avewright/proofwiki-math huggingface dataset:
Theorem: Equivalence of Definitions of Polynomial Ring in One Variable

Let $R$ be a commutative ring with unity.
The following definitions of polynomial ring are equivalent in the following sense:
: For every two constructions, there exists an $R$-isomorphism which sends indeterminates to indeterminates.
{{explain|this statement has to be made more precise}}
=== Definition 1: As a Ring of Sequences ===
{{:Definition:Polynomial Ring/Sequences}}
=== Definition 2: As a Monoid Ring on the Natural Numbers ===
{{:Definition:Polynomial Ring/Monoid Ring on Natural Numbers}}

Proof:
Use Polynomial Ring of Sequences Satisfies Universal Property
{{ProofWanted}}
Theorem: Smallest Triplet of Integers whose Product with Divisor Count are Equal

Let $\map {\sigma_0} n$ denote the divisor count function: the number of divisors of $n$.
The smallest set of $3$ integers $T$ such that $m \, \map {\sigma_0} m$ is equal for each $m \in T$ is:
:$\set {168, 192, 224}$

Proof:
{{begin-eqn}}
{{eqn | l = 168 \times \map {\sigma_0} {168}
      | r = 168 \times 16
      | c = {{DCFLink|168}}
}}
{{eqn | r = \paren {2^3 \times 3 \times 7} \times 2^4
      | c = 
}}
{{eqn | r = 2^7 \times 3 \times 7
      | c = 
}}
{{eqn | r = 2688
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 192 \times \map {\sigma_0} {192}
      | r = 192 \times 14
      | c = {{DCFLink|192}}
}}
{{eqn | r = \paren {2^6 \times 3} \times \paren {2 \times 7}
      | c = 
}}
{{eqn | r = 2^7 \times 3 \times 7
      | c = 
}}
{{eqn | r = 2688
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 224 \times \map {\sigma_0} {224}
      | r = 224 \times 12
      | c = {{DCFLink|224}}
}}
{{eqn | r = \paren {2^5 \times 7} \times \paren {2^2 \times 3}
      | c = 
}}
{{eqn | r = 2^7 \times 3 \times 7
      | c = 
}}
{{eqn | r = 2688
      | c = 
}}
{{end-eqn}}
{{ProofWanted|This needs to be identified as the smallest such.}}
Theorem: Pairs of Integers whose Product with Divisor Count are Equal

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:
{{begin-eqn}}
{{eqn | l = 18 \times \map {\sigma_0} {18}
      | r = 18 \times 6
      | c = {{DCFLink|18}}
}}
{{eqn | r = \paren {2 \times 3^2} \times \paren {2 \times 3}
      | c = 
}}
{{eqn | r = 2^2 \times 3^3
      | c = 
}}
{{eqn | r = 108
      | c = 
}}
{{eqn | l = 27 \times \map {\sigma_0} {27}
      | r = 27 \times 4
      | c = {{DCFLink|27}}
}}
{{eqn | r = 2^2 \times 3^3
      | c = 
}}
{{eqn | r = 108
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 24 \times \map {\sigma_0} {24}
      | r = 24 \times 8
      | c = {{DCFLink|24}}
}}
{{eqn | r = \paren {2^3 \times 3} \times 2^3
      | c = 
}}
{{eqn | r = 2^6 \times 3
      | c = 
}}
{{eqn | r = 192
      | c = 
}}
{{eqn | l = 32 \times \map {\sigma_0} {32}
      | r = 32 \times 6
      | c = {{DCFLink|32}}
}}
{{eqn | r = 2^5 \times \paren {2 \times 3}
      | c = 
}}
{{eqn | r = 192
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 56 \times \map {\sigma_0} {56}
      | r = 56 \times 8
      | c = {{DCFLink|56}}
}}
{{eqn | r = \paren {2^3 \times 7} \times 2^3
      | c = 
}}
{{eqn | r = 2^6 \times 7
      | c = 
}}
{{eqn | r = 448
      | c = 
}}
{{eqn | l = 64 \times \map {\sigma_0} {64}
      | r = 64 \times 7
      | c = {{DCFLink|64}}
}}
{{eqn | r = 2^6 \times 7
      | c = 
}}
{{eqn | r = 448
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Order Isomorphism forms Galois Connection

Let $L_1 = \struct {S_1, \preceq_1}$, $L_2 = \struct {S_2, \preceq_2}$ be ordered sets.
Let $f:S_1 \to S_2$ be an order isomorphism between $L_1$ and $L_2$.
Then $\struct {f, f^{-1} }$ is a Galois connection.

Proof:
Let $t \in S_2$, $s \in S_1$.
We will prove that:
:$t \preceq_2 \map f s \implies \map {f^{-1} } t \preceq_1 s$
Assume that:
:$t \preceq_2 \map f s$
By Inverse of Order Isomorphism is Order Isomorphism:
:$f^{-1}$ is an order isomorphism.
By definition of order isomorphism:
:$f^{-1}$ is an order embedding.
By definition of order embedding:
:$\map {f^{-1} } t \preceq_1 \map {f^{-1} } {\map f s}$
Thus by definition of bijection:
:$\map {f^{-1} } t \preceq_1 s$
{{qed|lemma}}
We will prove that:
:$\map {f^{-1} } t \preceq_1 s \implies t \preceq_2 \map f s$
Assume that:
:$\map {f^{-1} } t \preceq_1 s$
By definition of order embedding:
:$\map f {\map {f^{-1} } t} \preceq_2 \map f s$
Thus by definition of bijection:
:$t \preceq_2 \map f s$
{{qed}}
Theorem: Sequence of Square Centered Hexagonal Numbers

The sequence of centered hexagonal numbers which are also square begins:
:$1, 169, 32 \, 761, 6 \, 355 \, 441, \ldots$
{{OEIS|A006051}}

Proof:
{{:Definition:Centered Hexagonal Number/Sequence}}
We have that:
{{begin-eqn}}
{{eqn | l = 1
      | r = 1^2
      | c = {{Defof|Square Number}}
}}
{{eqn | r = 3 \times 1 \times \paren {1 - 1} + 1
      | c = Closed Form for Centered Hexagonal Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 169
      | r = 13^2
      | c = {{Defof|Square Number}}
}}
{{eqn | r = 3 \times 8 \times \paren {8 - 1} + 1
      | c = Closed Form for Centered Hexagonal Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 32 \, 761
      | r = 181^2
      | c = {{Defof|Square Number}}
}}
{{eqn | r = 3 \times 105 \times \paren {105 - 1} + 1
      | c = Closed Form for Centered Hexagonal Numbers
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 6 \, 355 \, 441
      | r = 2521^2
      | c = {{Defof|Square Number}}
}}
{{eqn | r = 3 \times 1456 \times \paren {1456 - 1} + 1
      | c = Closed Form for Centered Hexagonal Numbers
}}
{{end-eqn}}
{{qed}}
Theorem: 169 as Sum of up to 155 Squares

$169$ can be expressed as the sum of $n$ non-zero squares for all $n$ from $1$ to $155$.

Proof:
We note the following:
{{begin-eqn}}
{{eqn | l = 169
      | r = 13^2
      | c = 
}}
{{eqn | r = 12^2 + 5^2
      | c = 
}}
{{eqn | r = 12^2 + 4^2 + 3^2
      | c = 
}}
{{eqn | r = 8^2 + 8^2 + 5^2 + 4^2
      | c = 
}}
{{eqn | r = 8^2 + 8^2 + 4^2 + 4^2 + 3^2
      | c = 
}}
{{end-eqn}}
Let $41 \le n \le 155$.
Let $n \equiv 0 \pmod 3$.
Then $169$ can be represented as:
:$169 = 2 \times 3^2 + \left({37 - \dfrac {n - 42} 3}\right) \times 2^2 + \left({\dfrac {4 \left({n - 42}\right)} 3 + 3}\right) \times 1^2$
Let $n \equiv 1 \pmod 3$.
Then $169$ can be represented as:
:$169 = \left({42 - \dfrac {n - 43} 3}\right) \times 2^2 + \left({\dfrac {4 \left({n - 43}\right)} 3 + 1}\right) \times 1^2$
Let $n \equiv 2 \pmod 3$.
Then $169$ can be represented as:
:$169 = 3^2 + \left({40 - \dfrac {n - 41} 3}\right) \times 2^2 + \left({\dfrac {4 \left({n - 41}\right)} 3}\right) \times 1^2$
Thus in each case $169$ has been expressed in terms of $n$ squares: a combination of varying numbers of $1^2$, $2^2$ and $3^2$.
For $n > 155$ this construction does not work, because it results in a negative number of instances of $2^2$.
Similarly, for $n < 41$ this construction does not work, because it results in a negative number of instances of $1^2$.
For $n < 41$, we have:
:$169 = 25 + k$
where $k$ equals one of:
:$1 \times 12^2$
:$9 \times 4^2$
:$16 \times 3^2$
:$14 \times 3^2 + 3 \times 2^2 + 6 \times 1^2$
:$10 \times 3^2 + 11 \times 2^2 + 10 \times 1^2$
That is, $169$ equals $25$ plus $1$, $9$, $16$, $23$ or $31$ other squares.
But from 25 as Sum of 4 to 11 Squares, $25$ can also be written as a sum of $n$ squares for $4 \le n \le 11$.
Thus, by using judicious combinations of:
: sums of squares totalling $25$
: sums of squares totalling $169 - 25$
we can obtain a representation of $169$ as the sum of $n$ squares where $5 \le n \le 40$.
In summary, therefore, there exists a representation of $169$ as the sum of $n$ squares for $1 \le n \le 155$.
{{qed}}
Theorem: Lattice of Power Set is Arithmetic

Let $X$ be a set.
Let $P = \struct {\powerset X, \cup, \cap, \subseteq}$ be a lattice of power set.
Then $P$ is an arithmetic ordered set.

Proof:
Define $C = \struct {\map K P, \preceq}$ as an ordered subset of $P$
where $\map K P$ denotes the compact subset of $P$.
Thus by Lattice of Power Set is Algebraic:
:$P$ is algebraic.
It remains to prove that:
:$\map K P$ is meet closed.
Let $x, y \in \map K P$.
By definition of compact subset:
:$x$ is compact.
By Element is Finite iff Element is Compact in Lattice of Power Set
:$x$ is finite.
By Intersection is Subset:
:$x \cap y \subseteq x$
By Subset of Finite Set is Finite:
:$x \cap y$ is finite.
By Element is Finite iff Element is Compact in Lattice of Power Set
:$x \cap y$ is compact.
Thus by definition of compact subset:
:$x \cap y \in \map K P$
{{qed}}
Theorem: Nth Derivative of Natural Logarithm

The $n$th derivative of $\map \ln x$ for $n \ge 1$ is:
:$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
Theorem: Sum of Factorials of Digits of 169

Let the factorials of the digits of $169$ be added.
Let the same process be done on the result.
Repeat.
After $3$ iterations, the result will be $169$.

Proof:
{{begin-eqn}}
{{eqn | l = 1! + 6! + 9!
      | r = 1 + 720 + 362 \, 880
      | c = 
}}
{{eqn | r = 363 \, 601
      | c = 
}}
{{eqn | l = 3! + 6! + 3! + 6! + 0! + 1!
      | r = 6 + 720 + 6 + 720 + 1 + 1
      | c = 
}}
{{eqn | r = 1454
      | c = 
}}
{{eqn | l = 1! + 4! + 5! + 4!
      | r = 1 + 24 + 120 + 24
      | c = 
}}
{{eqn | r = 169
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Squares which are Difference between Two Cubes

$169$ is the smallest square number which is the difference between two cubes:
:$169 = 8^3 - 7^3$
{{expand|Add the rest of the sequence, having found out what they are.}}

Proof:
{{begin-eqn}}
{{eqn | l = 8^3 - 7^3
      | r = 512 - 343
      | c = 
}}
{{eqn | r = 169
      | c = 
}}
{{eqn | r = 13^2
      | c = 
}}
{{end-eqn}}
{{ProofWanted|Establish that this is the smallest such.}}
Theorem: Cube of 180 is Sum of Sequence of Consecutive Cubes

:$180^3 = \ds \sum_{k \mathop = 6}^{69} k^3$
That is:
:$180^3 = 6^3 + 7^3 + \cdots + 67^3 + 68^3 + 69^3$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{69} k^3
      | r = \paren {\dfrac {69 \paren {69 + 1} } 2}^2
      | c = Sum of Sequence of Cubes
}}
{{eqn | r = 5 \, 832 \, 225
      | c = 
}}
{{eqn | l = \sum_{k \mathop = 1}^5 k^3
      | r = \paren {\dfrac {5 \paren {5 + 1} } 2}^2
      | c = Sum of Sequence of Cubes
}}
{{eqn | r = 225
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 6}^{69} k^3
      | r = \sum_{k \mathop = 1}^{69} k^3 - \sum_{k \mathop = 1}^5 k^3
      | c = 
}}
{{eqn | r = 5 \, 832 \, 225 - 225
      | c = 
}}
{{eqn | r = 5 \, 832 \, 000
      | c = 
}}
{{eqn | r = 180^3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: 3-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 $3$-digit integers $m$ which need the largest number of iterations before reaching a palindromic number are:
:$187, 286, 385, 583, 682, 781, 880$
all of which need $23$ iterations.
:$869$ and $968$ are the result of the first iteration for each of these, and so take $22$ iterations to reach a palindromic number.

Proof:
The sequence obtained by iterating $r$ on $187$ is:
:$187, 968, 1837, 9218, 17347, 91718, 173437, 907808, 1716517, 8872688,$
:$17735476, 85189247, 159487405, 664272356, 1317544822, 3602001953, 7193004016, 13297007933,$
:$47267087164, 93445163438, 176881317877, 955594506548, 170120002107, 8713200023178$
{{OEIS|A033670}}
Note that $r \left({187}\right) = r \left({781}\right) = 968$, so the sequence obtained by iterating $r$ on $781$ is the same.
The rest of the numbers converge on the same sequence immediately:
:$286, 968, \ldots$
:$385, 968, \ldots$
:$880, 88, \ldots$
:$869, 1837, \ldots$
as do their reversals.
{{qed}}
Theorem: Numbers Expressible as Sum of Five Distinct Squares

The largest positive integer which cannot be expressed as the sum of no more than $5$ distinct squares is $188$.
Both $188$ and $124$ require as many as $6$ distinct squares to represent them:
{{begin-eqn}}
{{eqn | l = 124
      | r = 1 + 4 + 9 + 25 + 36 + 49
      | c = 
}}
{{eqn | r = 1^2 + 2^2 + 3^2 + 5^2 + 6^2 + 7^2
      | c = 
}}
{{eqn | l = 188
      | r = 1 + 4 + 9 + 25 + 49 + 100
      | c = 
}}
{{eqn | r = 1^2 + 2^2 + 3^2 + 5^2 + 7^2 + 10^2
      | c = 
}}
{{end-eqn}}

Proof:
From Numbers not Sum of Distinct Squares, the following positive integers cannot be expressed as the sum of distinct squares at all:
:$2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128$
{{OEIS|A001422}}
{{ProofWanted|It remains to be shown that all other positive integers less than $188$ can be expressed as the sum of no more than 5 distinct squares, and all numbers over $188$ can also be so expressed.}}
Theorem: Closed Form for Heptagonal Pyramidal Numbers

The closed-form expression for the $n$th heptagonal pyramidal number is:
:$Q_n = \dfrac {n \paren {n + 1} \paren {5 n - 2} } 6$

Proof:
{{begin-eqn}}
{{eqn | l = Q_n
      | r = \sum_{k \mathop = 1}^n H_n
      | c = {{Defof|Heptagonal Pyramidal Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \dfrac {k \paren {5 k - 3} } 2
      | c = Closed Form for Heptagonal Numbers
}}
{{eqn | r = \dfrac 1 2 \paren {5 \sum_{k \mathop = 1}^n k^2 - 3 \sum_{k \mathop = 1}^n k}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {5 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \sum_{k \mathop = 1}^n k}
      | c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac 1 2 \paren {5 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \dfrac {n \paren {n + 1} } 2}
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \dfrac 1 2 \paren {\frac {5 n \paren {n + 1} \paren {2 n + 1} - 9 n \paren {n + 1} } 6}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {5 \paren {2 n + 1} - 9} } 6}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {10 n + 5 - 9} } 6}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\frac {n \paren {n + 1} \paren {10 n - 4} } 6}
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {5 n - 2} } 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Heptagonal Pyramidal Numbers which are Square

The sequence of heptagonal pyramidal numbers which also have the property of being square begins:
:$0, 1, 196, 99 \, 225$
{{expand|Probably no more, related to elliptic curves}}

Proof:
{{begin-eqn}}
{{eqn | r = \dfrac {0 \paren {0 + 1} \paren {5 \times 0 - 2} } 6
      | o = 
      | c = Closed Form for Heptagonal Pyramidal Numbers
}}
{{eqn | r = \dfrac {0 \times 1 \times \paren {- 2} } 6
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{eqn | r = 0^2
      | c = {{Defof|Square Number}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | r = \dfrac {1 \paren {1 + 1} \paren {5 \times 1 - 2} } 6
      | o = 
      | c = Closed Form for Heptagonal Pyramidal Numbers
}}
{{eqn | r = \dfrac {1 \times 2 \times 3} 6
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = 1^2
      | c = {{Defof|Square Number}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | r = \dfrac {6 \paren {6 + 1} \paren{5 \times 6 - 2} } 6
      | o = 
      | c = Closed Form for Heptagonal Pyramidal Numbers
}}
{{eqn | r = \dfrac {6 \times 7 \times 28} 6
      | c = 
}}
{{eqn | r = 7 \times \paren {2^2 \times 7}
      | c = 
}}
{{eqn | r = \paren {2 \times 7}^2
      | c = {{Defof|Square Number}}
}}
{{eqn | r = 196
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | r = \dfrac {49 \paren {49 + 1} \paren{5 \times 49 - 2} } 6
      | o = 
      | c = Closed Form for Heptagonal Pyramidal Numbers
}}
{{eqn | r = \dfrac {49 \times 50 \times 243} 6
      | c = 
}}
{{eqn | r = 7^2 \times \paren {5^2 \times 9^2}
      | c = 
}}
{{eqn | r = \paren {5 \times 7 \times 9}^2
      | c = {{Defof|Square Number}}
}}
{{eqn | r = 99 \, 225
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Conditions for Limit Function to be Limit Minimizing Function of Functional

Let $y$ be a real function.
Let $J \sqbrk y$ be a functional.
Let $\sequence {y_n}$ be a minimizing sequence of $J$.
Let:
:$\ds \lim_{n \mathop \to \infty} y_n = \hat y$
Suppose $J$ is lower semicontinuous at $y = \hat y$.
Then:
:$\ds J \sqbrk {\hat y} = \lim_{n \mathop \to \infty} J \sqbrk {y_n}$

Proof:
By definition of minimizing sequence:
:$\ds \inf_y J \sqbrk y = \lim_{n \mathop \to \infty} J \sqbrk {y_n}$
Any mapping from this sequence either minimises the functional or not.
This is true for the limit mapping as well:
:$\ds J \sqbrk {\hat y} \ge \inf_y J \sqbrk y$
By assumption, $J$ is lower semicontinuous at $\hat y$:
:$\ds \forall \epsilon_1 > 0: \forall n \in \N: \exists N \in \N: \paren {n > N} \implies \paren {J \sqbrk {\hat y} - J \sqbrk {y_n} < \epsilon_1}$
Hence, for sufficiently large $n$:
:$\ds \inf_y J \sqbrk y \le J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon_1$
By the definition of minimizing sequence, where the label $n$ has been replaced with $m$:
:$\ds \forall \epsilon_2: \forall m \in \N: \exists M \in \N: \paren {m > M} \implies \paren {\size {J \sqbrk {y_m} - \inf_y J \sqbrk y} < \epsilon_2}$
From this, for sufficiently large $m$:
:$\ds J \sqbrk {y_m} - \epsilon_2 < \inf_y J \sqbrk y$
Then:
:$J \sqbrk {y_m} - \epsilon_2 < J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon_1$
Here $\epsilon_1$, $n$ have similar properties like $\epsilon_2$, $m$, but are otherwise arbitrary and independent.
Let $\epsilon_1 = \epsilon_2 = \epsilon$, $n = m$, $M = N$.
Arbitrariness is still not affected. 
Then:
:$J \sqbrk {y_n} - \epsilon < J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon$
Subtract $J \sqbrk {y_n}$ from all the terms.
Hence:
:$-\epsilon < J \sqbrk {\hat y} - J \sqbrk {y_n} < \epsilon$
or:
:$\size {J \sqbrk {y_n} - J \sqbrk {\hat y} } < \epsilon$
Therefore this relation inherits all the constraints on its values $n$, $N$, $\epsilon$, and by definition is a limit:
:$\ds \lim_{n \mathop \to \infty} J \sqbrk {y_n} = J \sqbrk {\hat y}$
{{qed}}
{{Proofread|may contain mistakes, since original was not satisfactory enough}}
Theorem: Smallest 10 Primes in Arithmetic Sequence

The smallest $10$ primes in arithmetic sequence are:
:$199 + 210 n$
for $n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
These are also the smallest $8$ and $9$ primes in arithmetic sequence.

Proof:
{{begin-eqn}}
{{eqn | l = 199 + 0 \times 210
      | r = 199
      | c = which is the $46$th prime
}}
{{eqn | l = 199 + 1 \times 210
      | r = 409
      | c = which is the $80$th prime
}}
{{eqn | l = 199 + 2 \times 210
      | r = 619
      | c = which is the $114$th prime
}}
{{eqn | l = 199 + 3 \times 210
      | r = 829
      | c = which is the $145$th prime
}}
{{eqn | l = 199 + 4 \times 210
      | r = 1039
      | c = which is the $175$th prime
}}
{{eqn | l = 199 + 5 \times 210
      | r = 1249
      | c = which is the $204$th prime
}}
{{eqn | l = 199 + 6 \times 210
      | r = 1459
      | c = which is the $232$nd prime
}}
{{eqn | l = 199 + 7 \times 210
      | r = 1669
      | c = which is the $263$rd prime
}}
{{eqn | l = 199 + 8 \times 210
      | r = 1879
      | c = which is the $289$th prime
}}
{{eqn | l = 199 + 9 \times 210
      | r = 2089
      | c = which is the $316$th prime
}}
{{end-eqn}}
{{OEIS|A033168}}
But note that $199 + 10 \times 210 = 2299 = 11^2 \times 19$ and so is not prime.
Now we show that this is the smallest $10$ primes in arithmetic sequence.
By Divisibility of Common Difference of Arithmetic Sequence of Primes, the common difference $d$ of any $10$ primes in arithmetic sequence must be divisible by all primes less that $10$.
That is, the common difference is a multiple of $2 \times 3 \times 5 \times 7 = 210$.
Suppose $d = 210$.
Then the first term $p_1$ of the arithmetic sequence cannot be $11$, as $11 + 210 = 221 = 13 \times 17$.
Note that $210 \equiv 1 \pmod {11}$.
Hence if the $p_1$ gives a remainder $r > 1$ upon divison by $11$, the $\paren {12 - r}^{th}$ term of the sequence is equivalent to:
:$r + \paren {12 - r - 1} \times 210 \equiv 0 \pmod {11}$
and is divisible by $11$, which is impossible as all terms are primes.
Therefore $p_1 \equiv 1 \pmod {11}$.
The first few primes of this form are:
:$23, 67, 89, 199, \dots$
and we eliminate the first three candidates because:
:$23 + 5 \times 210 = 1073 = 29 \times 37$
:$67 + 3 \times 210 = 697 = 17 \times 41$
:$89 + 1 \times 210 = 229 = 13 \times 23$
which are not primes, and we have checked that $199 + 210 n$ are indeed $10$ primes in arithmetic sequence.
Now suppose $d > 210$.
Then $d \ge 420$ and the last term is greater than $420 \times 9 = 3780$.
This shows that the one we found here is the smallest $10$ primes in arithmetic sequence.
{{qed|lemma}}
Theorem: Numbers that cannot be made Prime by changing 1 Digit

The following positive integers cannot be made into prime numbers by changing just one digit:
:$200, 202, 204, 205, 206, 208, \ldots$
{{OEIS|A192545}}

Proof:
In order to make any one of these positive integers into a prime number one would have to change the last digit.
Otherwise the number it was changed into would be either even, or divisible by $5$, and so not prime.
But all the other integers between $200$ and $209$ are composite:
{{begin-eqn}}
{{eqn | l = 201
      | r = 3 \times 67
      | c = 
}}
{{eqn | l = 203
      | r = 7 \times 29
      | c = 
}}
{{eqn | l = 207
      | r = 3^2 \times 23
      | c = 
}}
{{eqn | l = 209
      | r = 11 \times 19
      | c = 
}}
{{end-eqn}}
In order for there to be a smaller number with this property, a prime gap would need to be found which spans an entire decade.
There are none such.
On the other hand, each of $201, 203, 207, 209$ can themselves be turned into a prime number by changing the initial $2$ into a $1$:
:$101, 103, 107,109$
are all prime.
{{qed}}
Theorem: Sum of Cubes of 3 Consecutive Integers which is Square

The following sequences of $3$ consecutive (strictly) positive integers have cubes that sum to a square:
:$1, 2, 3$
:$23, 24, 25$
No other such sequence of $3$ consecutive positive integers has the same property.
However, if we allow sequences containing zero and negative integers, we also have:
:$-1, 0, 1$
:$0, 1, 2$

Proof:
{{begin-eqn}}
{{eqn | l = 1^3 + 2^3 + 3^3
      | r = 1 + 8 + 27
      | c = 
}}
{{eqn | r = 36
      | c = 
}}
{{eqn | r = 6^2
      | c = 
}}
{{eqn | l = 23^3 + 24^3 + 25^3
      | r = 12 \, 167 + 13 \, 824 + 15 \, 625
      | c = 
}}
{{eqn | r = 41 \, 616
      | c = 
}}
{{eqn | r = 204^2
      | c = 
}}
{{eqn | l = \paren {-1}^3 + 0^3 + 1^3
      | r = -1 + 0 + 1
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{eqn | r = 0^2
      | c = 
}}
{{eqn | l = 0^3 + 1^3 + 2^3
      | r = 0 + 1 + 8
      | c = 
}}
{{eqn | r = 9
      | c = 
}}
{{eqn | r = 3^2
      | c = 
}}
{{end-eqn}}
Any sequence of $3$ consecutive integers that have cubes that sum to a square would satisfy:
:$m^2 = \paren {n - 1}^3 + n^3 + \paren {n + 1}^3$
where $n$ is the middle number of the sequence, with $m, n \in \Z$.
Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | l = m^2
      | r = \paren {n - 1}^3 + n^3 + \paren {n + 1}^3
}}
{{eqn | r = n^3 - 3 n^2 + 3 n - 27 + n^3 + n^3 + 3 n^2 + 3 n + 27
      | c = Cube of Sum, Cube of Difference
}}
{{eqn | r = 3 n^3 + 6 n
}}
{{end-eqn}}
Substituting $y = 3 m$ and $x = 3 n$:
{{begin-eqn}}
{{eqn | l = \paren {\frac y 3}^2
      | r = 3 \paren {\frac x 3}^3 + 6 \paren {\frac x 3}
}}
{{eqn | ll= \leadsto
      | l = \frac {y^2} 9
      | r = \frac {x^3} 9 + 2 x
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = x^3 + 18 x
}}
{{end-eqn}}
which is an elliptic curve.
According to [https://www.lmfdb.org/EllipticCurve/Q/2304/a/2 LMFDB], this elliptic curve has exactly $7$ lattice points:
:$\tuple {0, 0}, \tuple {3, \pm 9}, \tuple {6, \pm 18}, \tuple {72, \pm 612}$
which correspond to these values of $n$:
:$0, 1, 2, 24$
Hence there are no more solutions.
{{qed}}