avewright/proofwiki-math huggingface dataset:
Theorem: Numbers Appearing 8 Times in Pascal's Triangle

Excluding $1$, the number $3003$ is the smallest integer to appear $8$ times in Pascal's triangle.
No other number below $2^{23}$ appears as often.

Proof:
{{begin-eqn}}
{{eqn | l = 3003
      | m = \frac {3003!} {3002! \times 1!}
      | mo= =
      | r = \dbinom {3003} 1
      | c = 
}}
{{eqn | m = \frac {78!} {76! \times 2!}
      | mo= =
      | r = \dbinom {78} 2
      | c = 
}}
{{eqn | m = \frac {15!} {10! \times 5!}
      | mo= =
      | r = \dbinom {15} 5
      | c = 
}}
{{eqn | m = \frac {14!} {8! \times 6!}
      | mo= =
      | r = \dbinom {14} 6
      | c = 
}}
{{eqn | m = \frac {14!} {6! \times 8!}
      | mo= =
      | r = \dbinom {14} 8
      | c = 
}}
{{eqn | m = \frac {15!} {5! \times 10!}
      | mo= =
      | r = \dbinom {15} {10}
      | c = 
}}
{{eqn | m = \frac {78!} {2! \times 76!}
      | mo= =
      | r = \dbinom {78} {76}
      | c = 
}}
{{eqn | m = \frac {3003!} {1! \times 3002!}
      | mo= =
      | r = \dbinom {3003} {3002}
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that there are no other occurrences of $3003$, and that there are no other numbers less than $2^{23}$ with this property.}}
Theorem: Squares of 3...34

The following pattern holds:
{{begin-eqn}}
{{eqn | l = 4^2
      | r = 16
}}
{{eqn | l = 34^2
      | r = 1156
}}
{{eqn | l = 334^2
      | r = 111 \, 556
}}
{{eqn | l = 3334^2
      | r = 11 \, 115 \, 556
}}
{{eqn | l = 33334^2
      | r = 1 \, 111 \, 155\, 556
}}
{{end-eqn}}
and so on.

Proof:
{{ProofWanted|Simple but tedious.}}
Theorem: 3367 Multiplied by 2-Digit Number

In order to multiply $3367$ by a $2$-digit integer $\sqbrk {xy}$:
:divide the $6$-digit integer $\sqbrk {xyxyxy}$ by $3$.

Proof:
We have that:
:$10101 = 3367 \times 3$
Then:
:$10101 \times \sqbrk {xy} = \sqbrk {xyxyxy}$
The result follows.
{{qed}}
Theorem: Integer as Sum of 2 Cubes in 3 Ways

$728$ is the smallest natural number which can be expressed as the sum of $2$ cubes in $3$ different ways:
{{begin-eqn}}
{{eqn | l = 728
      | r = 6^3 + 8^3
      | c = 
}}
{{eqn | r = \paren {-1}^3 + 9^3
      | c = 
}}
{{eqn | r = \paren {-10}^3 + 12^3
      | c = 
}}
{{end-eqn}}

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 6^3 + 8^3
      | r = 216 + 512
      | c = 
}}
{{eqn | r = 728
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {-1}^3 + 9^3
      | r = -1 + 729
      | c = 
}}
{{eqn | r = 728
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {-10}^3 + 12^3
      | r = -1000 + 1728
      | c = 
}}
{{eqn | r = 728
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown there are no smaller.}}
Category:Sums of Cubes
6i0wm6f2ns52sp0ryzop5qqkvs04c93
Theorem: Composite Fibonacci Numbers with Prime Index

The sequence of composite Fibonacci numbers with a prime index begins:
:$4181, 1 \, 346 \, 269, 24 \, 157 \, 817, 165 \, 580 \, 141, \ldots$
{{OEIS|A050937}}
The corresponding sequence of prime indices begins:
:$19, 31, 37, 41, 53, 59, 61, 67, 71, 73, 79, \ldots$
{{OEIS|A038672}}

Proof:
By observation:
{{begin-eqn}}
{{eqn | l = F_{19}
      | r = 4181
      | c = 
}}
{{eqn | r = 37 \times 113
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = F_{31}
      | r = 1 \, 346 \, 269
      | c = 
}}
{{eqn | r = 557 \times 2417
      | c = 
}}
{{end-eqn}}
{{finish}}
Theorem: Closed Subset is Upper Section in Lower Topology

Let $T = \struct {S, \preceq, \tau}$ be a transitive relational structure with lower topology.
Let $A \subseteq S$ such that
:$A$ is closed.
Then $A$ is an upper section of $S$.

Proof:
By definition of closed set:
:$S \setminus A$ is open.
By Open Subset is Lower Section in Lower Topology:
:$S \setminus A$ is a lower section.
Thus by Complement of Lower Section is Upper Section and Relative Complement of Relative Complement:
:$A$ is an upper section of $S$.
{{qed}}
Theorem: 5040 is Product of Consecutive Numbers in Two Ways

:$5040 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8 \times 7$

Proof:
Follows from Factorial as Product of Two Factorials:
:$10! = 6! \times 7!$
and so:
:$\dfrac {10!} {6!} = 10 \times 9 \times 8 \times 7 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
Hence the result.
{{qed}}
Theorem: Products of Consecutive Integers in 2 Ways

The following integers are the product of consecutive integers in $2$ ways:
:$-720, 720, 5040$

Proof:
From 720 is Product of Consecutive Numbers in Two Ways:
:$720 = 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8$
From 5040 is Product of Consecutive Numbers in Two Ways:
:$5040 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8 \times 7$
Then:
:$-720 = \left({-6}\right) \left({-5}\right) \left({-4}\right) \left({-3}\right) \left({-2}\right) = \left({-10}\right) \left({-9}\right) \left({-8}\right)$
The same trick cannot be used for $-5040$ because there are four divisors in $10 \times 9 \times 8 \times 7$, and negating them makes the product positive.
Hence the result.
{{qed}}
Category:Factorials
htuoyjqsfoqjoxcauiwslkjcxp6eitx
Theorem: 8 Mutually Non-Attacking Rooks on Chessboard

On a standard chessboard, it is possible to arrange a maximum of $8$ rooks so that no rook is attacking any other rook.
There are $5282$ such arrangements, up to rotation and reflection.

Proof:
{{ProofWanted|No doubt we will eventually progress to chess problems of various styles.}}
Theorem: Triangular Lucas Numbers

The only Lucas numbers which are also triangular are:
:$1, 3, 5778$
{{OEIS|A248506}}

Proof:
{{begin-eqn}}
{{eqn | l = 1
      | r = \dfrac {1 \times 2} 2
}}
{{eqn | l = 3
      | r = \dfrac {2 \times 3} 2
      | rr= = 2 + 1
}}
{{eqn | l = 5778
      | r = \dfrac {107 \times 108} 2
      | rr= = 2207 + 3571
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that these are the only ones.}}
Theorem: Sum of Two Rational 4th Powers but not Two Integer 4th Powers

$5906$ is the smallest integer which can be expressed as the sum of two rational $4$th powers, but not two integer $4$th powers.

Proof:
:$5906 = \paren {\dfrac {149} {17} }^4 + \paren {\dfrac {25} {17} }^4$
Suppose $5906$ is a sum of two integer $4$th powers.
We have:
:$9^4 = 6561 > 5906$
which shows that no $4$th power greater than $8^4$ is in the sum.
:$7^4 + 7^4 = 4802 < 5906$
which shows that some $4$th power greater than $7^4$ is in the sum.
So the sum must contain $8^4$.
We have:
:$5906 - 8^4 = 1810$
but $1810$ is not an integer $4$th power.
Therefore $5906$ is not a sum of two integer $4$th powers.
{{finish|It remains to show that it is the smallest such. It is, acccording to A NEW CHARACTERIZATION OF THE INTEGER 5906 by A. Bremner and P. Morton}}
Theorem: Subspace of Subspace is Subspace

Let $T = \struct{S, \tau}$ be a topological space.
Let $H \subseteq S$ be a non-empty subset of $S$ and $\tau_H$ be the subspace topology on $H$.
Let $K \subseteq S$ be a non-empty subset of $S$.
Then the subspace topology on $K$ induced by $\tau$ equals the subspace topology on $K$ induced by $\tau_H$.
{{explain|Why is it necessary to state that $H$ and $K$ are non-empty?}}

Proof:
Let $\tau_K$ be the subspace topology on $K$ induced by $\tau$.
Let $\tau'_K$ be the subspace topology on $K$ induced by $\tau_H$.
Then
{{begin-eqn}}
{{eqn | l = V \in \tau'_K
      | o = \leadstoandfrom
      | r = \exists U' \in \tau_H : V = U' \cap K 
      | c = {{Defof|Subspace Topology}} $\tau'_K$
}}
{{eqn | o = \leadstoandfrom
      | r = \exists U \in \tau : V = \paren {U \cap H} \cap K 
      | c = {{Defof|Subspace Topology}} $\tau_H$
}}
{{eqn | o = \leadstoandfrom
      | r = \exists U \in \tau : V = U \cap \paren {H \cap K} 
      | c = Intersection is Associative
}}
{{eqn | o = \leadstoandfrom
      | r = \exists U \in \tau : V = U \cap K 
      | c = Intersection with Subset is Subset
}}
{{eqn | o = \leadstoandfrom
      | r = V \in \tau_K 
      | c = {{Defof|Subspace Topology}} $\tau_K$
}}
{{end-eqn}}
{{qed}}
Theorem: Positive Integer Sum of 3 Fourth Powers in 2 Ways

The smallest positive integer which can be expressed as the sum of $3$ fourth powers in $2$ different ways is $6578$:
{{begin-eqn}}
{{eqn | l = 1^4 + 2^4 + 9^4
      | r = 1 + 16 + 6561
      | c = 
}}
{{eqn | r = 6578
      | c = 
}}
{{eqn | r = 81 + 2401 + 4096
      | c = 
}}
{{eqn | r = 3^4 + 7^4 + 8^4
      | c = 
}}
{{end-eqn}}

Proof:
The fact that this is the smallest can be demonstrated by calculation.
{{qed}}
Theorem: Mapping Preserves Infima implies Mapping is Continuous in Lower Topological Lattice

Let $T = \struct {S, \preceq, \tau}$ and $Q = \struct {X, \preceq', \tau'}$ be complete topological lattices with lower topologies.
Let $f: S \to X$ be a mapping such that
:$f$ preserves all infima.
Then $f$ is continuous mapping.

Proof:
By assumption:
:for all non-empty subsets $Y$ of $S$: $f$ preserves the infimum of $Y$.
Thus by Mapping Preserves Non-Empty Infima implies Mapping is Continuous in Lower Topological Lattice:
:$f$ is continuous mapping.
{{qed}}
Theorem: Compact in Subspace is Compact in Topological Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $K \subseteq S$ be a subset.
Let $\tau_K$ be the subspace topology on $K$.
Let $T' = \struct {K, \tau_K}$ be the topological subspace of $T$ determined by $K$.
Let $H \subseteq K$ be compact in $T'$.
Then $H$ is compact in $T$.

Proof:
Suppose that $H$ is compact in $T'$.
Let $\set {W_i}_{i \mathop \in J}$ be an open cover of $H$ in $T$.
Then $\ds H \subseteq \bigcup_{i \mathop \in J} W_i$.
Then: 
{{begin-eqn}}
{{eqn | l = H
      | r = H \cap K
      | c = Intersection with Subset is Subset: from $H \subseteq K$
}}
{{eqn | o = \subseteq
      | r = \paren {\bigcup_{i \mathop \in J} W_i} \cap K
      | c = Set Intersection Preserves Subsets: Corollary: from $\ds H \subseteq \bigcup_{i \mathop \in J} W_i$
}}
{{eqn | r = \bigcup_{i \mathop \in J} \paren {W_i \cap K}
      | c = Intersection Distributes over Union of Family
}}
{{end-eqn}}
Since $W_i \cap K \in \tau_K$, $\set {W_i \cap K}_{i \mathop \in J}$ is an open cover of $H$ in $T'$.
Since $H$ is compact in $T'$, $\set {W_i \cap K}_{i \mathop \in J}$ has some finite subcover $\set {W_i \cap K}_{i \mathop = 1}^r$.
Therefore:
{{begin-eqn}}
{{eqn | l = H
      | o = \subseteq
      | r = \set {W_i \cap K}_{i \mathop = 1}^r
      | c = {{Defof|Finite Subcover}}
}}
{{eqn | o = \subseteq
      | r = \paren {\bigcup_{i \mathop = 1}^r W_i} \cap K
      | c = Intersection Distributes over Union of Family
}}
{{eqn | r = \bigcup_{i \mathop = 1}^r W_i
      | c = Intersection is Subset
}}
{{end-eqn}}
So $\set {W_i}_{i \mathop = 1}^r$ is an open cover of $H$ in $T$, which is a finite subcover.
As $\set {W_i}_{i \mathop \in J}$ is arbitrary:
Any open cover of $H$ has a finite subcover in $T$.
So $H$ is compact in $T$.
{{qed}}
Category:Compact Topological Spaces
Category:Topological Subspaces
efu10c8o6wu82oq57uduptxel9nwomb
Theorem: If Infimum of Filtered Subset belongs to Element of Sub-Basis then Subset and Element Intersect implies Infimum of Subset belongs to Closure of Subset

Let $T = \struct {S, \preceq, \tau}$ be a complete topological lattice with lower topology.
Let $B$ be an analytic sub-basis of $T$.
Let $F$ be a filtered subset of $S$ such that
:$\forall A \in B: \inf F \in A \implies F \cap A \ne \O$
Then $\inf F \in F^-$
where $F^-$ denotes the topological closure of $F$.

Proof:
We will prove that
:$\forall A \in B, x \in F \cap A, y \in F: y \preceq x \implies y \in A$
Let $A \in B$, $x \in F \cap A$, $y \in F$.
By definition of sub-basis:
:$A$ is open.
By Open Subset is Lower Section in Lower Topology:
:$A$ is a lower section.
By definition of intersection:
:$x \in A$.
Thus by definition of lower section:
:$y \preceq x \implies y \in A$.
{{qed|lemma}}
Define $\ds H := \set {\bigcap G: G \subseteq B, G \text { is finite} }$
By definitions of sub-basis and basis:
:$H$ is basis of $T$.
We will prove that
:$\forall A \in H: \inf F \in A \implies F \cap A \ne \O$
Let $A \in H$ such that
:$\inf F \in A$
By definition of $H$:
:$\exists G \subseteq B: A = \bigcap G \land G$ is finite.
By definition of intersection:
:$\forall C \in G: \inf F \in C \in B$
By assumption:
:$\forall C \in G: F \cap C \ne \O$
By definition of non-empty set:
:$\forall C \in G: \exists x: x \in F \cap C$
By Axiom of Choice:
:$\exists f: G \to F: \forall C \in G: \map f C \in F \cap C$
By Image of Mapping from Finite Set is Finite:
:$f \sqbrk G$ is finite.
By Filtered iff Finite Subsets have Lower Bounds:
:$\exists h \in F: \forall x \in f \sqbrk G: h \preceq x$
Then
:$\forall C \in G: h \in C$
By definition of intersection:
:$h \in \bigcap G$
Thus by definitions of intersection and non-empty set:
:$F \cap A \ne \O$
{{qed|lemma}}
Thus by Characterization of Closure by Basis:
:$\inf F \in F^-$
{{qed}}
Theorem: Square of Repdigit Number consisting of Instances of 6

The following pattern holds:
{{begin-eqn}}
{{eqn | l = 6^2
      | r = 36
}}
{{eqn | l = 3 + 6
      | r = 9
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 66^2
      | r = 4356
}}
{{eqn | l = 43 + 56
      | r = 99
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 666^2
      | r = 443 \, 556
}}
{{eqn | l = 443 + 556
      | r = 999
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 6666^2
      | r = 44 \, 435 \, 556
}}
{{eqn | l = 4443 + 5556
      | r = 9999
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 66666^2
      | r = 4 \, 444 \, 355\, 556
}}
{{eqn | l = 44 \, 443 + 55 \, 556
      | r = 99 \, 999
}}
{{end-eqn}}
and so on.

Proof:
{{ProofWanted|Simple but tedious.}}
Theorem: Square of Repdigit Number consisting of Instances of 3

The following pattern holds:
{{begin-eqn}}
{{eqn | l = 3^2
      | r = 09
}}
{{eqn | l = 0 + 9
      | r = 9
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 33^2
      | r = 1089
}}
{{eqn | l = 10 + 89
      | r = 99
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 333^2
      | r = 110 \, 889
}}
{{eqn | l = 110 + 889
      | r = 999
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 3333^2
      | r = 11 \, 108 \, 889
}}
{{eqn | l = 1110 + 8889
      | r = 9999
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 33 \, 333^2
      | r = 1 \, 111 \, 088 \, 889
}}
{{eqn | l = 11 \, 110 + 88 \, 889
      | r = 99 \, 999
}}
{{end-eqn}}
and so on.

Proof:
{{ProofWanted|Simple but tedious.}}
Theorem: Product with Repdigit can be Split into Parts which Add to Repdigit

Let $n$ be a positive integer with $d_1$ digits.
Let $m$ be a repdigit number with $d_2$ digits such that $d_2 > d_1$.
Let $r$ consist of the result when the rightmost $d_2$ digits of $m n$ is cut off and added to the remaining left hand portion.
Then $r$ is a repdigit number.

Proof:
Let $b > 1$ be the base we are working on.
Let $m = \sqbrk {aaa \dots a}_b$.
Let $R = \dfrac m a = \sqbrk {111 \dots 1}_b = \dfrac {b^{d_2} - 1} {b - 1}$.
Let the rightmost $d_2$ digits of $m n$ be $y$ and the remaining left hand portion be $x$.
Then we have:
{{begin-eqn}}
{{eqn | l = 0
      | o = \equiv
      | r = m n
      | rr= \pmod R
}}
{{eqn | o = \equiv
      | r = b^{d_2} x + y
      | rr= \pmod R
}}
{{eqn | o = \equiv
      | r = \paren {\paren {b - 1} R + 1} x + y
      | rr= \pmod R
}}
{{eqn | o = \equiv
      | r = x + y
      | rr= \pmod R
}}
{{eqn | o = \equiv
      | r = r
      | rr= \pmod R
}}
{{end-eqn}}
so $r$ is divisible by $R$.
It remains to show that:
:$r \le b^{d_2} - 1 = \paren {b - 1} R$
because $\paren {b - 1} R$ is the largest $d_2$-digit integer.
This condition forces $r$ to have $d_2$ digits, so it remains a repdigit number.
Since $r$ is divisible by $R$, this is equivalent to:
:$r < b R$
We have:
{{begin-eqn}}
{{eqn | l = r
      | r = x + y
}}
{{eqn | o = <
      | r = b^{d_1} + \paren {b - 1} R
      | c =
}}
{{eqn | o = \le
      | r = b^{d_2 - 1} + \paren {b - 1} R
      | c = because $d_2 > d_1$
}}
{{eqn | o = \le
      | r = R + \paren {b - 1} R
      | c = because $R$ has $d_2$ digits
}}
{{eqn | r = b R
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Sum of Reciprocals of Squares of Odd Integers as Double Integral

:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}
      | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1} \paren {2 n - 1} }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \int_0^1 x^{2 n - 2} \rd x \int_0^1 y^{2 n - 2} \rd y
      | c = Primitive of Power
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \int_0^1 \int_0^1 \paren {x^2 y^2}^{n - 1} \rd x \rd y
      | c = Product of Powers, Power of Product
}}
{{eqn | r = \int_0^1 \int_0^1 \sum_{n \mathop = 0}^\infty \paren {x^2 y^2}^n \rd x \rd y
      | c = Fubini's Theorem, reindexing the sum
}}
{{eqn | r = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y
      | c = Sum of Infinite Geometric Sequence
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Reciprocals of Squares of Odd Integers as Double Integral

:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y$

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y
      | r = \int_0^1 \int_0^1 \paren {1 + x^2 y^2 + x^4 y^4 + x^6 y^6 + \cdots} \rd x \rd y
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \int_0^1 \intlimits {y + \frac {x^2y^3} 3 + \frac {x^4y^5} 5 + \frac {x^6y^7} 7 + \cdots } {y \mathop = 0} {y \mathop = 1} \rd x
}}
{{eqn | r = \int_0^1 \paren {1 + \frac 1 3 x^2 + \frac 1 5 x^4 + \frac 1 7 x^6 + \cdots} \rd x
}}
{{eqn | r = \intlimits {x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots } {x \mathop = 0} {x \mathop = 1}
}}
{{eqn | r = \paren {1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Three Tri-Automorphic Numbers for each Number of Digits

Let $d \in \Z_{>0}$ be a (strictly) positive integer.
Then there exist exactly $3$ tri-automorphic numbers with exactly $d$ digits.
These tri-automorphic numbers all end in $2$, $5$ or $7$.

Proof:
Let $n$ be a tri-automorphic number with $d$ digits.
Let $n = 10 a + b$.
Then:
:$3 n^2 = 300a^2 + 60 a b + 3 b^2$
As $n$ is tri-automorphic, we have:
:$(1): \quad 300 a^2 + 60 a b + 3 b^2 = 1000 z + 100 y + 10 a + b$
and:
:$(2): \quad 3 b^2 - b = 10 x$
where $x$ is an integer.
This condition is only satisfied by $b = 2$, $b = 5$, or $b = 7$
{{ProofWanted|Guess: Try proving for $n {{=}} 10 a + b$ and then by induction.}}
Substituting $b = 2$ in equation $(1)$:
:$a = 9$
Substituting $b = 5$ in equation $(1)$:
:$a = 7$
Substituting $b = 7$ in equation $(1)$:
:$a = 6$
{{qed}}
<!--
Corollary: 
For di-automorphic numbers, equation (5) becomes
2b^2-b=10x and b=8 
For tetra-automorphic numbers, equation (5) becomes
4b^2-b=10x and b=4 
For penta-automorphic numbers, equation (5) becomes
5b^2-b=10x and b=5
-->
Theorem: Fourier Series/x squared over Minus Pi to Pi

:$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$

Proof:
From Even Power is Even Function, $x^2$ is an even function.
By Fourier Series for Even Function over Symmetric Range, we have:
:$\ds x^2 \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac 2 \pi \int_0^\pi x^2 \map \cos {n x} \rd x
}}
{{eqn | ll= \leadsto
      | l = \frac 1 2 a_0
      | r = \frac 2 {2 \pi} \int_0^\pi x^2 \rd x
      | c = Cosine of Zero is One
}}
{{eqn | r = \frac 1 \pi \cdot \frac {\pi^3} 3
      | c = Primitive of Power, Fundamental Theorem of Calculus
}}
{{eqn | r = \frac {\pi^2} 3
}}
{{end-eqn}}
Then: 
{{begin-eqn}}
{{eqn | l = \frac 2 \pi \int_0^\pi x^2 \map \cos {n x} \rd x
      | r = \frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} + \paren {\frac {x^2} n - \frac 2 {n^3} } \sin n x} 0 \pi
      | c = Primitive of $x^2 \cos a x$, Fundamental Theorem of Calculus
}}
{{eqn | r = \frac 2 \pi \intlimits {\frac {2 x \cos n x} {n^2} } 0 \pi
      | c = Sine of Multiple of Pi
}}
{{eqn | r = \frac 2 \pi \cdot \frac {2 \pi \cos n \pi} {n^2}
}}
{{eqn | r = \frac {4 \cos n \pi} {n^2}
}}
{{eqn | r = \paren {-1}^n \frac 4 {n^2}
      | c = Cosine of Multiple of Pi
}}
{{end-eqn}}
Substituting for $a_n$ in $(1)$: 
:$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$
as required.
{{qed}}
Theorem: Integer and its Double forming Pandigital Pair

$6729$ and its double contain all the digits from $1$ to $9$ between them.

Proof:
$2 \times 6729 = 13 \, 458$
{{qed}}
Theorem: 4-Digit Numbers forming Longest Reverse-and-Add Sequence

Let $m \in \Z_{>0}$ be a positive integer expressed in decimal notation.
Let $r \left({m}\right)$ be the reverse-and-add process on $m$.
Let $r$ be applied iteratively to $m$.
The $4$-digit integers $m$ which need the largest number of iterations before reaching a palindromic number are:
:$6999, 7998, 8997, 9996$
all of which need $20$ iterations.

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | o = 
      | m = 6999 + 9996
      | mo= =
      | r = 16995
}}
{{eqn | n = 2
      | o = 
      | m = 16995 + 59961
      | mo= =
      | r = 76956
}}
{{eqn | n = 3
      | o = 
      | m = 76956 + 65967
      | mo= =
      | r = 142923
}}
{{eqn | n = 4
      | o = 
      | m = 142923 + 329241
      | mo= =
      | r = 472164
}}
{{eqn | n = 5
      | o = 
      | m = 472164 + 461274
      | mo= =
      | r = 933438
}}
{{eqn | n = 6
      | o = 
      | m = 933438 + 834339
      | mo= =
      | r = 1767777
}}
{{eqn | n = 7
      | o = 
      | m = 1767777 + 7777671
      | mo= =
      | r = 9545448
}}
{{eqn | n = 8
      | o = 
      | m = 9545448 + 8445459
      | mo= =
      | r = 17990907
}}
{{eqn | n = 9
      | o = 
      | m = 17990907 + 70909971
      | mo= =
      | r = 88900878
}}
{{eqn | n = 10
      | o = 
      | m = 88900878 + 87800988
      | mo= =
      | r = 176701866
}}
{{eqn | n = 11
      | o = 
      | m = 176701866 + 668107671
      | mo= =
      | r = 844809537
}}
{{eqn | n = 12
      | o = 
      | m = 844809537 + 735908448
      | mo= =
      | r = 1580717985
}}
{{eqn | n = 13
      | o = 
      | m = 1580717985 + 5897170851
      | mo= =
      | r = 7477888836
}}
{{eqn | n = 14
      | o = 
      | m = 7477888836 + 6388887747
      | mo= =
      | r = 13866776583
}}
{{eqn | n = 15
      | o = 
      | m = 13866776583 + 38567766831
      | mo= =
      | r = 52434543414
}}
{{eqn | n = 16
      | o = 
      | m = 52434543414 + 41434543425
      | mo= =
      | r = 93869086839
}}
{{eqn | n = 17
      | o = 
      | m = 93869086839 + 93868096839
      | mo= =
      | r = 187737183678
}}
{{eqn | n = 18
      | o = 
      | m = 187737183678 + 876381737781
      | mo= =
      | r = 1064118921459
}}
{{eqn | n = 19
      | o = 
      | m = 1064118921459 + 9541298114601
      | mo= =
      | r = 10605417036060
}}
{{eqn | n = 20
      | o = 
      | m = 10605417036060 + 06063071450601
      | mo= =
      | r = 16668488486661
}}
{{end-eqn}}
which is palindromic.
$7998$ and its reversal converge on the same sequence immediately:
{{begin-eqn}}
{{eqn | l = 7998 + 8997
      | r = 16995
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that these are the longest.}}
Theorem: Fourier Series/Identity Function over Minus Pi to Pi

For $x \in \openint {-\pi} \pi$:
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$

Proof:
From Odd Power is Odd Function, $x$ is an odd function.
By Fourier Series for Odd Function over Symmetric Range, we have:
:$\ds x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where:
{{begin-eqn}}
{{eqn | l = b_n
      | r = \frac 2 \pi \int_0^\pi x \sin n x \rd x
      | c = 
}}
{{eqn | r = \frac 2 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi
      | c = Primitive of $x \sin n x$, Fundamental Theorem of Calculus
}}
{{eqn | r = -\frac 2 \pi \intlimits {\frac{x \cos n x} n} 0 \pi
      | c = Sine of Multiple of Pi
}}
{{eqn | r = -\frac 2 \pi \paren {\frac{\pi \cos n \pi} n - \frac {0 \cos 0} n}
      | c = 
}}
{{eqn | r = -\frac {2 \pi \cos n \pi} {\pi n}
}}
{{eqn | r = -\frac {2 \paren {-1}^n} n
      | c = Cosine of Multiple of Pi
}}
{{eqn | r = \frac {2 \paren {-1}^{n + 1} } n
}}
{{end-eqn}}
Substituting for $b_n$ in $(1)$: 
:$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
as required. 
{{qed}}