nyuuzyou/wikis huggingface dataset:
P-adic Valuation Extends to P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\nu_p^\Q: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on the set of rational numbers.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ be defined by:
:$\forall x \in \Q_p : \map {\nu_p} x = \begin {cases}
-\log_p \norm x_p : x \ne 0 \\
+\infty : x = 0
\end{cases}$
Then $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ is a valuation that extends $\nu_p^\Q$ from $\Q$ to $\Q_p$.
== Proof ==
It needs to be shown that $\nu_p$:
:$(1): \quad \nu_p$ is a mapping into $\Z \cup \set {+\infty}$
:$(2): \quad \nu_p$ satisfies the valuation axioms $\text V 1$, $\text V 2$ and $\text V 3$
:$(3): \quad \nu_p$ extends $\nu_p^\Q$.
Let $x, y \in \Q_p$.
=== $\nu_p$ is a mapping into $\Z \cup \set {+\infty}$ ===
If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition.
Let $x \ne 0$.
By P-adic Norm of p-adic Number is Power of p then:
:$\exists v \in \Z: \norm x_p = p^{-v}$
Hence:
=== $\nu_p$ satisfies $(\text V 1)$ ===
If $x = 0$ then:
Similarly, if $y = 0$ then:
If $x \ne 0, y \ne 0$ then:
=== $\nu_p$ satisfies $(\text V 2)$ ===
If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition.
If $x \ne 0$ then $\map {\nu_p} x \in \Z$ by the above.
Hence:
:$\map {\nu_p} x = +\infty \iff x = 0$ 
=== $\nu_p$ satisfies $(\text V 3)$ ===
Suppose $x = 0$.
Then:
Similarly, if $y = 0$ then:
Suppose $x + y = 0$.
Then:
Suppose $x \ne 0, y \ne 0, x + y \ne 0$.
Then:
=== $\nu_p$ extends $\nu_p^\Q$ ===
Let $x \in \Q$.
If $x = 0$ then $\map {\nu_p} 0 = +\infty = \map {\nu_p^\Q} 0$.
Let $x \ne 0$.
From Rational Numbers are Dense Subfield of P-adic Numbers:
:the $p$-adic norm $\norm {\,\cdot\,}_p$ on $p$-adic numbers $\Q_p$ is an extension of the $p$-adic norm $\norm {\,\cdot\,}_p$ on rational numbers $\Q$ by definition.
Hence:
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.2$
Category:P-adic Number Theory
P-adic Valuation is Valuation

== Theorem ==
The $p$-adic valuation $\nu_p: \Q \to \Z \cup \set {+\infty}$ is a valuation on $\Q$.
== Proof ==
To prove that $\nu_p$ is a valuation it is necessary to demonstrate:
Let $q := \dfrac a b, r := \dfrac c d \in \Q$.
=== Axiom $(\text V 1)$ ===
=== Axiom $(\text V 2)$ ===
=== Axiom $(\text V 3)$ ===
From Restricted P-adic Valuation is Valuation follows that:
Hence:
:$\map {\nu_p} {\dfrac a b + \dfrac c d} \ge \min \set {\map {\nu_p} {\dfrac a b}, \map {\nu_p} {\dfrac c d} }$
Thus $\nu_p: \Q \to \Z \cup \set {+\infty}$ is a valuation on $\Q$ by definition.
== Sources ==
* : $\S 2.1$: Absolute Values on a Field: Lemma $2.1.3$
Category:P-adic Valuations
P-adic Valuation of Difference of Powers with Coprime Exponent

== Theorem ==
Let $x, y \in \Z$ be distinct integers.
Let $n \ge 1$ be a natural number.
Let $p$ be a prime number.
Let:
:$p \divides x - y$
and:
:$p \nmid x y n$.
Then 
:$\map {\nu_p} {x^n - y^n} = \map {\nu_p} {x - y}$
where $\nu_p$ denotes $p$-adic valuation.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Lifting The Exponent Lemma
== Sources ==
* : Theorem $\text{II}$
Category:Lifting The Exponent Lemma
Category:P-adic Valuation of Difference of Powers with Coprime Exponent
P-adic Valuation of Difference of Powers with Coprime Exponent/Proof 1

== Theorem ==
== Proof ==
We have:
:$x^n - y^n = \paren {x - y} \paren {x^{n - 1} + \cdots + y^{n - 1} }$
We have to show that:
:$p \nmid x^{n - 1} + \cdots + y^{n - 1}$
Because $x \equiv y \pmod p$:
:$x^{n - 1} + \cdots + y^{n - 1} \equiv x^{n - 1} + x^{n - 1} + \cdots + x^{n - 1} = n x^{n - 1} \pmod p$
Because $p \nmid x$ and $p \nmid n$:
:$p \nmid n x^{n - 1}$
Category:P-adic Valuation of Difference of Powers with Coprime Exponent
P-adic Valuation of Difference of Powers with Coprime Exponent/Proof 2

== Lemma ==
== Proof ==
We have:
:$x^n - y^n = \paren {x - y} \paren {x^{n - 1} + \cdots + y^{n - 1} }$
We have to prove that:
:$p \nmid x^{n - 1} + \cdots + y^{n - 1}$
Let $\map P u = u^n - y^n$.
If $p \divides x^{n - 1} + \cdots + y^{n - 1}$, then $x$ would be a double root of $P$ modulo $p$.
By Double Root of Polynomial is Root of Derivative (or a version of this in modular arithmetic):
:$p \divides \map {P'} x = n x^{n - 1}$
which is impossible.
Therefore:
:$p \nmid x^{n - 1} + \cdots + y^{n - 1}$
Category:P-adic Valuation of Difference of Powers with Coprime Exponent
P-adic Valuation of Rational Number is Well Defined

== Theorem ==
The $p$-adic valuation:
:$\nu_p: \Q \to \Z \cup \set {+\infty}$
is well defined.
== Proof ==
Let $\dfrac a b = \dfrac c d \in \Q$.
Thus:
:$a d = b c \in \Z$
By :
:$b, d \ne 0$
By :
:$\map {\nu_p^\Z} b, \map {\nu_p^\Z} d < +\infty$
=== Case 1 : $a \ne 0$ ===
Let $a \ne 0$.
It follows:
:$c \ne 0$
By :
:$\map {\nu_p^\Z} a, \map {\nu_p^\Z} c < +\infty$
Then:
=== Case 2 : $a = 0$ ===
Let $a = 0$.
It follows:
:$c = 0$
By :
:$\map {\nu_p^\Z} a = \map {\nu_p^\Z} c = +\infty$
Then:
In either case:
So:
Thus, by definition, $\nu_p: \Q \to \Z \cup \set {+\infty}$ is well defined.
Category:P-adic Valuations
PGF of Sum of Independent Discrete Random Variables

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be independent discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $Z$ be a discrete random variable such that $Z = X + Y$.
Then:
:$\map {\Pi_Z} s = \map {\Pi_X} s \, \map {\Pi_Y} s$
where $\map {\Pi_Z} s$ is the probability generating function of $Z$.
=== General Result ===
== Proof ==
== Sources ==
* : $\S 4.4$: Sums of independent random variables: Theorem $4 \ \text{C}$
Category:Probability Generating Functions
PGF of Sum of Independent Discrete Random Variables/General Result

== Theorem ==
Let:
:$Z = X_1 + X_2 + \cdots + X_n$
where each of $X_1, X_2, \ldots, X_n$ are independent discrete random variables with PGFs $\map {\Pi_{X_1} } s, \map {\Pi_{X_2} } s, \ldots, \map {\Pi_{X_n} } s$.
Then:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^n \map {\Pi_{X_j} } s$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$
whenever $m \le n$
$\map P 1$ is true, as this just says $\map {\Pi_{X_1} } s = \map {\Pi_{X_1} } s$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\map {\Pi_{X + Y} } s = \map {\Pi_X} s \, \map {\Pi_Y} s$
which is proved in PGF of Sum of Independent Discrete Random Variables.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true, where $j \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$
whenever $m \le k$.
Then we need to show:
:$\ds \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$
whenever $m \le {k + 1}$.
=== Induction Step ===
This is our induction step:
Let $Z = X_1 + X_2 + \cdots + X_k + X_{k + 1}$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: \map {\Pi_Z} s = \ds \prod_{j \mathop = 1}^n \map {\Pi_{X_j} } s$
== Sources ==
* : $\S 4.4$: Sums of independent random variables: Theorem $4 \ \text{C}$
Category:Probability Generating Functions
PGF of Sum of Random Number of Independent Discrete Random Variables

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let:
:$N, X_1, X_2, \ldots$
be independent discrete random variables such that the $X$'s have the same probability distribution.
Let:
:$\map {\Pi_N} s$ be the PGF of $N$
:$\map {\Pi_X} s$ be the PGF of each of the $X$'s.
Let:
:$Z = X_1 + X_2 + \ldots + X_N$
Then:
:$\map {\Pi_Z} s = \map {\Pi_N} {\map {\Pi_X} s}$
== Proof ==
== Sources ==
* : $\S 4.4$: Sums of independent random variables: Theorem $4 \text{D}$
Category:Probability Generating Functions
P versus NP

== Open Question ==
The class of problems for which an algorithm can find a solution in polynomial time is termed $P$.
The class of problems for which an algorithm can verify a solution in nondeterministic polynomial time is termed $NP$.
The $P$ versus $NP$ question is:
:Are all problems in $NP$ also in $P$?
== Progress ==
Most mathematicians and computer scientists expect the answer to the question is: no, they are not.
However, there has been no proof one way or the other so far.
== Also known as ==
== Sources ==
* 
* 
* 
* 
* 
* : Appendix $18$: Millennium Prize problems: $1$.
* : Appendix $23$: Millennium Prize problems: $1$.
Category:P versus NP
Category:Millennium Problems
Category:Open Questions
Packing of Circles into Plane

== Theorem ==
The most space-efficient way to pack unit circles into the plane is to place their centers at the lattice points of the point lattice in $\R^n$ with a basis $\set {\tuple {2, 0}, \tuple {1, \sqrt 3} }$.
== Proof ==
== Sources ==
* 
* 
Category:Unit Circles
Category:Point Lattices
Padé Approximant/Examples/Exponential

== Example of Padé Approximant ==
Let $f: \R \to \R$ denote the real function $f: x \mapsto e^x$.
A Padé approximant to $f$ with numerator and denominator of degree $2$ is:
:$\dfrac {x^2 + 6 x + 12} {x^2 - 6 x + 12}$
== Proof ==
== Sources ==
* 
Category:Examples of Padé Approximants
Pair of Consecutive Powerful Numbers whose First is Odd

== Theorem ==
The only known pair of consecutive integers which are both powerful numbers such that the first of the pair is odd is:
:$\tuple {675, 676}$
== Proof ==
By investigation:
That there are no smaller ones can be determined again by investigation.
== Sources ==
* 
* : $675$
Category:Powerful Numbers
Category:675
Category:676
Pair of Large Twin Primes

== Theorem ==
The integers defined as:
:$1\,159\,142\,985 \times 2^{2304} \pm 1$
are a pair of twin primes each with $703$ digits.
== Proof ==
$1\,159\,142\,985 \times 2^{2304} - 1$:
$1\,159\,142\,985 \times 2^{2304} + 1$:
== Historical Note ==
== Sources ==
* 
* 
* : $1159142985 \times 2^{2304} \pm 1$
Category:Twin Primes/Examples
Pair of Large Twin Primes/Historical Note

== Historical Note on Pair of Large Twin Primes ==
 reports in his of $1986$ that this pair of twin primes is the largest reported on by in his of $1981$.
It is reported that they were discovered by and in $1979$.
Apparently they also discovered the pair $694\,513\,810 \times 2^{2304} \pm 1$ at around the same time.
== Sources ==
* 
* 
* : $1159142985 \times 2^{2304} \pm 1$
Category:Historical Notes
Pair of Titanic Twin Primes

== Theorem ==
The integers defined as:
:$190 \, 116 \times 3003 \times 10^{5120} \pm 1$
are a pair of titanic twin primes.
That is:
:$570 \, 918 \, 347 \paren 9_{5820}$
and:
:$570 \, 918 \, 348 \paren 0_{5819} 1$
where $\paren a_b$ means $b$ instances of $a$ in a string.
== Proof ==
It is noted that these integers have $9 + 5820 = 5829$ digits, making them titanic.
== Historical Note ==
== Sources ==
* 
* : $190,116 \times 3003 \times 10^{5120} - 1$ and $190,116 \times 3003 \times 10^{5120} + 1$
Category:Titanic Primes
Category:Twin Primes/Examples
Pair of Titanic Twin Primes/Historical Note

== Historical Note on Pair of Titanic Twin Primes ==
According to in his of $1997$, this pair of titanic twin primes was discovered by on $5$ October $1995$, but this has not been corroborated.
== Sources ==
* 
* : $190,116 \times 3003 \times 10^{5120} - 1$ and $190,116 \times 3003 \times 10^{5120} + 1$
Category:Historical Notes
Pair of Ulam Numbers differing by One/Historical Note

== Historical Note on Pair of Ulam Numbers differing by One ==
According to in his of $1997$, the Ulam numbers were investigated up to $20 \, 000$ terms by a , whom this contributor has not been able to identify.
Notwithstanding the scarcity of pairs of Ulam numbers which differ by $1$, more than $60\%$ of pairs differ by $2$.
== Sources ==
* : $47$
Category:Historical Notes
Pairs of Consecutive Integers with 6 Divisors

== Theorem ==
The following sequence of integers are those $n$ which fulfil the equation:
:$\map {\sigma_0} n = \map {\sigma_0} {n + 1} = 6$
where $\map {\sigma_0} n$ denotes the divisor count function.
That is, they are the first of pairs of consecutive integers which each have $6$ divisors:
:$44, 75, 98, 116, 147, 171, 242, 243, 244, 332, \ldots$
== Proof ==
From Divisor Count Function from Prime Decomposition:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where:
:$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
:$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.
== Sources ==
* : $44$
Category:Divisor Count Function
Pairs of Integers whose Product with Divisor Count are Equal

== Theorem ==
Let $\map {\sigma_0} n$ denote the divisor count function: the number of divisors of $n$.
The following pairs of integers $T$ have the property that $m \, \map {\sigma_0} m$ is equal for each $m \in T$:
:$\set {18, 27}$
:$\set {24, 32}$
:$\set {56, 64}$
== Proof ==
== Sources ==
* 
* : $168$
Category:Divisor Count Function
Pairwise Disjoint Sets Partition Any Sets in Semiring

== Lemma ==
Let $\mathbb S$ be a semiring of sets.
Let $A_1, A_2, \ldots, A_n$ all belong to $\mathbb S$.
Then there exists a finite system of pairwise disjoint sets:
:$B_1, B_2, \ldots, B_t \in \mathbb S$
such that every $A_k$ where $k \in \closedint 1 n$ has a finite expansion:
:$\ds A_k = \bigcap_{s \mathop \in M_k} B_s$
where $M_k \subseteq \closedint 1 s$.
That is, the nature of a semiring of sets $\mathbb S$ is such that for every collection $\mathbb T$ of elements of $\mathbb S$, there exists a finite collection of disjoint sets of $\mathbb S$ from which you can pick sets from that will build any of the elements of $\mathbb T$.
== Proof ==
If $n = 1$, the lemma follows trivially from the definition of semiring of sets.
Now we suppose that the lemma holds for $n = m$, and we attempt to show it consequently holds for $n = m + 1$.
We consider a system of sets $A_1, A_2, \ldots, A_m, A_{m + 1}$ in $\mathbb S$.
Let $B_1, B_2, \ldots, B_t \in \mathbb S$ satisfy the terms of the lemma for $A_1, A_2, \ldots, A_m$.
Now let $B_{s 1} = A_{m + 1} \cap B_s$.
By Pairwise Disjoint Subsets in Semiring Part of Partition, there exists an expansion:
:$\ds A_{m + 1} = \paren {\bigcup_{s \mathop = 1}^t B_{s 1} } \cup \paren {\bigcup_{p \mathop = 1}^q B_p'}$ where $B_p' \in \mathbb S$.
By definition of a semiring of sets, there is an expansion:
:$B_s = B_{s 1} \cup B_{s 2} \cup \cdots \cup B_{s r_s}$ where $B_{s j} \in \mathbb S$.
It can be seen that:
:$\ds A_k = \bigcup_{s \mathop \in M_k} \paren {\bigcup_{j \mathop = 1}^{r_s}B_{sj} }$ for $k \in \closedint 1 m$
for some $M_k$.
Also, the sets $B_{s j}, B_p'$ are pairwise disjoint.
So, the sets $B_{s j}, B_p'$ satisfy the conditions of the lemma with respect to the sets $A_1, A_2, \ldots, A_m, A_{m + 1}$.
Hence the result by induction.
Category:Semirings of Sets
Pairwise Disjoint Subsets in Semiring Part of Partition

== Lemma ==
Let $\mathbb S$ be a semiring of sets.
Let $A, A_1, A_2, \ldots, A_n$ all belong to $\mathbb S$.
Let $A_1, A_2, \ldots, A_n$ all be pairwise disjoint subsets of $A$.
Then there exists a finite expansion of $A$:
:$\ds \exists s \ge n: A = \bigcup_{k \mathop = 1}^s A_k$
with $A_1, \ldots, A_n$ as its first $n$ elements, such that:
:$(1): \quad \forall k, 1 \le k \le s: A_k \in \mathbb S$
:$(2): \quad \forall k, l, 1 \le k \le s, 1 \le l \le s: k \ne l \implies A_k \cap A_l = \O$.
That is, the nature of a semiring is such that every collection of pairwise disjoint subsets of a given set $A$ of that semiring is part of a larger collection of pairwise disjoint subsets of $A$ which forms a complete partition of $A$.
== Proof ==
=== Basis for the Induction ===
By definition of semiring of sets, the lemma holds for $n = 1$.
=== Induction Hypothesis ===
Now we suppose that the lemma holds for $n = m$, and we attempt to show it consequently holds for $n = m + 1$.
=== Induction Step ===
So, let $A_1, A_2, \ldots, A_m, A_{m + 1}$ all be pairwise disjoint subsets of $A$.
By hypothesis:
:$A = A_1 \cup A_2 \cup \cdots \cup A_m \cup B_1 \cup \cdots \cup B_p$
where $A_1, A_2, \ldots, A_m, B_1, \ldots, B_p$ are pairwise disjoint subsets of $A$, all belonging to $\mathbb S$.
Let $B_{q 1} = A_{m + 1} \cap B_q$.
By definition of semiring of sets:
:$B_q = B_{q 1} \cup \cdots \cup B_{q r_q}$
where all the $B_{q j}$ are pairwise disjoint subsets of $B_q$, all belonging to $\mathbb S$.
But then we see that:
:$\ds A = A_1 \cup A_2 \cup \cdots \cup A_m \cup A_{m + 1} \cup \bigcup_{q \mathop = 1}^p \paren {\bigcup_{j \mathop = 2}^{r_q} B_{q j} }$
and so the lemma is true for $m + 1$.
The result follows by induction.
Category:Semirings of Sets
Pairwise Independence does not imply Independence

== Theorem ==
Just because all the events in a family of events in a probability space are pairwise independent, it does not mean that the family is independent.
== Proof ==
Consider throwing a fair four-sided die.
This gives us an event space $\Omega = \set {1, 2, 3, 4}$, with each $\omega \in \Omega$ equally likely to occur:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 4$
Consider the set of events:
:$\SS = \set {A, B, C}$
where:
:$A = \set {1, 2}, B = \set {1, 3}, C = \set {1, 4}$
We have that:
:$\map \Pr A = \map \Pr B = \map \Pr C = \dfrac 1 2$
We also have that:
:$\map \Pr {A \cap B} = \map \Pr {A \cap C} = \map \Pr {B \cap C} = \map \Pr {\set 1} = \dfrac 1 4$
Thus:
:$\map \Pr A \map \Pr B = \map \Pr {A \cap B}$
:$\map \Pr A \map \Pr C = \map \Pr {A \cap C}$
:$\map \Pr B \map \Pr C = \map \Pr {B \cap C}$
Thus the events $A, B, C$ are pairwise independent.
Now, consider:
:$\map \Pr {A \cap B \cap C} = \map \Pr {\set 1} = \dfrac 1 4$
But:
:$\map \Pr A \map \Pr B \map \Pr C = \dfrac 1 8 \ne \map \Pr {A \cap B \cap C}$
So, although $\SS$ is pairwise independent, it is not independent.
== Sources ==
* : $\S 1.7$: Independent Events: Example $23$
Category:Independent Events
Palindromes Formed by Multiplying by 55

== Theorem ==
$55$ multiplied by any of the odd integers between $91$ and $109$ inclusive produces a palindromic number.
== Proof ==
We have that:
:$55 \times 91 = 5005$
Then we have that:
:$55 \times 2 = 110$
Thus:
:$\forall k: 1 \le k \le 9: 55 \times 2 k = 110k = 110, 220, 330, \ldots, 990$
Thus for $1 \le k \le 9$:
:$55 \times \paren {91 + 2 k} = \sqbrk {5kk5}_{10}$
That is:
== Sources ==
* : $55$
Category:Recreational Mathematics
Category:55
Palindromes in Base 10 and Base 2

== Theorem ==
The following $n \in \Z$ are palindromic in both decimal and binary:
:$0, 1, 3, 5, 7, 9, 33, 99, 313, 585, 717, 7447, 9009, 15 \, 351, 32 \, 223, 39 \, 993, \ldots$
== Proof ==
:
$n_{10}$ 	$n_2$
$0$ 	$0$
$1$ 	$1$
$3$ 	$11$
$5$ 	$101$
$7$ 	$111$
$9$ 	$1001$
$33$ 	$100 \, 001$
$99$ 	$1 \, 100 \, 011$
$313$ 	$100 \, 111 \, 001$
$585$ 	$1 \, 001 \, 001 \, 001$
$717$ 	$1 \, 011 \, 001 \, 101$
$7447$ 	$1 \, 110 \, 100 \, 010 \, 111$
$15 \, 351$	$11 \, 101 \, 111 \, 110 \, 111$
$32 \, 223$	$111 \, 110 \, 111 \, 011 \, 111$
$39 \, 993$	$1 \, 001 \, 110 \, 000 \, 111 \, 001$
== Sources ==
* 
* 
* : $33$
Category:Palindromic Numbers
Category:2
Category:10
Palindromes in Base 10 and Base 3

== Theorem ==
The following $n \in \Z$ are palindromic in both decimal and ternary:
:$0, 1, 2, 4, 8, 121, 151, 212, 242, 484, 656, 757, \ldots$
== Proof ==
:
$n_{10}$	$n_3$
$0$ 	$0$
$1$ 	$1$
$2$ 	$2$
$4$ 	$11$
$8$ 	$22$
$121$ 	$11 \, 111$
$151$ 	$12 \, 121$
$212$ 	$21 \, 212$
$242$ 	$22 \, 222$
$484$ 	$122 \, 221$
$656$ 	$220 \, 022$
$757$ 	$1 \, 001 \, 001$
== Sources ==
* 
* : $121$
Category:Palindromic Numbers
Category:3
Category:10
Palindromic Cube with Non-Palindromic Root

== Theorem ==
The only known palindromic cube with a root that is not itself palindromic is $10 \, 662 \, 526 \, 601$.
== Proof ==
We have that:
:$10 \, 662 \, 526 \, 601 = 2201^3$
There are no others whose cube root is below $10^{15}$.
== Sources ==
* 
* 
* : $10,662,526,601$
* : $10,662,526,601$
Category:Cube Numbers
Category:Palindromic Numbers
Category:10,662,526,601
Palindromic Primes in Base 10 and Base 2

== Theorem ==
The following $n \in \Z$ are prime numbers which are palindromic in both decimal and binary:
:$3, 5, 7, 313, 7 \, 284 \, 717 \, 174 \, 827, 390 \, 714 \, 505 \, 091 \, 666 \, 190 \, 505 \, 417 \, 093, \ldots$
It is not known whether there are any more.
== Proof ==
:
$n_{10}$ 	$n_2$
$3$ 	$11$
$5$ 	$101$
$7$ 	$111$
$313$ 	$100 \, 111 \, 001$
$7 \, 284 \, 717 \, 174 \, 827$ 	$1 \, 101 \, 010 \, 000 \, 000 \, 011 \, 010 \, 111 \, 110 \, 101 \, 100 \, 000 \, 000 \, 101 \, 011$
$390 \, 714 \, 505 \, 091 \, 666 \, 190 \, 505 \, 417 \, 093$	$10 \, 100 \, 001 \, 100 \, 110 \, 001 \, 000 \, 000 \, 111 \, 100 \, 001 \, 100 \, 011 \, 000 \, 111 \, 011 \, 100 \, 011 \, 000 \, 110 \, 000 \, 111 \, 100 \, 000 \, 010 \, 001 \, 100 \, 110 \, 000 \, 101$
The last two numbers have $43$ and $89$ digits (in binary) respectively .
== Sources ==
* : $313$
Category:Palindromic Numbers
Category:Prime Numbers
Category:2
Category:10
Palindromic Smith Number

== Sequence ==
The sequence of palindromic Smith numbers begins:
:$4, 22, 121, 202, 454, 535, 636, 666, 1111, 1881, 3663, 7227, 7447, 9229, 10 \, 201, 17 \, 271, 22 \, 522, 24 \, 142, \ldots$
== Examples ==
== Sources ==
* 
* : $12,345,554,321$
Category:Smith Numbers
Category:Palindromic Numbers
Palindromic Smith Number/Examples/123,455,554,321

== Example of Palindromic Smith Number ==
$123 \, 455 \, 554 \, 321$ is a palindromic Smith number:
:$123 \, 455 \, 554 \, 321 = 11 \times 41 \times 73 \times 101 \times 137 \times 271$
and:
:$1 + 2 + 3 + 4 + 5 + 5 + 5 + 5 + 4 + 3 + 2 + 1 = 40 = 1 + 1 + 4 + 1 + 7 + 3 + 1 + 0 + 1 + 1 + 3 + 7 + 2 + 7 + 1$
== Sources ==
* 
* : $12,345,554,321$
Category:Smith Numbers
Category:Palindromic Numbers
Category:123,455,554,321
Palindromic Smith Number/Examples/123,455,554,321/Mistake

== Source Work ==
:
:The Dictionary
::$12,345,554,321$
== Mistake ==
:[$12,345,554,321$ is] a palindromic Smith number.
== Correction ==
In fact it is not.
:$12,345,554,321 = 41 \times 239 \times 271 \times 4649$
while:
:$1 + 2 + 3 + 4 + 5 + 5 + 5 + 4 + 3 + 2 + 1 = 35$
but:
:$4 + 1 + 2 + 3 + 9 + 2 + 7 + 1 + 4 + 6 + 4 + 9 = 52$
The mistake can be seen in the article by , where he states:
:There we find palindromic Smith numbers, as $12345554321$, ...
In turn, he is quoting an article in by , which states:
:We find that $R_2^2 = 121, R_5 R_8 = 123455554321, R_4 R_{10}, R_7 R_{12}, R_4 R_{20}, R_5 R_{24}, R_5 R_{31}$, and $R_4 R_{55}$ are palindromic Smith Numbers.
Indeed, $123 \, 455 \, 554 \, 321$ is a palindromic Smith number:
:$1 + 2 + 3 + 4 + 5 + 5 + 5 + 5 + 4 + 3 + 2 + 1 = 40 = 1 + 1 + 4 + 1 + 7 + 3 + 1 + 0 + 1 + 1 + 3 + 7 + 2 + 7 + 1$
as:
:$123 \, 455 \, 554 \, 321 = 11 \times 41 \times 73 \times 101 \times 137 \times 271$
Hence it appears that the transcription error was indeed made by .
== Sources ==
* 
* 
* : $12,345,554,321$
Category:Mistakes/Curious and Interesting Numbers
Category:123,455,554,321
Palindromic Squares with Non-Palindromic Roots

== Theorem ==
The sequence of palindromic squares with non-palindromic square roots begins:
:$676, 69 \, 696, 94 \, 249, 698 \, 896, 5 \, 221 \, 225, 6 \, 948 \, 496, 522 \, 808 \, 225, \ldots$
This sequence is not explicitly given in .
The sequence of those corresponding non-palindromic square roots begins:
:$26, 264, 307, 836, 2285, 2636, 22 \, 865, 24 \, 846, 30 \, 693, \ldots$
== Proof ==
By investigating all square numbers which are palindromic.
and so on.
== Sources ==
* : $676$
* : $676$
Category:Square Numbers
Category:Palindromic Numbers
Palindromic Triangular Numbers

== Sequence of Palindromic Triangular Numbers ==
The sequence of triangular numbers which are also palindromic begins:
:$1, 3, 6, 55, 66, 171, 595, 666, 3003, 5995, 8778, 15 \, 051, 66 \, 066, 617 \, 716, 828 \, 828, 1 \, 269 \, 621, 1 \, 680 \, 861, 3 \, 544 \, 453, \ldots$
Their indices are:
:$1, 2, 3, 10, 11, 18, 34, 36, 77, 109, 132, 173, 363, 1111, 1287, 1593, 1833, 2662, \ldots$
== Historical Note ==
== Sources ==
* 
* : $15$
* : $15$
* : $55$
Category:Triangular Numbers
Category:Palindromic Numbers