avewright/proofwiki-math huggingface dataset:
Theorem: Magic Hexagon of Order 3 is Unique

Apart from the trivial order $1$ magic hexagon, there exists only one magic hexagon: the order $3$ magic hexagon:
{{:Order 3 Magic Hexagon}}

Proof:
We first prove that only order $3$ magic hexagons exist apart from the order $1$ magic hexagon, by calculating the magic constant for each order.
If we ignore the center tile, the rest of the hexagon can be divided into $3$ parallelogram-like parts:
For example, the above can be divided into $\set {3, 17, 18, 19, 7, 1}$, $\set {11, 6, 8, 9, 14, 15}$ and $\set {2, 4, 13, 16, 12, 10}$.
Each part has $n \paren {n - 1}$ tiles.
Thus the order $n$ hexagon has $3 n \paren {n - 1} + 1$ tiles.
Since we require each row to have the same sum, and there are $2 n - 1$ rows:
:The sum of numbers in the hexagon $= \paren {2 n - 1} \times$ magic constant
By Sum of Arithmetic Sequence, the sum of numbers in the tiles is:
:$\dfrac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2}} 2$
Hence the magic constant is:
{{begin-eqn}}
{{eqn | o =
      | r = \frac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2} } {2 \paren {2 n - 1} }
}}
{{eqn | r = \frac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1} }
}}
{{end-eqn}}
Since the magic constant is a sum of integers, it must be an integer.
Then:
:$32 \paren {\dfrac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1}}} = \dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1}$ is also an integer.
Since:
:$\dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1} = 72 n^3 - 108 n^2 + 90 n - 27 + \dfrac 5 {2 n - 1}$
We must require $\dfrac 5 {2 n - 1}$ to be an integer.
This happens only if $n = 1$ or $n = 3$.
Thus there are no magic hexagons that are not of order $1$ or $3$.
It remains to show that the order $3$ magic hexagon is unique.
{{ProofWanted}}
Theorem: Magic Constant of Order 3 Magic Hexagon

The magic constant of the order 3 magic hexagon is $38$.
It is noted that the central cell is $5$, the same as that of the order 3 magic square.

Proof:
{{ProofWanted|Plenty background work needed}}
Theorem: Prime to Own Power minus 1 over Prime minus 1 being Prime

Let $n \in \Z_{>1}$ be an integer greater than $1$.
Then $\dfrac {n^n - 1} {n - 1}$ is a prime for $n$ equal to:
:$2, 3, 19, 31$
{{OEIS|A088790}}

Proof:
{{ProofWanted|Can easily be proved that $n$ must itself be prime for the expression to be prime. Then it's a matter of checking them all.}}
Note that if $4 p + 1$ is prime for prime $p$, then $\dfrac {p^p - 1} {p - 1}$ is divisible by $4 p + 1$:
Let $q = 4 p + 1$ be prime.
By First Supplement to Law of Quadratic Reciprocity:
:$\paren {\dfrac {-1} q} = 1$
that is, there exists some integer $I$ such that $I^2 \equiv -1 \pmod q$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {1 + I}^4
      | r = \paren {1 + 2 I + I^2}^2
      | c = Square of Sum
}}
{{eqn | o = \equiv
      | r = \paren {2 I}^2
      | rr= \pmod q
      | c = Congruence of Powers
}}
{{eqn | o = \equiv
      | r = -4
      | rr= \pmod q
}}
{{end-eqn}}
and thus:
{{begin-eqn}}
{{eqn | l = p^p
      | o = \equiv
      | r = p^p \paren {1 + I}^{q - 1}
      | rr= \pmod q
      | c = Fermat's Little Theorem
}}
{{eqn | o = \equiv
      | r = p^p \paren {1 + I}^{4 p}
      | rr= \pmod q
}}
{{eqn | o = \equiv
      | r = p^p \paren {-4}^p
      | rr= \pmod q
}}
{{eqn | o = \equiv
      | r = \paren {-4 p}^p
      | rr= \pmod q
}}
{{eqn | o = \equiv
      | r = 1^p
      | rr= \pmod q
}}
{{eqn | o = \equiv
      | r = 1
      | rr= \pmod q
}}
{{end-eqn}}
Hence $q \divides \paren {p^p - 1}$.
Obviously $q > p - 1$.
Therefore $q \divides \dfrac {p^p - 1} {p - 1}$.
Theorem: Squares Ending in 5 Occurrences of 2-Digit Pattern

Let $n$ be a square number whose decimal representation ends in the pattern $\mathtt {xyxyxyxyxy}$.
Then $\mathtt {xy}$ is one of:
:$21, 29, 61, 69, 84$
The smallest examples of such numbers are:
{{begin-eqn}}
{{eqn | l = 508 \, 853 \, 989^2
      | r = \phantom {0 \,} 258 \, 932 \, 38 \mathbf {2 \, 121 \, 212 \, 121}
}}
{{eqn | l = 162 \, 459 \, 327^2
      | r = \phantom {0 \, 0} 26 \, 393 \, 03 \mathbf {2 \, 929 \, 292 \, 929}
}}
{{eqn | l = 1 \, 318 \, 820 \, 881^2
      | r = 1 \, 739 \, 288 \, 51 \mathbf {6 \, 161 \, 616 \, 161}
}}
{{eqn | l = 541 \, 713 \, 187^2
      | r = \phantom {0 \,} 293 \, 453 \, 17 \mathbf {6 \, 969 \, 696 \, 969}
}}
{{eqn | l = 509 \, 895 \, 478^2
      | r = \phantom {0 \,} 259 \, 993 \, 39 \mathbf {8 \, 484 \, 848 \, 484}
}}
{{end-eqn}}

Proof:
=== Case $1$: $\mathtt {xy}$ is odd ===
Consider the last $3$ digits of $n$: $\mathtt {yxy}$.
By Odd Square Modulo 8:
:$n \equiv \mathtt {yxy} \equiv 1 \pmod 8$
By Square Modulo 5:
:$n \equiv \mathtt {yxy} \equiv 0, 1, 4 \pmod 5$
We have:
{{begin-eqn}}
{{eqn | l = \mathtt {yxy}
      | o = \equiv
      | r = 25, 1, 9
      | rr = \pmod {40}
      | c = Chinese Remainder Theorem
}}
{{eqn | o = \equiv
      | r = 1, 9, 25, 41, 49, 65, 81, 89, 105, 121, 129, 145, 161, 169, 185
      | rr = \pmod {200}
}}
{{end-eqn}}
Since $y$ is odd, we have $\mathtt {yxy} \equiv 1 \mathtt {xy} \pmod {200}$.
Hence $\mathtt {xy}$ can only be $05, 21, 29, 45, 61, 69, 85$.
Suppose $n$ ends in $5$ in decimal representation.
By Divisibility by 5, $5 \divides n$.
Since $5$ is a prime, $5 \divides \sqrt n$.
By Odd Square Modulo 8 and Chinese Remainder Theorem:
:$n \equiv 25 \pmod {200}$
for which none of $\mathtt {xy} = 05, 45, 85$ satisfy.
This shows that $\mathtt {xy}$ can only be $21, 29, 61, 69$.
{{qed|lemma}}
Theorem: Ideals are Continuous Lattice Subframe of Power Set

Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $I = \paren {\map {\operatorname{Ids} } L, \precsim}$ be an inclusion ordered set
where
:$\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$,
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$
Let $P = \struct {\powerset S, \precsim'}$ be an inclusion ordered set
where
:$\powerset S$ denotes the power set of $S$,
:$\mathord \precsim' = \mathord \subseteq \cap \paren {\powerset S \times \powerset S}$
Then $I$ is continuous lattice subframe of $P$.

Proof:
By definition of subset:
:$\map {\operatorname{Ids} } L \subseteq \powerset S$
Then
:$\mathord \precsim = \mathord \precsim' \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$
Hence $I$ is ordered subset of $P$.
Theorem: Perfect Square Dissection of Lowest Order

The perfect square dissection of lowest order is of an integer square of side $112$ into $21$ parts.

Proof:
:700px
{{ProofWanted|That this is the smallest still needs to be proved.}}
Theorem: Smallest Number Expressible as Sum of at most Three Triangular Numbers in 4 ways

$21$ is the smallest number which can be expressed as the sum of at most $3$ triangular numbers in $4$ ways.

Proof:
By inspection:
{{begin-eqn}}
{{eqn | l = 1
      | r = T_1
      | c = $1$ way
}}
{{eqn | l = 2
      | r = T_1 + T_1
      | c = $1$ way
}}
{{eqn | l = 3
      | r = T_1 + T_1 + T_1
      | c = 
}}
{{eqn | r = T_2
      | c = $2$ ways
}}
{{eqn | l = 4
      | r = T_2 + T_1
      | c = $1$ way
}}
{{eqn | l = 5
      | r = T_2 + T_1 + T_1
      | c = $1$ way
}}
{{eqn | l = 6
      | r = T_2 + T_2
}}
{{eqn | r = T_3
      | c = $2$ ways
}}
{{eqn | l = 7
      | r = T_2 + T_2 + T_1
}}
{{eqn | r = T_3 + T_1
      | c = $2$ ways
}}
{{eqn | l = 8
      | r = T_3 + T_1 + T_1
      | c = $1$ way
}}
{{eqn | l = 9
      | r = T_3 + T_2
}}
{{eqn | r = T_2 + T_2 + T_2
      | c = $2$ ways
}}
{{eqn | l = 10
      | r = T_4
}}
{{eqn | r = T_3 + T_2 + T_1
      | c = $2$ ways
}}
{{eqn | l = 11
      | r = T_4 + T_1
      | c = $1$ way
}}
{{eqn | l = 12
      | r = T_4 + T_1 + T_1
}}
{{eqn | r = T_3 + T_3
      | c = 
}}
{{eqn | r = T_3 + T_2 + T_2
      | c = $3$ ways
}}
{{eqn | l = 13
      | r = T_4 + T_2
      | c = 
}}
{{eqn | r = T_3 + T_3 + T_1
      | c = $2$ ways
}}
{{eqn | l = 14
      | r = T_4 + T_2 + T_1
      | c = $1$ way
}}
{{eqn | l = 15
      | r = T_5
}}
{{eqn | r = T_3 + T_3 + T_2
      | c = $2$ ways
}}
{{eqn | l = 16
      | r = T_5 + T_1
}}
{{eqn | r = T_4 + T_3
      | c = 
}}
{{eqn | r = T_4 + T_2 + T_2
      | c = $3$ ways
}}
{{eqn | l = 17
      | r = T_5 + T_1 + T_1
}}
{{eqn | r = T_4 + T_3 + T_1
      | c = $2$ ways
}}
{{eqn | l = 18
      | r = T_5 + T_2
}}
{{eqn | r = T_3 + T_3 + T_3
      | c = $2$ ways
}}
{{eqn | l = 19
      | r = T_5 + T_2 + T_1
}}
{{eqn | r = T_4 + T_3 + T_2
      | c = $2$ ways
}}
{{eqn | l = 20
      | r = T_4 + T_4
      | c = $1$ way
}}
{{eqn | l = 21
      | r = T_6
}}
{{eqn | r = T_5 + T_3
      | c = 
}}
{{eqn | r = T_5 + T_2 + T_2
      | c = 
}}
{{eqn | r = T_4 + T_4 + T_1
      | c = $4$ ways
}}
{{end-eqn}}
{{OEIS|A002636}}{{qed}}
Theorem: Necessary Condition for Twice Differentiable Functional to have Minimum

Let $J \sqbrk y$ be a twice differentiable functional.
Let $\delta J \sqbrk {\hat y; h} = 0$.
Suppose, for $y = \hat y$ and all admissible $h$:
:$\delta^2 J \sqbrk {y; h} \ge 0$ 
Then $J$ has a minimum for $y=\hat y$ if 
{{explain|if what?}}

Proof:
By definition, $ \Delta J \sqbrk y$ can be expressed as:
:$\Delta J \sqbrk {y; h} = \delta J \sqbrk {y; h} + \delta^2 J \sqbrk {y; h} + \epsilon \size h^2$
By assumption:
:$\delta J \sqbrk {\hat y; h} = 0$
Hence:
:$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$
Therefore, for sufficiently small $\size h$ both $\Delta J \sqbrk {\hat y; h}$ and $\delta^2 J \sqbrk {\hat y; h}$ will have the same sign.
{{qed|lemma}}
{{AimForCont}} there exists $h = h_0$ such that:
:$\delta^2 J \sqbrk {\hat y; h_0} < 0$ 
Then, for any $\alpha \ne 0$:
{{begin-eqn}}
{{eqn | l = \delta^2 J \sqbrk {\hat y; \alpha h_0}
      | r = \alpha^2 \delta^2 J \sqbrk {\hat y; h_0}
}}
{{eqn | o = <
      | r = 0
}}
{{end-eqn}}
Therefore, $\Delta J \sqbrk {\hat y; h}$ can be made negative for arbitrary small $\size h$.
However, by assumption $\Delta J \sqbrk {\hat y; h}$ is a minimum of $\Delta J \sqbrk {y; h}$ for all sufficiently small $\size h$.
This is a contradiction.
Thus, a function $h_0: \delta^2 J \sqbrk {\hat y; h_0} < 0$ does not exist.
In other words:
:$\delta^2 J \sqbrk {\hat y; h} \ge 0$ 
for all $h$.
{{qed}}
Theorem: Numbers Equal to Number of Digits in Factorial

For $n \in \set {1, 22, 23, 24}$ the number of digits in the decimal representation of $n!$ is equal to $n$.

Proof:
First we note that:
:$1! = 1$
which has $1$ digit.
Then from Examples of Factorials:
:$2! = 2$
which has $1$ digit.
Multiplying $n$ by a $1$-digit number increases the number of digits of $n$ by no more than $1$.
Thus from $n = 3$ to $n = 9$, $n!$ has no more than $n - 1$ digits.
From $21$ Factorial:
:$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$
and so $21!$ has $20$ digits.
Multiplying $n$ by a $2$-digit number increases the number of digits of $n$ by at least $1$.
Thus from $n = 10$ to $n = 20$, $n!$ has no more than $n - 1$ digits.
It is noted that:
:From $22$ Factorial:
::$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$
:which has $22$ digits.
:From $23$ Factorial:
::$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$
:which has $23$ digits.
:From $24$ Factorial:
::$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$
:which has $24$ digits.
But from $25$ Factorial:
:$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$
which has $26$ digits.
As above, multiplying $n$ by a number with $2$ digits or more increases the number of digits of $n$ by at least $1$.
So if $n!$ has more than $n$ digits, $\paren {n + 1}!$ has more than $n + 1$ digits.
It follows by induction that for all $n \ge 25$, $n!$ has more than $n$ digits.
Hence the result.
{{qed}}
Theorem: Square of Small-Digit Palindromic Number is Palindromic

Let $n$ be an integer such that the sum of the squares of the digits of $n$ in decimal representation is less than $10$.
Let $n$ be palindromic.
Then $n^2$ is also palindromic.
The sequence of such numbers begins:
:$0, 1, 2, 3, 11, 22, 101, 111, 121, 202, 212, 1001, 1111, \dots$
{{OEIS|A057135}}

Proof:
Let $\ds n = \sum_{k \mathop = 0}^m a_k 10^k$ be a number satisfying the conditions above.
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = \sum_{k \mathop = 0}^m a_k^2
      | o = <
      | r = 10
}}
{{eqn | n = 2
      | l = a_k
      | r = a_{m - k}
      | rr = \forall k: 0 \le k \le m
}}
{{end-eqn}}
Consider $\ds n^2 = \paren {\sum_{k \mathop = 0}^m a_k 10^k}^2$.
From definition of Multiplication of Polynomials, the coefficient of $10^l$ in the product is:
{{begin-eqn}}
{{eqn | o =
      | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k
}}
{{eqn | o = \le
      | r = \sqrt {\paren {\sum_{j \mathop = 0}^l a_j^2} \paren {\sum_{k \mathop = 0}^l a_k^2} }
      | c = Cauchy's Inequality
}}
{{eqn | r = \sum_{j \mathop = 0}^l a_j^2
}}
{{eqn | o = \le
      | r = \sum_{j \mathop = 0}^m a_j^2
}}
{{eqn | o = <
      | r = 10
      | c = From (1)
}}
{{end-eqn}}
so no carries occur in the multiplication, and this form satisfies Basis Representation Theorem.
Moreover:
{{begin-eqn}}
{{eqn | l = \sum_{\substack {j \mathop + k \mathop = 2 m \mathop - l \\ j, k \mathop \in \Z} } a_j a_k
      | r = \sum_{\substack {m \mathop - j \mathop + m \mathop - k \mathop = 2 m \mathop - l \\ m \mathop - j, m \mathop - k \mathop \in \Z} } a_{m - j} a_{m - k}
}}
{{eqn | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_{m - j} a_{m - k}
}}
{{eqn | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k
      | c = From (2)
}}
{{end-eqn}}
so the coefficient of $10^{2 m - l}$ is equal to the coefficient of $10^l$ in the expansion of $n^2$.
This shows that $n^2$ is palindromic.
{{qed}}
Theorem: Sufficient Condition for Twice Differentiable Functional to have Minimum

Let $J$ be a twice differentiable functional.
Let $J$ have an extremum for $y=\hat y$.
Let the second variation $\delta^2 J \sqbrk {\hat y; h}$ be strongly positive {{WRT}} $h$.
Then $J$ acquires the minimum for $y = \hat y$ .

Proof:
By assumption, $J$ has an extremum for $y = \hat y$:
:$\delta J \sqbrk {\hat y; h} = 0$
The increment is expressible then as:
:$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$
where $\epsilon \to 0$ as $\size h \to 0$.
By assumption, the second variation is strongly positive:
:$\delta^2 J \sqbrk {\hat y; h} \ge k \size h^2, \quad k \in \R_{>0}$
Hence:
:$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2$
What remains to be shown is that there exists a set of $h$ such that $\epsilon$ is small enough so that {{RHS}} is always positive.
Since $\epsilon \to 0$ as $\size h \to 0$, there exist $c \in \R_{>0}$, such that:
:$\size h < c \implies \size \epsilon < \dfrac 1 2 k$
Choose $h$ such that this inequality holds.
Then
{{begin-eqn}}
{{eqn | l = \frac 1 2 k
      | o = >
      | r = \epsilon > -\frac 1 2 k
      | c = $\big \vert + k$, by Membership is Left Compatible with Ordinal Addition
}}
{{eqn | l = \frac 3 2 k
      | o = >
      | r = k + \epsilon > \frac 1 2 k
      | c = $\big \vert \cdot \size h^2$, by Membership is Left Compatible with Ordinal Multiplication
}}
{{eqn | l = \frac 3 2 k \size h^2
      | o = >
      | r = \paren {k + \epsilon} \size h^2 > \frac 1 2 k \size h^2
}}
{{end-eqn}}
{{Explain|What does this mean? : $\big \vert + k$ and $\big \vert \cdot \size h^2$}}
Therefore:
:$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2 > \dfrac 1 2 k \size h^2 $
For $k \in \R_{>0}$ and $\size h \ne 0$ {{RHS}} is always positive.
Thus, there exists a neighbourhood around $y = \hat y$ where the increment is always positive: 
:$\exists c \in \R_{>0}: \size h < c \implies \Delta J \sqbrk {\hat y; h} > 0$ 
and $J$ has a minimum for $y = \hat y$.
{{qed}}
Theorem: Three times Number whose Divisor Sum is Square

Let $n \in \Z_{>0}$ be a positive integer.
Let the divisor sum $\map {\sigma_1} n$ of $n$ be square.
Let $3$ not be a divisor of $n$.
Then the divisor sum of $3 n$ is square.

Proof:
Let $\map {\sigma_1} n = k^2$.
We have from {{DSFLink|3}}:
:{{:Numbers whose Divisor Sum is Square/Examples/3}}
As $3$ is not a divisor of $n$, it follows that $3$ and $n$ are coprime.
Thus:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {3 n}
      | r = \map {\sigma_1} {3 n} \map {\sigma_1} {3 n}
      | c = Divisor Sum Function is Multiplicative
}}
{{eqn | r = 2^2 k^2
      | c = from above
}}
{{eqn | r = \paren {2 k}^2
      | c = from above
}}
{{end-eqn}}
{{qed}}
Theorem: Three times Number whose Divisor Sum is Square

Let $n \in \Z_{>0}$ be a positive integer.
Let the divisor sum $\map {\sigma_1} n$ of $n$ be square.
Let $3$ not be a divisor of $n$.
Then the divisor sum of $3 n$ is square.

Proof:
From Numbers whose Divisor Sum is Square:
:{{:Numbers whose Divisor Sum is Square/Examples/3}}
The result follows as a specific instance of Product of Coprime Numbers whose Divisor Sum is Square has Square Divisor Sum.
{{qed}}
Theorem: Product of Coprime Numbers whose Divisor Sum is Square has Square Divisor Sum

Let $m, n \in \Z_{>0}$ be a positive integer.
Let the divisor sum of $m$ and $n$ both be square.
Let $m$ and $n$ be coprime.
Then the divisor sum of $m n$ is square.

Proof:
Let $\map {\sigma_1} m = k^2$.
Let $\map {\sigma_1} n = l^2$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {m n}
      | r = \map {\sigma_1} m \map {\sigma_1} n
      | c = Divisor Sum Function is Multiplicative
}}
{{eqn | r = k^2 l^2
      | c = from above
}}
{{eqn | r = \paren {k l}^2
      | c = from above
}}
{{end-eqn}}
{{qed}}
Category:Numbers whose Divisor Sum is Square
Category:Coprime Integers
q8a6xmqtdl3ojghf4ep5jz4e5azf6h2
Theorem: Sequence of Differences on Generalized Pentagonal Numbers

Recall the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$
Consider the sequence defined as $\Delta_n = GP_{n + 1} - GP_n$:
:$1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, \ldots$
{{OEIS|A026741}}
Then:
:The values of $\Delta_n$ for odd $n$ consist of the odd numbers
:The values of $\Delta_n$ for even $n$ consist of the natural numbers.

Proof:
Recall the definition of the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$
:$GP_n = \begin{cases} \dfrac {m \paren {3 m + 1} } 2 & : n = 2 m \\
\dfrac {m \paren {3 m - 1} } 2 & : n = 2 m - 1 \end{cases}$
for $n = 0, 1, 2, \ldots$
Hence:
{{begin-eqn}}
{{eqn | l = \Delta_{2 n - 1}
      | r = GP_{2 n} - GP_{2 n - 1}
      | c = 
}}
{{eqn | r = \dfrac {n \paren {3 n + 1} } 2 - \dfrac {n \paren {3 n - 1} } 2
      | c = 
}}
{{eqn | r = \dfrac {3 n^2 + n - 3 n^2 + n} 2
      | c = 
}}
{{eqn | r = \dfrac {2 n} 2
      | c = 
}}
{{eqn | r = n
      | c = 
}}
{{end-eqn}}
which defines the sequence of natural numbers.
Then:
{{begin-eqn}}
{{eqn | l = \Delta_{2 n}
      | r = GP_{2 n + 1} - GP_{2 n}
      | c = 
}}
{{eqn | r = GP_{2 \paren {n + 1} - 1} - GP_{2 n}
      | c = 
}}
{{eqn | r = \dfrac {\paren {n + 1} \paren {3 \paren {n + 1} - 1} } 2 - \dfrac {n \paren {3 n + 1} } 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {3 n^2 + 6 n + 3} - \paren {n + 1} - 3 n^2 - n} 2
      | c = 
}}
{{eqn | r = \dfrac {4 n + 2} 2
      | c = 
}}
{{eqn | r = 2 n + 1
      | c = 
}}
{{end-eqn}}
which defines the sequence of odd numbers.
{{qed}}
Theorem: Intersection of Semilattice Ideals is Ideal/Set of Sets

Let $\struct {S, \preceq}$ be a bounded below join semilattice.
Let $\II$ be a set of ideals in $\struct {S, \preceq}$.
Then $\bigcap \II$ is an ideal in $\struct {S, \preceq}$.

Proof:
=== Non-Empty Set ===
By Bottom in Ideal:
:$\forall I \in \II: \bot \in I$
where $\bot$ denotes the smallest element in $S$.
By definition of intersection:
:$\bot \in \bigcap \II$
Hence $\bigcap \II$ is non-empty.
{{qed|lemma}}
Theorem: Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset

Let $X, E$ be sets.
Let $P = \struct {\powerset X, \precsim}$ be an inclusion ordered set
where
:$\powerset X$ denotes the power set of $X$
:$\mathord \precsim = \mathord \subseteq \cap \paren {\powerset X \times \powerset X}$
Let $L = \struct {S, \preceq}$ be a continuous lattice subframe of $P$.
Then $E$ is compact element in $L$ {{iff}}:
:$\exists F \in \map {\mathit {Fin} } X: E = \bigcap \set {I \in S: F \subseteq I} \land F \subseteq E$
where $\map {\mathit {Fin} } X$ denotes the set of all finite subsets of $X$.

Proof:
By Power Set is Complete Lattice:
:$P$ is a complete lattice.
By Infima Inheriting Ordered Subset of Complete Lattice is Complete Lattice:
:$L$ is a complete lattice.
By Image of Operator Generated by Closure System is Set of Closure System:
:$\map {\operatorname {operator} } L \sqbrk {\powerset X} = S$
where $\map {\operatorname {operator} } L$ denotes the operator generated by $L$.
By Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema:
:$\map {\operatorname {operator} } L$ preserves directed suprema.
By Image of Compact Subset under Directed Suprema Preserving Closure Operator:
:$\map {\operatorname {operator} } L \sqbrk {\map K P} = \map K {\paren {\map {\operatorname {operator} } L \sqbrk {\powerset X}, \precsim'} }$
where $\map K P$ denotes the compact subset of $P$.
By Image of Operator Generated by Closure System is Set of Closure System:
::$ = \map K L$
Theorem: Legendre's Condition

Let $y =\map y x$ be a real function, such that:
:$\map y a = A,\quad \map y b = B$
Let $J \sqbrk y$ be a functional, such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
where
:$F \in C^2 \closedint a b$ 
{{WRT}} all its variables, and $C$ stands for differentiability class.
Then a necessary condition for $J \sqbrk y$ to have a minimum at $y = \hat y$ is:
:$\bigintlimits {F_{y'y'} } {y \mathop = \hat y} {} \ge 0 \quad \forall x \in \closedint a b$

Proof:
=== Lemma 1 ===
Let $y = \map y x$ be a real function, such that:
:$\map y a = A, \quad \map y b = B$
Let $J \sqbrk y$ be a functional, such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
where:
:$F \in C^2 \closedint a b$ 
{{WRT}} all its variables.
Then:
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x } h'^2 + \map Q {x, \map y x} h^2} \rd x$
where:
:$\map P {x, \map y x} = \dfrac 1 2 F_{y' y'}, \quad \map Q {x, \map y x} = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$
Theorem: Euler's Pentagonal Numbers Theorem/Corollary 1

Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\map {\sigma_1} n$ denote the divisor sum of $n$.
Then:
:$\map {\sigma_1} n = \ds \sum_{1 \mathop \le n - GP_k \mathop < n} -\paren {-1}^{\ceiling {k / 2} } \map {\sigma_1} {n - GP_k} + n \sqbrk {\exists k \in \Z: GP_k = n}$

Proof:
{{ProofWanted|Follows somehow from Euler's Pentagonal Numbers Theorem, but at this time of night I have not a clue how.}}
Theorem: Euler's Pentagonal Numbers Theorem/Corollary 2

Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\map p n$ denote the number of partitions on $n$.
Then:
:$\map p n = \ds \sum_{1 \mathop \le n - GP_k \mathop < n} -\paren {-1}^{\ceiling {k / 2} } \map p {n - GP_k} + \sqbrk {\exists k \in \Z: GP_k = n}$

Proof:
{{ProofWanted|Follows somehow from Euler's Pentagonal Numbers Theorem, but at this time of night I have not a clue how.}}
Theorem: Long Period Prime/Examples/23

The prime number $23$ is a long period prime:
:$\dfrac 1 {23} = 0 \cdotp \dot 04347 \, 82608 \, 69565 \, 21739 \, 1 \dot 3$

Proof:
From Reciprocal of $23$:
{{:Reciprocal of 23}}
Counting the digits, it is seen that this has a period of recurrence of $22$.
Hence the result.
{{qed}}