} } k$ is even.
And thus all the equations above are never satisfied.
So our original equation:
:$x^p = y^2 + 1$
has no solution in the integers for $x, y, p > 1$.
== Historical Note ==
== Sources ==
* 
* : Chapter $2$: Even Exponents
Category:Number Theory
2,097,152

== Number ==
$2 \, 097 \, 152$ (two million, ninety-seven thousand, one hundred and fifty-two) is:
:$2^{21}$
:The $7$th power of $8$ after $(1)$, $8$, $64$, $512$, $4096$, $32 \, 768$, $262 \, 144$:
::$2 \, 097 \, 152 = 8^7$
:The $8$th $7$th power after after $1$, $128$, $2187$, $16 \, 384$, $78 \, 125$, $279 \, 936$, $823 \, 543$:
::$2 \, 097 \, 152 = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \times 8$
:The $21$st power of $2$ after $(1)$, $2$, $4$, $8$, $16$, $\ldots$, $65 \, 536$, $131 \, 072$, $262 \, 144$, $524 \, 288$, $1 \, 048 \, 576$:
::$2 \, 097 \, 152 = 2^{21}$
:The $22$nd almost perfect number after $1$, $2$, $4$, $8$, $16$, $\ldots$, $65 \, 536$, $131 \, 072$, $262 \, 144$, $524 \, 288$, $1 \, 048 \, 576$:
::$\map {\sigma_1} {2 \, 097 \, 152} = 4 \, 194 \, 303 = 2 \times 2 \, 097 \, 152 - 1$
:The $128$th cube number:
::$2 \, 097 \, 152 = 128 \times 128 \times 128$
== Also see ==
* 
* 
* 
* 
* 
Category:Specific Numbers
Category:2,097,152
2,124,679

== Number ==
$2 \, 124 \, 679$ (two million, one hundred and twenty-four thousand, six hundred and seventy-nine) is:
:The $157 \, 504$th prime number
:The larger of the two known Wolstenholme primes, the other being $16 \, 843$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2,124,679
2,147,483,647

== Number ==
$2 \, 147 \, 483 \, 647$ is:
:A prime number
:The $8$th Mersenne prime after $3$, $7$, $31$, $127$, $8191$, $131 \, 071$, $524 \, 287$:
::$2 \, 147 \, 483 \, 647 = 2^{31} - 1$
:The $11$th Mersenne number after $3$, $7$, $31$, $127$, $2047$, $8191$, $131 \, 071$, $524 \, 287$, $8 \, 388 \, 607$, $536 \, 870 \, 911$:
::$2 \, 147 \, 483 \, 647 = 2^{31} - 1$
== Historical Note ==
== Also see ==
* 
* 
* 
Category:Specific Numbers
Category:2,147,483,647
2,147,483,647/Historical Note

== Historical Note on $2 \, 147 \, 483 \, 647$ ==
$2 \, 147 \, 483 \, 647$ was one of the Mersenne numbers that predicted to be prime in his of $1644$.
It was who demonstrated it to be prime, which he did in $1772$.
=== Barlow's Prediction ===
== Sources ==
* : Preface
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
2,185,095

== Number ==
$2 \, 185 \, 095$ (two million, one hundred and eighty-five thousand and ninety-five) is:
:$3 \times 5 \times 11 \times 17 \times 19 \times 41$
:The $2$nd of the $26$th pair of triangular numbers whose sum and difference are also both triangular:
::$1 \, 747 \, 515 = T_{1869}$, $2 \, 185 \, 095 = T_{2090}$, $2 \, 185 \, 095 + 1 \, 747 \, 515 = T_{2804}$, $2 \, 185 \, 095 - 1 \, 747 \, 515 = T_{935}$
:The $2090$th triangular number:
::$1 \, 747 \, 515 = \ds \sum_{k \mathop = 1}^{2090} k = \dfrac {2090 \times \paren {2090 + 1} } 2$
== Also see ==
* 
* 
== Sources ==
* : $1,747,515$
* : $1,747,515$
Category:Specific Numbers
Category:2,185,095
2,235,197,406,895,366,368,301,560,000

== Number ==
$2 \, 235 \, 197 \, 406 \, 895 \, 366 \, 368 \, 301 \, 560 \, 000$ is:
:$2^6 \times 3^2 \times 5^4 \times 7^4 \times 17^3 \times 19^2 \times 23^2 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47$
:The odds against all $4$ players each being dealt a complete suit in a single deal at bridge
== Also see ==
* Probability of All Players receiving Complete Suit at Bridge
== Sources ==
* : $2,235,197,406,895,366,368,301,560,000$
* : $2,235,197,406,895,366,368,301,560,000$
Category:Specific Numbers
Category:2,235,197,406,895,366,368,301,560,000
2,236,133,941

== Number ==
$2\,236\,133\,941$ is:
:The $1$st of an arithmetic sequence of $16$ prime numbers:
::$2\,236\,133\,941 = 2\,236\,133\,941 + 0 \times 223\,092\,870$ 
== Also see ==
* 
* Arithmetic Sequence of 16 Primes
== Sources ==
* : $2,236,133,941$
Category:Specific Numbers
Category:2,236,133,941
2,245,319

== Number ==
$2 \, 245 \, 319$ (two million, two hundred and forty-five thousand, three hundred and nineteen) is:
:The $165 \, 760$th prime number
:The $2$nd term of the smallest Cunningham chain of the first kind of length $7$:
::$1 \, 122 \, 659$, $2 \, 245 \, 319$, $4 \, 490 \, 639$, $8 \, 981 \, 279$, $17 \, 962 \, 559$, $35 \, 925 \, 119$, $71 \, 850 \, 239$
== Also see ==
* 
* 
Category:Specific Numbers
Category:2,245,319
2,300,000/Historical Note

== Historical Note on $2 \, 300 \, 000$ (Two Million, Three Hundred Thousand) ==
According to , this number appears on the Columna Rostrata of Gaius Duilius, in the form of the symbol for $100 \, 000$ repeated $23$ times.
However, the inscription is worn due to age, and the precise details of what is exactly written there are disputed.
== Sources ==
* : $2,300,000$
* : $2,300,000$
Category:Historical Notes
2,305,843,009,213,693,951

== Number ==
$2 \, 305 \, 843 \, 009 \, 213 \, 693 \, 951$ is:
:A prime number
:The $9$th Mersenne prime after $3$, $7$, $31$, $127$, $8191$, $131 \, 071$, $524 \, 287$, $2 \, 147 \, 483 \, 647$:
::$2 \, 305 \, 843 \, 009 \, 213 \, 693 \, 951 = 2^{61} - 1$
== Historical Note ==
== Also see ==
* 
* 
Category:Specific Numbers
Category:2,305,843,009,213,693,951
2,305,843,009,213,693,951/Historical Note

== Historical Note on $2 \, 305 \, 843 \, 009 \, 213 \, 693 \, 951$ ==
$2 \, 305 \, 843 \, 009 \, 213 \, 693 \, 951 = 2^{61} - 1$ was discovered to be prime in $1883$ by .
It was discovered independently in $1886$ by , and confirmed by in $1887$.
It was the $10$th Mersenne prime to be discovered, although the $9$th in sequence.
 had demonstrated in $1876$ that $2^{127} - 1$ (actually the $12$th in sequence) is prime.
== Sources ==
* : Preface
Category:Historical Notes
2,380,642,361

== Number ==
$2 \, 380 \, 642 \, 361$ is:
:$7507 \times 317 \, 123$
:The smallest integer after $1 \, 111 \, 111 \, 111$ whose square ends in $987 \, 654 \, 321$ :
::$2 \, 380 \, 642 \, 361^2 = 5 \, 667 \, 458 \, 050 \, 987 \, 654 \, 321$
== Also see ==
* Square Number ending in 9 Digits in Reverse Order
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Specific Numbers
Category:2,380,642,361
2,438,195,760

== Number ==
$2 \, 438 \, 195 \, 760$ is:
:$2^4 \times 3^2 \times 5 \times 7 \times 11 \times 13 \times 17 \times 199$
:The $1$st of $4$ pandigital integers which are divisible by all the positive integers up to $18$
== Also see ==
* 
== Sources ==
* : $2,438,195,760$
* : $2,438,195,760$
Category:Specific Numbers
Category:2,438,195,760
2,646,798

== Number ==
$2 \, 646 \, 798$ is:
:$2 \times 3 \times 7 \times 11 \times 17 \times 337$
:The $19$th number after $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $89$, $135$, $175$, $518$, $598$, $1306$, $1676$, $2427$ which is the sum of the increasing powers of its digits taken in order:
::$2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7 = 2 \, 646 \, 798$
== Also see ==
* 
Category:Specific Numbers
Category:2,646,798
2,674,440

== Number ==
$2 \, 674 \, 440$ (two million, six hundred and seventy-four thousand, four hundred and forty) is:
:$2^3 \times 3^2 \times 5 \times 17 \times 19 \times 23$
:The $14$th Catalan number after $(1)$, $1$, $2$, $5$, $14$, $42$, $132$, $429$, $1430$, $4862$, $16 \, 796$, $58 \, 786$, $208 \, 012$, $742 \, 900$:
::$2 \, 674 \, 440 = \dfrac 1 {14 + 1} \dbinom {2 \times 14} {14} = \dfrac 1 {15} \times 40 \, 116 \, 600$
== Also see ==
* 
Category:Specific Numbers
Category:2,674,440
2,759,640

== Number ==
$2 \, 759 \, 640$ (two million, seven hundred and fifty-nine thousand, six hundred and forty) is:
:$2^3 \times 3 \times 5 \times 13 \times 29 \times 61$
:The $3$rd and last positive integer $n$ after $15$, $22 \, 440$ such that $\pm n$ allows $x^5 - x + n$ to be factorized into the product of an irreducible quadratic and an irreducible cubic:
::$x^5 - x + 2 \, 759 \, 640 = \paren {x^2 + 12 x + 377} \paren {x^3 - 12 x^2 - 233 x + 7320}$
== Also see ==
* 
== Sources ==
* : $2,759,640$
Category:Specific Numbers
Category:2,759,640