avewright/proofwiki-math huggingface dataset:
Theorem: Factor of Mersenne Number equivalent to +-1 mod 8

:$q \equiv \pm 1 \pmod 8$

Proof:
Suppose $q \divides M_p$, where $\divides$ denotes divisibility.
From Factor of Mersenne Number $M_p$ is of form $2 k p + 1$: 
:$q - 1 = 2 k p$
From above:
:$2^{\paren {q - 1} / 2} \equiv 2 k p \equiv 1 \pmod q$
and so $2$ is a quadratic residue $\pmod q$.
From Second Supplement to Law of Quadratic Reciprocity:
:$q \equiv \pm 1 \pmod 8$
{{qed}}
Theorem: Numbers with 7 or more Prime Factors

The sequence of positive integers with $7$ or more prime factors (not necessarily distinct) begins:
:$128, 192, 256, 288, 320, 384, 432, 448, 480, 512, \ldots$
{{OEIS|A046307}}

Proof:
{{begin-eqn}}
{{eqn | l = 128
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2
      | c = 
}}
{{eqn | l = 192
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3
      | c = 
}}
{{eqn | l = 256
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \left({\times 2}\right)
      | c = 
}}
{{eqn | l = 288
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3
      | c = 
}}
{{eqn | l = 320
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5
      | c = 
}}
{{eqn | l = 384
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \left({\times 3}\right)
      | c = 
}}
{{eqn | l = 432
      | r = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3
      | c = 
}}
{{eqn | l = 448
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7
      | c = 
}}
{{eqn | l = 480
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5
      | c = 
}}
{{eqn | l = 512
      | r = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \left({\times 2 \times 2}\right)
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Numbers not Sum of Distinct Squares

The positive integers which are not the sum of $1$ or more distinct squares are:
:$2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128$
{{OEIS|A001422}}

Proof:
{{finish|a) Demonstration that these cannot be so expressed, b) demonstration that all others below 324 can be so expressed}}
It will be proved that the largest integer which cannot be expressed as the sum of distinct squares is $128$.
The remaining integers in the sequence can be identified by inspection.
We prove this using a variant of Second Principle of Mathematical Induction.
Let $\map P n$ be the proposition:
:$n$ can be expressed as the sum of distinct squares.
Theorem: 132 is Sum of all 2-Digit Numbers formed from its Digits

$132$ is the smallest sum of all the $2$-digit (positive) integers formed from its own digits.

Proof:
It is necessary to postulate that such (positive) integers have $3$ digits or more, as the $2$ digit solution is trivial.
Let $n = \sqbrk {abc}$ be a $3$-digit number.
Let $\map s n$ denote the sum of all the $2$-digit (positive) integers formed from the digits of $n$.
Then:
{{begin-eqn}}
{{eqn | l = \map s n
      | r = \sqbrk {ab} + \sqbrk {ac} + \sqbrk {bc} + \sqbrk {ba} + \sqbrk {ca} + \sqbrk {cb}
      | c = 
}}
{{eqn | r = \paren {10 a + b} + \paren {10 a + c} + \paren {10 b + c} + \paren {10 b + a} + \paren {10 c + a} + \paren {10 c + b}
      | c = 
}}
{{eqn | r = 2 \times \paren {10 a + a} + 2 \times \paren {10 b + b} + 2 \times \paren {10 c + c}
      | c = 
}}
{{eqn | r = 22 \times \paren {a + b + c}
      | c = 
}}
{{end-eqn}}
So for $n = \map s n$, $n = \sqbrk {abc}$ needs to be divisible by both $22$ and $a + b + c$.
{{begin-eqn}}
{{eqn | l = 22 \times 4
      | r = 88
      | c = too small
}}
{{eqn | l = 22 \times 5
      | r = 110
      | c = but $\map s {110} = 11 + 10 + 01 = 22$
}}
{{eqn | l = 22 \times 6
      | r = 132
      | c = and $\map s {132} = 12 + 13 + 21 + 23 + 31 + 32 = 132$
}}
{{end-eqn}}
and the result is apparent.
{{qed}}
Theorem: Numbers which are Sum of Increasing Powers of Digits

The following integers are the sum of the increasing powers of their digits taken in order begin as follows:
:$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2 \, 646 \, 798, 12 \, 157 \, 692 \, 622 \, 039 \, 623 \, 539$
{{OEIS|A032799}}

Proof:
Single digit integers are trivial:
:$d^1 = d$
for all $d \in \Z$.
Then we have:
{{begin-eqn}}
{{eqn | l = 8^1 + 9^2
      | r = 8 + 81
      | rr= = 89
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1^1 + 3^2 + 5^3
      | r = 1 + 9 + 125
      | rr= = 135
      | c = 
}}
{{eqn | l = 1^1 + 7^2 + 5^3
      | r = 1 + 49 + 125
      | rr= = 175
      | c = 
}}
{{eqn | l = 5^1 + 1^2 + 8^3
      | r = 5 + 1 + 512
      | rr= = 518
      | c = 
}}
{{eqn | l = 5^1 + 9^2 + 8^3
      | r = 5 + 81 + 512
      | rr= = 598
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1^1 + 3^2 + 0^3 + 6^4
      | r = 1 + 9 + 0 + 1296
      | rr= = 1306
      | c = 
}}
{{eqn | l = 1^1 + 6^2 + 7^3 + 6^4
      | r = 1 + 36 + 343 + 1296
      | rr= = 1676
      | c = 
}}
{{eqn | l = 2^1 + 4^2 + 2^3 + 7^4
      | r = 2 + 16 + 8 + 2401
      | rr= = 2427
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | r = 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7 
      | o = 
      | c = 
}}
{{eqn | r = 2 + 36 + 64 + 1296 + 16 \, 807 + 531 \, 441 + 1 \, 097 \, 152
      | c = 
}}
{{eqn | r = 2 \, 646 \, 798
      | c = 
}}
{{end-eqn}}
and finally:
{{begin-eqn}}
{{eqn | r = 1 + 2^2 + 1^3 + 5^4 + 7^5 + 6^6 + 9^7 + 2^8 + 6^9 + 2^{10} + 2^{11} + 0^{12} + 3^{13} + 9^{14} + 6^{15} + 2^{16} + 3^{17} + 5^{18} + 3^{19} + 9^{20}
      | o = 
      | c = 
}}
{{eqn | r = 1 + 4 + 1 + 625 + 16 \, 807 + 46 \, 656 + 4 \, 782 \, 969 + 256 + 10 \, 077 \, 696 + 1024 + 2048 + 0 + 1 \, 594 \, 323
      | c = 
}}
{{eqn | o = 
      | ro= + 
      | r = 22 \, 876 \, 792 \, 454 \, 961 + 470 \, 184 \, 984 \, 576 + 65 \, 536 + 129 \, 140 \, 163 + 3 \, 814 \, 697 \, 265 \, 625
      | c = 
}}
{{eqn | o = 
      | ro= + 
      | r = 1 \, 162 \, 261 \, 467 + 12 \, 157 \, 665 \, 459 \, 056 \, 928 \, 801
      | c = 
}}
{{eqn | r = 12 \, 157 \, 692 \, 622 \, 039 \, 623 \, 539
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be proved that there are no more like this.}}
Theorem: Prime Gaps of 10

The following pairs of consecutive prime numbers are those whose difference is $10$:
:$\tuple {139, 149}, \tuple {181, 191}, \tuple {241, 251}, \tuple {283, 293}, \ldots$
{{OEIS|A031928|order = lower}}

Proof:
Demonstrated by listing the prime gaps.
{{qed}}
Theorem: Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum

Let $n \in \Z_{>0}$ be a positive integer.
The harmonic mean of the divisors of $n$ is given by:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\map H n}
      | r = \frac 1 {\map {\sigma_0} n} \paren {\sum_{d \mathop \divides n} \frac 1 d}
      | c = {{Defof|Harmonic Mean}}
}}
{{eqn | l = \sum_{d \mathop \divides n} \frac 1 d
      | r = \frac {\map {\sigma_1} n} n
      | c = Sum of Reciprocals of Divisors equals Abundancy Index
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\map H n}
      | r = \frac {\map {\sigma_1} n} {n \, \map {\sigma_0} n}
}}
{{eqn | ll= \leadsto
      | l = \map H n
      | r = \frac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}
}}
{{end-eqn}}
{{qed}}
Theorem: Perfect Number is Ore Number

Let $n \in \Z_{>0}$ be a perfect number.
Then $n$ is an Ore number.

Proof:
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum, the harmonic mean of the divisors of $n$ is given by:
:$\map H n = \dfrac {n \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Let $n$ be a perfect number.
By definition of perfect number:
:$\dfrac {\map {\sigma_1} n} n = 2$
From Divisor Count Function is Odd Iff Argument is Square:
:$\map {\sigma_0} n = 2 k$
for some $k \in \Z$.
Hence:
:$\map H n = \dfrac {2 k} 2 = k$
Hence the result.
{{qed}}
Theorem: Sequence of Numbers with Integer Arithmetic and Harmonic Means of Divisors

The following sequence of integers have the property that both the harmonic mean and arithmetic mean of their divisors are integers:
:$1, 6, 140, 270, 672, \ldots$
{{OEIS|A007340}}

Proof:
Note the integers whose harmonic mean of their divisors are integers are the Ore numbers:
:$1, 6, 28, 140, 270, 496, 672, \ldots$
{{OEIS|A001599}}
It remains to calculate the arithmetic mean of their divisors.
Let $\map A n$ denote the arithmetic mean of the divisors of $n$.
Then we have:
:$\map A n = \dfrac {\map {\sigma_1} n} {\map {\sigma_0} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} 6
      | r = 12
      | c = {{DSFLink|6}}
}}
{{eqn | l = \map {\sigma_0} 6
      | r = 4
      | c = {{DCFLink|6}}
}}
{{eqn | ll= \leadsto
      | l = \map A 6
      | r = \dfrac {12} 4 = 3
      | c = and so the arithmetic mean is an integer
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {28}
      | r = 56
      | c = {{DSFLink|28}}
}}
{{eqn | l = \map {\sigma_0} {28}
      | r = 4
      | c = {{DCFLink|28}}
}}
{{eqn | ll= \leadsto
      | l = \map A {28}
      | r = \dfrac {56} 6 = 9 \cdotp \dot 3
      | c = and so the arithmetic mean is not an integer
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {140}
      | r = 336
      | c = {{DSFLink|140}}
}}
{{eqn | l = \map {\sigma_0} {140}
      | r = 12
      | c = {{DCFLink|140}}
}}
{{eqn | ll= \leadsto
      | l = \map A {140}
      | r = \dfrac {336} {12} = 28
      | c = and so the arithmetic mean is an integer
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {270}
      | r = 720
      | c = {{DSFLink|270}}
}}
{{eqn | l = \map {\sigma_0} {270}
      | r = 16
      | c = {{DCFLink|270}}
}}
{{eqn | ll= \leadsto
      | l = \map A {270}
      | r = \dfrac {720} {16} = 45
      | c = and so the arithmetic mean is an integer
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {496}
      | r = 992
      | c = {{DSFLink|496}}
}}
{{eqn | l = \map {\sigma_0} {496}
      | r = 10
      | c = {{DCFLink|496}}
}}
{{eqn | ll= \leadsto
      | l = \map A {496}
      | r = \dfrac {992} {10} = 9 \cdotp 92
      | c = and so the arithmetic mean is not an integer
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {672}
      | r = 2016
      | c = {{DSFLink|672}}
}}
{{eqn | l = \map {\sigma_0} {672}
      | r = 24
      | c = {{DCFLink|672}}
}}
{{eqn | ll= \leadsto
      | l = \map A {672}
      | r = \dfrac {2016} {24} = 84
      | c = and so the arithmetic mean is an integer
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {1638}
      | r = 4368
      | c = {{DSFLink|1638}}
}}
{{eqn | l = \map {\sigma_0} {1638}
      | r = 24
      | c = {{DCFLink|1638}}
}}
{{eqn | ll= \leadsto
      | l = \map A {1638}
      | r = \dfrac {4368} {24} = 182
      | c = and so the arithmetic mean is an integer
}}
{{end-eqn}}
{{qed}}
Theorem: Carmichael's Theorem

Let $n \in \Z$ such that $n > 12$.
Then the $n$th Fibonacci number $F_n$ has at least one prime factor which does not divide any smaller Fibonacci number.
The exceptions for $n \le 12$ are:
:$F_1 = 1, F_2 = 1$: neither have any prime factors
:$F_6 = 8$ whose only prime factor is $2$ which is $F_3$
:$F_{12} = 144$ whose only prime factors are $2$ (which is $F_3$) and $3$ (which is $F_4$).

Proof:
We have that:
:$1$ has no prime factors.
Hence, vacuously, $1$ has no primitive prime factors.
:$8 = 2^3$
and $2 \divides 2 = F_3$
:$144 = 2^4 3^2$
and:
:$2 \divides 8 = F_6$
:$3 \divides 21 = F_8$
for example.
{{ProofWanted|It remains to be shown that these are the only examples.}}
Theorem: Weierstrass's Necessary Condition

Let $\mathbf y: \R \to \R^n$ be an $n$-dimensional vector-valued function such that $\map {\mathbf y} a = A$ and $\map {\mathbf y} b = B$.
Let $J$ be a functional such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let $\mathbf w$ be an $n$-dimensional vector such that $\mathbf w \in \R^n$.
Let $\gamma$ be a strong minimum of $J$.
{{explain|exactly what a ''strong'' minimum is, by means of a link to a definition page}}
Then along $\gamma$ and for every $\mathbf w$:
:$\map E {x, \mathbf y, \mathbf y', \mathbf w} \ge 0$
{{explain|exactly what ''along'' $\gamma$ means, by means of a link to a definition page}}
where $E$ stands for the Weierstrass E-Function.

Proof:
{{ProofWanted}}
{{Namedfor|Karl Theodor Wilhelm Weierstrass|cat = Weierstrass}}
Theorem: Magic Constant of Smallest Prime Magic Square with Consecutive Primes

The magic constant of the smallest prime magic square whose elements are consecutive odd primes is $4 \, 440 \, 084 \, 513$.

Proof:
The smallest prime magic square whose elements are consecutive odd primes is:
{{:Prime Magic Square/Examples/Order 3/Smallest with Consecutive Primes}}
As can be seen by inspection, the sums of the elements in the rows, columns and Diagonal of Array is $4 \, 440 \, 084 \, 513$:
{{begin-eqn}}
{{eqn | l = 1 \, 480 \, 028 \, 159 + 1 \, 480 \, 028 \, 153 + 1 \, 480 \, 028 \, 201
      | r = 4 \, 440 \, 084 \, 513
}}
{{eqn | l = 1 \, 480 \, 028 \, 213 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 129
      | r = 4 \, 440 \, 084 \, 513
}}
{{eqn | l = 1 \, 480 \, 028 \, 141 + 1 \, 480 \, 028 \, 189 + 1 \, 480 \, 028 \, 183
      | r = 4 \, 440 \, 084 \, 513
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1 \, 480 \, 028 \, 159 + 1 \, 480 \, 028 \, 213 + 1 \, 480 \, 028 \, 141
      | r = 4 \, 440 \, 084 \, 513
}}
{{eqn | l = 1 \, 480 \, 028 \, 153 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 189
      | r = 4 \, 440 \, 084 \, 513
}}
{{eqn | l = 1 \, 480 \, 028 \, 201 + 1 \, 480 \, 028 \, 129 + 1 \, 480 \, 028 \, 183
      | r = 4 \, 440 \, 084 \, 513
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 1 \, 480 \, 028 \, 159 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 183
      | r = 4 \, 440 \, 084 \, 513
}}
{{eqn | l = 1 \, 480 \, 028 \, 141 + 1 \, 480 \, 028 \, 171 + 1 \, 480 \, 028 \, 201
      | r = 4 \, 440 \, 084 \, 513
}}
{{end-eqn}}
{{ProofWanted|It remains to be demonstrated that there are no prime magic squares that are any smaller than this.}}
Category:Prime Magic Squares
Category:4,440,084,513
h04gygpinc4wf0ihrubo1lf8vghj185
Theorem: Magic Constant of Smallest Prime Magic Square with Consecutive Primes from 3

The magic constant of the smallest prime magic square whose elements are consecutive odd primes from $3$ upwards is $4514$.

Proof:
The smallest prime magic square whose elements are the first consecutive odd primes is:
{{:Prime Magic Square/Examples/Order 12/Smallest with Consecutive Primes from 3}}
The fact that this is the smallest is proven here.
The sum of the first $144$ prime numbers can either be calculated or looked up: it is $54 \, 169$.
We see that $2$ is not included in this prime magic square, but instead $1$ is used.
So the total of all the elements of this prime magic square is $54 \, 169 - 2 + 1 = 54 \, 168$.
There are $12$ rows, and $12$ columns, each with the same magic constant.
This magic constant must therefore be $\dfrac {54 \, 168} {12} = 4514$.
As can be seen by inspection, the sums of the elements in the rows, columns and diagonals is $4514$.
{{qed}}
Theorem: Factorions Base 10

The following positive integers are the only factorions base $10$:
:$1, 2, 145, 40 \, 585$

Proof:
From examples of factorials:
{{begin-eqn}}
{{eqn | l = 1
      | r = 1!
      | c = 
}}
{{eqn | l = 2
      | r = 2!
      | c = 
}}
{{eqn|| l = 145
      | r = 1 + 24 + 120
      | c = 
}}
{{eqn | r = 1! + 4! + 5!
      | c = 
}}
{{eqn|| l = 40 \, 585
      | r = 24 + 1 + 120 + 40 \, 320 + 120
      | c = 
}}
{{eqn | r = 4! + 0! + 5! + 8! + 5!
      | c = 
}}
{{end-eqn}}
A computer search can verify solutions under $2540160 = 9! \times 7$ in seconds.
Let $n$ be a $7$-digit number with $n > 2540160$.
Then the sum of the factorials of its digits is not more than $9! \times 7 = 2540160$.
So $n$ cannot be a factorion base $10$.
Now let $n$ be a $k$-digit number, for $k \ge 8$.
Then the sum of the factorials of its digits is not more than $9! \times k$.
But we have:
{{begin-eqn}}
{{eqn | l = n
      | o = \ge
      | r = 10^{k - 1}
}}
{{eqn | r = 10^7 \times 10^{k - 8}
}}
{{eqn | o = \ge
      | r = 10^7 \times \paren {1 + 9 \paren {k - 8} }
      | c = Bernoulli's Inequality
}}
{{eqn | o = >
      | r = 8 \times 9! \times \paren {9 k - 71}
      | c = $8 \times 9! = 2903040$
}}
{{eqn | r = 9! \paren {72 k - 71 \times 8}
}}
{{eqn | o = >
      | r = 9! \times k
      | c = $k \ge 8$
}}
{{eqn | o = \ge
      | r = \text{sum of the factorials of digits of } n
}}
{{end-eqn}}
So there are no more factorions base $10$.
{{qed}}
Theorem: Sum of 2 Squares in 2 Distinct Ways/Examples/145

$145$ can be expressed as the sum of two square numbers in two distinct ways:
{{begin-eqn}}
{{eqn | l = 145
      | r = 12^2 + 1^2
}}
{{eqn | r = 9^2 + 8^2
}}
{{end-eqn}}

Proof:
We have that:
:$145 = 5 \times 29$
Both $5$ and $29$ can be expressed as the sum of two distinct square numbers:
{{begin-eqn}}
{{eqn | l = 5
      | r = 1^2 + 2^2
}}
{{eqn | l = 29
      | r = 2^2 + 5^2
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 5^2}
      | c = 
}}
{{eqn | r = \paren {1 \times 2 + 2 \times 5}^2 + \paren {1 \times 5 - 2 \times 2}^2
      | c = Brahmagupta-Fibonacci Identity
}}
{{eqn | r = \paren {2 + 10}^2 + \paren {5 - 4}^2
      | c = 
}}
{{eqn | r = 12^2 + 1^2
      | c = 
}}
{{eqn | r = 144 + 1
      | c = 
}}
{{eqn | r = 145
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | r = \paren {1^2 + 2^2} \paren {2^2 + 5^2}
      | c = 
}}
{{eqn | r = \paren {1 \times 2 - 2 \times 5}^2 + \paren {1 \times 5 + 2 \times 2}^2
      | c = {{Corollary|Brahmagupta-Fibonacci Identity}}
}}
{{eqn | r = \paren {2 - 10}^2 + \paren {5 + 4}^2
      | c = 
}}
{{eqn | r = \paren {8 - 2}^2 + \paren {5 + 4}^2
      | c = 
}}
{{eqn | r = 8^2 + 9^2
      | c = 
}}
{{eqn | r = 64 + 81
      | c = 
}}
{{eqn | r = 145
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Representation of 1 as Sum of n Unit Fractions

Let $U \left({n}\right)$ denote the number of different ways of representing $1$ as the sum of $n$ unit fractions.
Then for various $n$, $U \left({n}\right)$ is given by the following table:
:{| border="1"
|-
! align="right" style = "padding: 2px 10px" | $n$ 
! align="right" style = "padding: 2px 10px" | $U \left({n}\right)$ 
|-
| align="right" style = "padding: 2px 10px" | $1$
| align="right" style = "padding: 2px 10px" | $1$
|-
| align="right" style = "padding: 2px 10px" | $2$
| align="right" style = "padding: 2px 10px" | $1$
|-
| align="right" style = "padding: 2px 10px" | $3$
| align="right" style = "padding: 2px 10px" | $3$
|-
| align="right" style = "padding: 2px 10px" | $4$
| align="right" style = "padding: 2px 10px" | $14$
|-
| align="right" style = "padding: 2px 10px" | $5$
| align="right" style = "padding: 2px 10px" | $147$
|-
| align="right" style = "padding: 2px 10px" | $6$
| align="right" style = "padding: 2px 10px" | $3462$
|}
{{OEIS|A002966}}

Proof:
Trivially:
:$1 = \dfrac 1 1$
and it follows that:
$U \left({1}\right) = 1$
Also trivially:
:$1 = \dfrac 1 2 + \dfrac 1 2$
and it follows that:
$U \left({2}\right) = 1$
From Sum of 3 Unit Fractions that equals 1:
:$U \left({3}\right) = 3$
From Sum of 4 Unit Fractions that equals 1:
:$U \left({4}\right) = 14$
From Sum of 5 Unit Fractions that equals 1:
:$U \left({5}\right) = 147$
{{ProofWanted|and for the rest}}
Theorem: Sum of 3 Unit Fractions that equals 1

There are $3$ ways to represent $1$ as the sum of exactly $3$ unit fractions.

Proof:
Let:
:$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$
where:
:$0 < a \le b \le c$
and:
{{AimForCont}} $a = 1$.
Then:
:$1 = \dfrac 1 1 + \dfrac 1 b + \dfrac 1 c$
and so:
:$\dfrac 1 b + \dfrac 1 c = 0$
which contradicts the stipulation that $b, c > 0$.
So there is no solution possible when $a = 1$.
Therefore $a \ge 2$.
Theorem: Repeated Sum of Cubes of Digits of Multiple of 3

Let $k \in \Z_{>0}$ be a positive integer.
Let $f: \Z_{>0} \to \Z_{>0}$ be the mapping defined as:
:$\forall m \in \Z_{>0}: \map f m = $ the sum of the cubes of the digits of $n$.
Let $n_0 \in \Z_{>0}$ be a (strictly) positive integer which is a multiple of $3$.
Consider the sequence:
:$s_n = \begin{cases} n_0 & : n = 0 \\
\map f {s_{n - 1} } & : n > 0 \end{cases}$
Then:
:$\exists r \in \N_{>0}: s_r = 153$
That is, by performing $f$ repeatedly on a multiple of $3$ eventually results in the pluperfect digital invariant $153$.

Proof:
We verify by brute force:
:Starting on $n_0 \le 2916 = 4 \times 9^3$, we will end on $153$.
{{qed|lemma}}
First we prove that if $3 \divides n_0$, then $3 \divides \map f {n_0}$.
From Divisibility by 3:
The sum of digits of $n_0$ is divisible by $3$.
By Fermat's Little Theorem:
:$\forall x \in \Z: x^3 \equiv x \pmod 3$
Therefore sum of the cubes of the digits of $n_0$ is also divisible by $3$.
That is the definition of $\map f {n_0}$.
Thus $3 \divides \map f {n_0}$.
{{qed|lemma}}
It remains to be shown that for every $n_0 > 2916$, $\map f {n_0} < n_0$.
For $n_0 \le 9999$:
:$\map f {n_0} \le 9^3 + 9^3 + 9^3 + 9^3 = 2916 < n_0$
Now suppose $n_0$ is a $k$-digit number, where $k \ge 5$.
Then:
{{begin-eqn}}
{{eqn | l = n_0
      | o = >
      | r = 10^{k - 1}
      | c = the smallest $k$-digit number
}}
{{eqn | r = 10^4 \times 10^{k - 5}
}}
{{eqn | o = \ge
      | r = 10^4 \paren {1 + 9 \paren {k - 5} }
      | c = Bernoulli's Inequality
}}
{{eqn | o = >
      | r = 10 \times 9^3 \paren {9 k - 44}
      | c = $10 > 9$
}}
{{eqn | o = >
      | r = 9^3 \paren {90 k - 440}
}}
{{eqn | o = \ge
      | r = 9^3 \paren {k + 445 - 440}
      | c = $k \ge 5$
}}
{{eqn | o = >
      | r = 9^3 \times k
}}
{{eqn | o = \ge
      | r = \map f {n_0}
}}
{{end-eqn}}
This shows that $\map f {n_0} < n_0$ for all $n_0 > 2916$.
Thus for any $n_0 > 2916$, $s_n$ is eventually less than $2916$.
Therefore we will eventually reach $153$.
{{qed}}
Theorem: 159 is not Expressible as Sum of Fewer than 19 Fourth Powers

:$159 = 14 \times 1^4 + 4 \times 2^4 + 3^4$

Proof:
We have:
:$4^4 = 256 > 159$
:$3^4 = 81$
:$2^4 = 16$
:$1^4 = 1$
Let us attempt to construct an expression of $159$ as the sum of fewer than $19$ fourth powers:
If no $3^4$ is used in our sum, the sum consists only of $2^4$ and $1^4$.
Using $2^4$ is more efficient than using $1^4$, since $2^4$ can replace $16 \times 1^4$.
So we have:
:$159 = 9 \times 2^4 + 15 \times 1^4$
which uses $24$ fourth powers.
If $3^4$ is used in our sum, the most efficient way is demonstrated above:
:$159 = 14 \times 1^4 + 4 \times 2^4 + 3^4$
which uses $19$ fourth powers.
We cannot use more than $2$ instances of $3^4$, as $2 \times 3^4 = 162 > 159$.
{{qed}}
Theorem: Numbers not Expressible as Sum of Distinct Pentagonal Numbers

The positive integers which cannot be expressed as the sum of distinct pentagonal numbers are:
:$2, 3, 4, 7, 8, 9, 10, 11, 14, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30,$
:$31, 32, 33, 37, 38, 42, 43, 44, 45, 46, 49, 50, 54, 55, 59, 60, 61, 65,$
:$66, 67, 72, 77, 80, 81, 84, 89, 94, 95, 96, 100, 101, 102, 107, 112, 116,$
:$124, 136, 137, 141, 142, 147, 159$

Proof:
It will be proved that the largest integer which cannot be expressed as the sum of distinct pentagonal numbers is $159$.
The remaining integers in the sequence can be identified by inspection.
We prove this using a variant of Second Principle of Mathematical Induction.
Let $\map P n$ be the proposition:
:$n$ can be expressed as the sum of distinct pentagonal numbers.
In the below, $P_n$ denotes the $n$th pentagonal number.