avewright/proofwiki-math huggingface dataset:
Theorem: 23 is Largest Integer not Sum of Distinct Perfect Powers

The largest integer which cannot be expressed as the sum of distinct perfect powers is $23$.

Proof:
The first few perfect powers are:
:$1, 4, 8, 9, 16, 25, 27, 32, \dots$
First we show that $23$ cannot be expressed as the sum of distinct perfect powers.
Only $1, 4, 8, 9, 16$ are perfect powers less than $23$.
Suppose $23$ can be so expressed.
Since $1 + 4 + 8 + 9 = 22 < 23$, $16$ must be used in the sum.
However $23 - 16 = 7$ cannot be expressed as a sum of $1$ and $4$.
Thus $23$ cannot be expressed as the sum of distinct perfect powers.
{{qed|lemma}}
Now we show that all numbers greater than $23$ can be so expressed.
By Richert's Theorem, we just need to show:
:For any $23 < n \le 23 + 32$, $n$ can be expressed as a sum of distinct elements in $\set {1, 4, 8, 9, 16, 25, 27}$
:$s_{i + 1} \le 2 s_i$ for every $i \ge 7$, where $s_i$ is the $i$th perfect power
Verification of the first statement is included in the bottom of this proof.
To verify the second statement:
Let $i \ge 7$.
Let $m$ be the integer satisfying:
:$2^{m + 1} > s_i \ge 2^m$
Note that $2^{m + 1}$ is also a perfect power.
There must be a perfect power greater than $s_i$ but not greater than $2^{m + 1}$.
Thus:
:$s_{i + 1} \le 2^{m + 1} = 2 \cdot 2^m \le 2 s_i$
Therefore $23$ is the largest integer that cannot be expressed as the sum of distinct perfect powers..
{{qed|lemma}}
Here is $23 < n \le 55$ expressed as a sum of distinct elements in $\set {1, 4, 8, 9, 16, 25, 27}$:
{{begin-eqn}}
{{eqn | l = 24
      | r = 16 + 8
}}
{{eqn | l = 25
      | r = 25
}}
{{eqn | l = 26
      | r = 25 + 1
}}
{{eqn | l = 27
      | r = 27
}}
{{eqn | l = 28
      | r = 27 + 1
}}
{{eqn | l = 29
      | r = 25 + 4
}}
{{eqn | l = 30
      | r = 25 + 4 + 1
}}
{{eqn | l = 31
      | r = 27 + 4
}}
{{eqn | l = 32
      | r = 27 + 4 + 1
}}
{{eqn | l = 33
      | r = 25 + 8
}}
{{eqn | l = 34
      | r = 25 + 9
}}
{{eqn | l = 35
      | r = 25 + 9 + 1
}}
{{eqn | l = 36
      | r = 27 + 9
}}
{{eqn | l = 37
      | r = 27 + 9 + 1
}}
{{eqn | l = 38
      | r = 25 + 9 + 4
}}
{{eqn | l = 39
      | r = 25 + 9 + 4 + 1
}}
{{eqn | l = 40
      | r = 27 + 9 + 4
}}
{{eqn | l = 41
      | r = 27 + 9 + 4 + 1
}}
{{eqn | l = 42
      | r = 25 + 9 + 8
}}
{{eqn | l = 43
      | r = 25 + 9 + 8 + 1
}}
{{eqn | l = 44
      | r = 27 + 9 + 8
}}
{{eqn | l = 45
      | r = 27 + 9 + 8 + 1
}}
{{eqn | l = 46
      | r = 25 + 9 + 8 + 4
}}
{{eqn | l = 47
      | r = 25 + 9 + 8 + 4 + 1
}}
{{eqn | l = 48
      | r = 27 + 9 + 8 + 4
}}
{{eqn | l = 49
      | r = 27 + 9 + 8 + 4 + 1
}}
{{eqn | l = 50
      | r = 25 + 16 + 9
}}
{{eqn | l = 51
      | r = 25 + 16 + 9 + 1
}}
{{eqn | l = 52
      | r = 27 + 25
}}
{{eqn | l = 53
      | r = 27 + 25 + 1
}}
{{eqn | l = 54
      | r = 25 + 16 + 9 + 4
}}
{{eqn | l = 55
      | r = 25 + 16 + 9 + 4 + 1
}}
{{end-eqn}}
{{qed}}
Theorem: Smallest Integer not Sum of Two Ulam Numbers

The smallest integer greater than $1$ which is not the sum of two Ulam numbers is $23$.

Proof:
Recall the Ulam numbers:
{{:Definition:Ulam Number/Sequence}}
We have:
{{begin-eqn}}
{{eqn | l = 2
      | r = 1 + 1
}}
{{eqn | l = 3
      | r = 2 + 1
}}
{{eqn | l = 4
      | r = 3 + 1
}}
{{eqn | r = 2 + 2
}}
{{eqn | l = 5
      | r = 4 + 1
}}
{{eqn | r = 3 + 2
}}
{{eqn | l = 6
      | r = 4 + 2
}}
{{eqn | r = 3 + 3
}}
{{eqn | l = 7
      | r = 6 + 1
}}
{{eqn | r = 4 + 3
}}
{{eqn | l = 8
      | r = 6 + 2
}}
{{eqn | r = 4 + 4
}}
{{eqn | l = 9
      | r = 8 + 1
}}
{{eqn | r = 6 + 3
}}
{{eqn | l = 10
      | r = 8 + 2
}}
{{eqn | r = 6 + 4
}}
{{eqn | l = 11
      | r = 8 + 3
}}
{{eqn | l = 12
      | r = 11 + 1
}}
{{eqn | r = 8 + 4
}}
{{eqn | r = 6 + 6
}}
{{eqn | l = 13
      | r = 11 + 2
}}
{{eqn | l = 14
      | r = 13 + 1
}}
{{eqn | r = 11 + 3
}}
{{eqn | r = 8 + 6
}}
{{eqn | l = 15
      | r = 13 + 2
}}
{{eqn | r = 11 + 4
}}
{{eqn | l = 16
      | r = 13 + 3
}}
{{eqn | r = 8 + 8
}}
{{eqn | l = 17
      | r = 16 + 1
}}
{{eqn | r = 13 + 4
}}
{{eqn | r = 11 + 6
}}
{{eqn | l = 18
      | r = 16 + 2
}}
{{eqn | l = 19
      | r = 18 + 1
}}
{{eqn | r = 16 + 3
}}
{{eqn | r = 13 + 6
}}
{{eqn | r = 11 + 8
}}
{{eqn | l = 20
      | r = 18 + 2
}}
{{eqn | r = 16 + 4
}}
{{eqn | l = 21
      | r = 18 + 3
}}
{{eqn | r = 13 + 8
}}
{{eqn | l = 22
      | r = 18 + 4
}}
{{eqn | r = 16 + 6
}}
{{eqn | r = 11 + 11
}}
{{end-eqn}}
Now consider the the difference between $23$ and successive Ulam numbers:
{{begin-eqn}}
{{eqn | l = 23 - 18
      | r = 5
      | c = not a Ulam number
}}
{{eqn | l = 23 - 16
      | r = 7
      | c = not a Ulam number
}}
{{eqn | l = 23 - 13
      | r = 10
      | c = not a Ulam number
}}
{{eqn | l = 23 - 11
      | r = 12
      | c = not a Ulam number
}}
{{end-eqn}}
and it is not necessary to go further back.
Hence the result.
{{qed}}
Theorem: Numbers with Square-Free Binomial Coefficients

For every $n$ greater than $23$, there exists a binomial coefficient $\dbinom n k$ that is not square-free.
More specifically, the list of numbers $n$ such that $\dbinom n k$ are square-free for all $k = 0, \dots, n$ is given by:
:$1, 2, 3, 5, 7, 11, 23$
{{OEIS|A048278}}

Proof:
=== Lemma ===
{{:Numbers with Square-Free Binomial Coefficients/Lemma}}{{qed|lemma}}
Theorem: Infima Inheriting Ordered Subset of Complete Lattice is Complete Lattice

Let $L = \struct {X, \preceq}$ be a complete lattice.
Let $S = \struct {T, \precsim}$ be an infima inheriting ordered subset of $L$.
Then $S$ is a complete lattice.

Proof:
Let $A$ be subset of $T$.
By definition of complete lattice:
:$A$ admits an infimum in $L$.
Thus by definition of infima inheriting:
:$A$ admits an infimum in $S$.
Hence by dual of Lattice is Complete iff it Admits All Suprema:
:$S$ is a complete lattice.
{{qed}}
Theorem: Sum of Reciprocals in Base 10 with Zeroes Removed

The infinite series
:$\ds \sum_{\map P n} \frac 1 n$
where $\map P n$ is the propositional function:
:$\forall n \in \Z_{>0}: \map P n \iff$ the decimal representation of $n$ contains no instances of the digit $0$
converges to the approximate limit $23 \cdotp 10345 \ldots$

Proof:
For each $k \in \N$, there are $9^k$ $k$-digit numbers containing no instances of the digit $0$.
Each of these numbers is at least $10^{k - 1}$.
Hence the reciprocals of each of these numbers is at most $\dfrac 1 {10^{k - 1}}$.
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{\map P n} \frac 1 n
      | o = <
      | r = \sum_{k \mathop = 1}^\infty \frac {9^k} {10^{k - 1} }
}}
{{eqn | r = \frac 9 {1 - \frac 9 {10} }
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 90
}}
{{end-eqn}}
showing that the sum converges.
{{finish|Finer approximations can be obtained (e.g. in virtue of Closed Form for Triangular Numbers/Direct Proof), but due to the slow growth of the harmonic series, many numbers must to summed to obtain the approximation in the theorem. Case in point: $H_{5 \times 10^9} < 23$}}
Theorem: Gelfond's Constant minus Pi

Gelfond's constant minus $\pi$ is very close to $20$:
:$e^\pi - \pi \approx 20$

Proof:
We have:
:$e^\pi \approx 23 \cdotp 14069 \, 26327 \, 79269 \ldots$
:$\pi \approx 3 \cdotp 14159 \, 26535 \, 89793$
Then:
<pre>
 23.14069 26327 79269
- 3.14159 26535 89793
---------------------
 19.99909 99791 89486
---------------------
</pre>{{qed}}
Theorem: Sums of Consecutive Sequences of Squares that equal Squares

Apart from $1$, the $24$th square pyramidal number is the only one which is square:
:$1^2 + 2^2 + 3^2 + \cdots + 24^2 = 70^2$
while there are several Sum of Sequence of Squares which are square, for example:
:$18^2 + 19^2 + \cdots + 28^2 = 77^2$
and:
:$25^2 + 26^2 + \cdots + 624^2 = 9010^2$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 1^2 + 2^2 + 3^2 + \cdots + 24^2
      | r = \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6
      | c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac {24 \times 25 \times 49} 6
      | c = 
}}
{{eqn | r = \dfrac {2^3 \times 3 \times 5^2 \times 7^2} {2 \times 3}
      | c = 
}}
{{eqn | r = 2^2 \times 5^2 \times 7^2
      | c = 
}}
{{eqn | r = \paren {2 \times 5 \times 7}^2
      | c = 
}}
{{eqn | r = 70^2
      | c = 
}}
{{end-eqn}}
{{finish|It remains to be shown that this is the only one}}
Finally we note:
{{begin-eqn}}
{{eqn | l = 18^2 + 19^2 + \cdots + 28^2
      | r = \dfrac {28 \times \paren {28 + 1} \times \paren {2 \times 28 + 1} } 6 - \dfrac {17 \times \paren {17 + 1} \times \paren {2 \times 17 + 1} } 6
      | c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac {28 \times 29 \times 57 - 17 \times 18 \times 35} 6
      | c = 
}}
{{eqn | r = \dfrac {\paren {2^2 \times 7} \times 29 \times \paren {3 \times 19} - 17 \times \paren {2 \times 3^2} \times \paren {5 \times 7} } {2 \times 3}
      | c = 
}}
{{eqn | r = 2 \times 7 \times 29 \times 19 - 17 \times 3 \times 5 \times 7
      | c = 
}}
{{eqn | r = 7 \times \paren {1102 - 255}
      | c = 
}}
{{eqn | r = 7 \times 847
      | c = 
}}
{{eqn | r = 7 \times 7 \times 11^2
      | c = 
}}
{{eqn | r = 77^2
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = 25^2 + 26^2 + \cdots + 624^2
      | r = \dfrac {624 \times \paren {624 + 1} \times \paren {2 \times 624 + 1} } 6 - \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6
      | c = Sum of Sequence of Squares
}}
{{eqn | r = \dfrac {624 \times 625 \times 1249 - 24 \times 25 \times 49} 6
      | c = 
}}
{{eqn | r = \dfrac {\paren {2^4 \times 3 \times 13} \times 5^4 \times 1249 - \paren {2^3 \times 3} \times 5^2 \times 7^2} {2 \times 3}
      | c = 
}}
{{eqn | r = 2^3 \times 5^4 \times 13 \times 1249 - 2^2 \times 5^2 \times 7^2
      | c = 
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {2 \times 5^2 \times 13 \times 1249 - 7^2}
      | c = 
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {811 \, 850 - 49}
      | c = 
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {811 \, 801}
      | c = 
}}
{{eqn | r = 2^2 \times 5^2 \times \paren {17^2 \times 53^2}
      | c = 
}}
{{eqn | r = 9010^2
      | c = 
}}
{{end-eqn}}
Theorem: Sum of Squares of Divisors of 24 and 26 are Equal

The sum of the squares of the divisors of $24$ equals the sum of the squares of the divisors of $26$:
:$\map {\sigma_2} {24} = \map {\sigma_2} {26}$
where $\sigma_\alpha$ denotes the divisor function.

Proof:
The divisors of $24$ are:
:$1, 2, 3, 4, 6, 8, 12, 24$
The divisors of $26$ are:
:$1, 2, 13, 26$
Then we have:
{{begin-eqn}}
{{eqn | r = 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 8^2 + 12^2 + 24^2
      | o = 
      | c = 
}}
{{eqn | r = 1 + 4 + 9 + 16 + 36 + 64 + 144 + 576
      | c = 
}}
{{eqn | r = 850
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | r = 1^2 + 2^2 + 13^2 + 26^2
      | o = 
      | c = 
}}
{{eqn | r = 1 + 4 + 169 + 676
      | c = 
}}
{{eqn | r = 850
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Smallest Scalene Obtuse Triangle with Integer Sides and Area

The smallest scalene obtuse triangle with integer sides and area has sides of length $4, 13, 15$.

Proof:
From Heron's Formula, the area $A$ of $\triangle ABC$ is given by:
:$A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
Here we have:
{{begin-eqn}}
{{eqn | l = s
      | r = \dfrac {4 + 13 + 15} 2
      | c = 
}}
{{eqn | r = 16
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = A
      | r = \sqrt {16 \paren {16 - 4} \paren {16 - 13} \paren {16 - 15} }
      | c = 
}}
{{eqn | r = \sqrt {16 \times 12 \times 3 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {2^4 \times 2^2 \times 3 \times 3 \times 1}
      | c = 
}}
{{eqn | r = \sqrt {2^6 \times 3^2}
      | c = 
}}
{{eqn | r = 2^3 \times 3
      | c = 
}}
{{eqn | r = 24
      | c = 
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that it is the smallest. This can be done by exhaustion: the number of triples defining an obtuse scalene triangle are not that many.}}
Theorem: Image of Compact Subset under Directed Suprema Preserving Closure Operator

Let $L = \struct {S, \preceq}$ be a bounded below algebric lattice.
Let $c: S \to S$ be a closure operator that preserves directed suprema.
Then:
:$c \sqbrk {\map K L} = \map K {\struct {c \sqbrk S, \precsim} }$
where
:$\map K L$ denotes the compact subset of $L$,
:$c \sqbrk S$ denotes the image of $S$ under $c$,
:$\mathord \precsim = \mathord \preceq \cap \paren {c \sqbrk S \times c \sqbrk S}$

Proof:
We will prove that:
:$\map K {\struct {c \sqbrk S, \precsim} } \subseteq c \sqbrk {\map K L}$
By Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset:
:$c \sqbrk {\map K L} \subseteq \map K {\struct {c \sqbrk S, \precsim} }$
Thus the result by definition of set equality.
{{qed}}
Theorem: Smallest Positive Integer with 5 Fibonacci Partitions

The smallest positive integer which can be partitioned into distinct Fibonacci numbers in $5$ different ways is $24$.

Proof:
{{begin-eqn}}
{{eqn | l = 1
      | r = 1
      | c = $1$ way
}}
{{eqn | l = 2
      | r = 2
      | c = $1$ way
}}
{{eqn | l = 3
      | r = 3
}}
{{eqn | r = 2 + 1
      | c = $2$ ways
}}
{{eqn | l = 4
      | r = 3 + 1
      | c = $1$ way
}}
{{eqn | l = 5
      | r = 5
}}
{{eqn | r = 3 + 2
      | c = $2$ ways
}}
{{eqn | l = 6
      | r = 5 + 1
}}
{{eqn | r = 3 + 2 + 1
      | c = $2$ ways
}}
{{eqn | l = 7
      | r = 5 + 2
      | c = $1$ way
}}
{{eqn | l = 8
      | r = 8
}}
{{eqn | r = 5 + 3
}}
{{eqn | r = 5 + 2 + 1
      | c = $3$ ways
}}
{{eqn | l = 9
      | r = 8 + 1
}}
{{eqn | r = 5 + 3 + 1
      | c = $2$ ways
}}
{{eqn | l = 10
      | r = 8 + 2
}}
{{eqn | r = 5 + 3 + 2
      | c = $2$ ways
}}
{{eqn | l = 11
      | r = 8 + 3
}}
{{eqn | r = 8 + 2 + 1
}}
{{eqn | r = 5 + 3 + 2 + 1
      | c = $3$ ways
}}
{{eqn | l = 12
      | r = 8 + 3 + 1
      | c = $1$ way
}}
{{eqn | l = 13
      | r = 13
}}
{{eqn | r = 8 + 5
}}
{{eqn | r = 8 + 3 + 2
      | c = $3$ ways
}}
{{eqn | l = 14
      | r = 13 + 1
}}
{{eqn | r = 8 + 5 + 1
}}
{{eqn | r = 8 + 3 + 2 + 1
      | c = $3$ ways
}}
{{eqn | l = 15
      | r = 13 + 2
}}
{{eqn | r = 8 + 5 + 2
      | c = $2$ ways
}}
{{eqn | l = 16
      | r = 13 + 3
}}
{{eqn | r = 13 + 2 + 1
}}
{{eqn | r = 8 + 5 + 3
}}
{{eqn | r = 8 + 5 + 2 + 1
      | c = $4$ ways
}}
{{eqn | l = 17
      | r = 13 + 3 + 1
}}
{{eqn | r = 8 + 5 + 3 + 1
      | c = $2$ ways
}}
{{eqn | l = 18
      | r = 13 + 5
}}
{{eqn | r = 13 + 3 + 2
}}
{{eqn | r = 8 + 5 + 3 + 2
      | c = $3$ ways
}}
{{eqn | l = 19
      | r = 13 + 5 + 1
}}
{{eqn | r = 13 + 3 + 2 + 1
}}
{{eqn | r = 8 + 5 + 3 + 2 + 1
      | c = $3$ ways
}}
{{eqn | l = 20
      | r = 13 + 5 + 2
      | c = $1$ way
}}
{{eqn | l = 21
      | r = 21
}}
{{eqn | r = 13 + 8
}}
{{eqn | r = 13 + 5 + 3
}}
{{eqn | r = 13 + 5 + 2 + 1
      | c = $4$ ways
}}
{{eqn | l = 22
      | r = 21 + 1
}}
{{eqn | r = 13 + 8 + 1
}}
{{eqn | r = 13 + 5 + 3 + 1
      | c = $3$ ways
}}
{{eqn | l = 23
      | r = 21 + 2
}}
{{eqn | r = 13 + 8 + 2
}}
{{eqn | r = 13 + 5 + 3 + 2
      | c = $3$ ways
}}
{{eqn | l = 24
      | r = 21 + 3
}}
{{eqn | r = 21 + 2 + 1
}}
{{eqn | r = 13 + 8 + 3
}}
{{eqn | r = 13 + 8 + 2 + 1
}}
{{eqn | r = 13 + 5 + 3 + 2 + 1
      | c = $5$ ways
}}
{{end-eqn}}
{{OEIS|A000119}}{{qed}}
Theorem: Numbers Divisible by Sum and Product of Digits

The sequence of positive integers which are divisible by both the sum and product of its digits begins:
:$1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 24, 36, 111, 112, 132, 135, \ldots$
{{OEIS|A038186}}

Proof:
Let $S$ be the set of all positive integers which are divisible by both the sum and product of their digits.
Trivially, the sum and product of the digits of a one-digit number $n$ are themselves $n$.
Thus from Integer Divides Itself, the positive integers from $1$ to $9$ are in $S$.
The product of any integer with a $0$ in it is $0$.
From Zero Divides Zero $0$ is not a divisor of $n$ unless $n$ is itself $0$.
So $10, 20, 30, \ldots$ are not in $S$.
For all prime numbers $p$ with $2$ or more digits, the sum of its digits is greater than $1$ and less than $p$.
Thus $p$ is not a multiple of the sum of its digits.
{{ProofWanted|Having eliminated a fair number of potential candidates, the remaining ones can be found by trial and error.}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$115 = 5 \times 23$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {115}
      | r = \frac {5^2 - 1} {5 - 1} \times \frac {23^2 - 1} {23 - 1}
      | c = 
}}
{{eqn | r =  \frac {4 \times 6} 4 \times \frac {22 \times 24} {22}
      | c = Difference of Two Squares
}}
{{eqn | r = 6 \times 24
      | c = 
}}
{{eqn | r = \paren {2 \times 3} \times \paren {2^3 \times 3}
      | c = 
}}
{{eqn | r = 2^4 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$115 = 5 \times 23$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {115}
      | r = \paren {5 + 1} \paren {23 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 6 \times 24
      | c = 
}}
{{eqn | r = \paren {2 \times 3} \times \paren {2^3 \times 3}
      | c = 
}}
{{eqn | r = 2^4 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2^2 \times 3}^2
      | c = 
}}
{{eqn | r = 12^2
      | c = 
}}
{{eqn | r = 144
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$14 = 2 \times 7$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1}  {14}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {7^2 - 1} {7 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac {48} 6
      | c = 
}}
{{eqn | r = 3 \times 8
      | c = 
}}
{{eqn | r = 24
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$14 = 2 \times 7$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {14}
      | r = \paren {2 + 1} \paren {7 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 3 \times 8
      | c = 
}}
{{eqn | r = 24
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$15 = 3 \times 5$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {15}
      | r = \frac {3^2 - 1} {3 - 1} \times \frac {5^2 - 1} {5 - 1}
      | c = 
}}
{{eqn | r = \frac 8 2 \times \frac {24} 4
      | c = 
}}
{{eqn | r = 4 \times 6
      | c = 
}}
{{eqn | r = 24
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$15 = 3 \times 5$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {15}
      | r = \paren {3 + 1} \paren {5 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 4 \times 6
      | c = 
}}
{{eqn | r = 24
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$206 = 2 \times 103$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {206}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {103^2 - 1} {103 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac {10 \, 608} {102}
      | c = 
}}
{{eqn | r = 3 \times 104
      | c = 
}}
{{eqn | r = 312
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$206 = 2 \times 103$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {206}
      | r = \paren {2 + 1} \paren {103 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 3 \times 104
      | c = 
}}
{{eqn | r = 312
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.
We have that:
:$22 = 2 \times 11$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {22}
      | r = \frac {2^2 - 1} {2 - 1} \times \frac {11^2 - 1} {11 - 1}
      | c = 
}}
{{eqn | r = \frac 3 1 \times \frac {120} {10}
      | c = 
}}
{{eqn | r = 3 \times 12
      | c = 
}}
{{eqn | r = 3 \times \paren {2^2 \times 3}
      | c = 
}}
{{eqn | r = 2^2 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2 \times 3}^2
      | c = 
}}
{{eqn | r = 6^2
      | c = 
}}
{{eqn | r = 36
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Non-Square Semiprime

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.
Then:
:$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$
where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof:
We have that:
:$22 = 2 \times 11$
and so by definition is a semiprime whose prime factors are distinct.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {22}
      | r = \paren {2 + 1} \paren {11 + 1}
      | c = Divisor Sum of Non-Square Semiprime
}}
{{eqn | r = 3 \times 12
      | c = 
}}
{{eqn | r = 3 \times \paren {2^2 \times 3}
      | c = 
}}
{{eqn | r = 2^2 \times 3^2
      | c = 
}}
{{eqn | r = \paren {2 \times 3}^2
      | c = 
}}
{{eqn | r = 6^2
      | c = 
}}
{{eqn | r = 36
      | c = 
}}
{{end-eqn}}
{{qed}}