hcju/proofwikiqa huggingface dataset:
Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in an arithmetic sequence. Then: {{begin-eqn}} {{eqn | l = \dbinom n {k + 2} - \dbinom n {k + 1}       | r = \dbinom n {k + 1} - \dbinom n k       | c = {{Defof|Arithmetic Sequence}} }} {{eqn | l = \frac {n!} {\paren {n - k - 2}! \paren {k + 2}!} - \frac {n!} {\paren {n - k - 1}! \paren {k + 1}!}       | r = \frac {n!} {\paren {n - k - 1}! \paren {k + 1}!} - \frac {n!} {\paren {n - k}! \paren k!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | l = \paren {n - k - 1} \paren {n - k} - \paren {n - k} \paren {k + 2}       | r = \paren {n - k} \paren {k + 2} - \paren {k + 1} \paren {k + 2}       | c = Multiply both sides by $\dfrac {\paren {n - k}! \paren {k + 2}!} {n!}$ }} {{eqn | l = n^2 - k n - k n + k^2 - n + k - n k - 2 n + k^2 + 2 k       | r = n k + 2 n - k^2 - 2 k - k^2 - 2 k - k - 2       | c =  }} {{eqn | l = n^2 - \paren {4 k + 5} n + \paren {4 k^2 + 8 k + 2}       | r = 0       | c =  }} {{eqn | l = n       | r = \frac {4 k + 5 \pm \sqrt {\paren {4 k + 5}^2 - 4 \paren {4 k^2 + 8 k + 2} } } 2       | c = Quadratic Formula }} {{eqn | r = \frac {4 k + 5 \pm \sqrt {16 k^2 + 40 k + 25 - 16 k^2 - 32 k - 8} } 2       | c =  }} {{eqn | r = \frac {4 k + 5 \pm \sqrt {8 k + 17} } 2       | c =  }} {{end-eqn}} Since $n$ is rational, we require $8 k + 17$ to be a square. Since $8 k + 17$ is odd, if $8 k + 17$ is square, then $\sqrt {8 k + 17}$ is odd. Write $\sqrt {8 k + 17} = 2 x + 1$. Notice that: {{begin-eqn}} {{eqn | l = \paren {2 x + 1}^2       | r = 4 x^2 + 4 x + 1 }} {{eqn | r = 8 \paren {\frac {x^2 + x - 4} 2} + 17 }} {{end-eqn}} Using the substitution $k = \dfrac {x^2 + x - 4} 2$: {{begin-eqn}} {{eqn | l = n       | r = \frac {2 x^2 + 2 x - 8 + 5 \pm \paren {2 x + 1} } 2 }} {{eqn | r = x^2 - 2 \text { or } x^2 + 2 x - 1 }} {{eqn | r = x^2 - 2 \text { or } \paren {x + 1}^2 - 2 }} {{end-eqn}} Each $x$ with $k = \dfrac {x^2 + x - 4} 2 > 0$ give a value for $n$. Therefore there are an infinite number of rows of Pascal's triangle which contain $3$ integers in arithmetic sequence. {{qed}}
Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in a geometric sequence. Then: {{begin-eqn}} {{eqn | l = \dbinom n {k + 2} / \dbinom n {k + 1}       | r = \dbinom n {k + 1} / \dbinom n k       | c = {{Defof|Geometric Sequence}} }} {{eqn | l = \paren {\frac {n!} {\paren {n - k - 2}! \paren {k + 2}!} } \paren {\frac {\paren {n - k - 1}! \paren {k + 1}!} {n!} }       | r = \paren {\frac {n!} {\paren {n - k - 1}! \paren {k + 1}!} } \paren {\frac {\paren {n - k}! \paren k!} {n!} }       | c = {{Defof|Binomial Coefficient}} }} {{eqn | l = \frac {n - k - 1} {k + 2}       | r = \frac {n - k} {k + 1}       | c =  }} {{eqn | l = \paren {n - k - 1} \paren {k + 1}       | r = \paren {n - k} \paren {k + 2}       | c =  }} {{eqn | l = n k - k^2 - k + n -k - 1       | r = n k + 2 n - k^2 - 2 k       | c =  }} {{eqn | l = n       | r = -1       | c =  }} {{end-eqn}} Since $n > 0$, no row of Pascal's triangle contains $3$ integers in geometric sequence. However, suppose one extends the definition of binomial coefficients to allow $n < 0$. Then by Negated Upper Index of Binomial Coefficient, we have:  :$\dbinom {-1} k = \paren {-1}^k$ which indeed forms a geometric sequence. {{qed}}
Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in a harmonic sequence. Then: {{begin-eqn}} {{eqn | l = \paren {\dbinom n {k + 2} }^{-1} - \paren {\dbinom n {k + 1} }^{-1}       | r = \paren {\dbinom n {k + 1} }^{-1} - \paren {\dbinom n k}^{-1}       | c = {{Defof|Harmonic Sequence}} }} {{eqn | l = \frac {\paren {n - k - 2}! \paren {k + 2}!} {n!} - \frac {\paren {n - k - 1}! \paren {k + 1}!} {n!}       | r = \frac {\paren {n - k - 1}! \paren {k + 1}!} {n!} - \frac {\paren {n - k}! \paren k!} {n!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | l = \paren {k + 2} \paren {k + 1} - \paren {n - k - 1} \paren {k + 1}       | r = \paren {n - k - 1} \paren {k + 1} - \paren {n - k} \paren {n - k - 1}       | c = Multiply both sides by $\dfrac {n!} {\paren {n - k - 2}! \paren k!}$ }} {{eqn | l = k^2 + 3 k + 2 - n k + k^2 + k - n + k + 1       | r = n k - k^2 - k + n - k - 1 - n^2 + n k + n + k n - k^2 - k       | c =  }} {{eqn | l = n^2 - \paren {4 k + 3} n + \paren {4 k^2 + 8 k + 4}       | r = 0       | c =  }} {{end-eqn}} Notice that for each $k \ge 0$: :$\paren {4 k + 3}^2 - 4 \paren {4 k^2 + 8 k + 4} = - \paren {8 k + 7} < 0$ By Solution to Quadratic Equation with Real Coefficients, there is no real solution for $n$. Therefore there is no row of Pascal's triangle which contain $3$ integers in harmonic sequence. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \dbinom {n - 1} k \dfrac 1 {k + 1}       | r = \frac 1 n \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom n {k + 1}       | c = Factors of Binomial Coefficient }} {{eqn | r = \frac 1 n \sum_{k \mathop = 1}^n \paren {-1}^{k + 1} \binom n k       | c = Translation of Index Variable of Summation }} {{eqn | r = \frac 1 n \paren {\sum_{k \mathop = 0}^n \paren {-1}^{k + 1} \binom n k + 1}       | c = $\paren {-1}^1 \dbinom n 0 = -1$ }} {{eqn | r = \frac 1 n \paren {-\paren {1 - 1}^n + 1}       | c = Binomial Theorem }} {{eqn | r = \frac 1 n }} {{end-eqn}} {{qed}}
By definition of Leibniz harmonic triangle, element $\tuple {n, m}$ is: :$\dfrac 1 {\paren {n + 1} \binom n m}$ Thus we have: {{begin-eqn}} {{eqn | r = \dfrac 1 {\paren {n + 2} \binom {n + 1} m} + \dfrac 1 {\paren {n + 2} \binom {n + 1} {m + 1} }       | o =        | c = Elements of Leibniz Harmonic Triangle immediately below }} {{eqn | r = \dfrac {\binom {n + 1} {m + 1} + \binom {n + 1} m} {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }       | c =  }} {{eqn | r = \dfrac {\binom {n + 2} {m + 1} } {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }       | c = Pascal's Rule }} {{eqn | r = \dfrac {\frac {\paren {n + 2}!} {\paren {m + 1}! \paren {n - m + 1}!} } {\paren {n + 2} \frac {\paren {n + 1}!} {m! \paren {n - m + 1}!} \frac {\paren {n + 1}!} {\paren {m + 1}! \paren {n - m}!} }       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \dfrac {\paren {n + 1}!} {\frac {\paren {n + 1}!} {m!} \frac {\paren {n + 1}!} {\paren {n - m}!} }       | c = some simplification }} {{eqn | r = \dfrac 1 {\frac {\paren {n + 1}!} {m! \paren {n - m}!} }       | c = further simplification }} {{eqn | r = \dfrac 1 {\paren {n + 1} \frac {n!} {m! \paren {n - m}!} }       | c = further simplification }} {{eqn | r = \dfrac 1 {\paren {n + 1} \binom n m}       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} Hence the result. {{qed}}
Let $Y, Z \in \tau$ such that :$Y \wedge Z \preceq \complement_S\left({X}\right)$ By definition of topological space: :$Y \cap Z \in \tau$ By Meet in Inclusion Ordered Set: :$Y \cap Z = Y \wedge Z$ By definition of inclusion ordered set: :$Y \cap Z \subseteq \complement_S\left({X}\right)$ By Relative Complement is Decreasing and Relative Complement of Relative Complement: :$X \subseteq \complement_S\left({Y \cap Z}\right)$ By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection: :$X \subseteq \complement_S\left({Y}\right) \cup \complement_S\left({Z}\right)$ By Intersection with Subset is Subset: :$X = \left({\complement_S\left({Y}\right) \cup \complement_S\left({Z}\right)}\right) \cap X$ By Intersection Distributes over Union: :$X = \left({\complement_S\left({Y}\right) \cap X}\right) \cup \left({\complement_S\left({Z}\right) \cap X}\right)$ By definition of closed set and Relative Complement of Relative Complement: :$X$, $\complement_S\left({Y}\right)$, and $\complement_S\left({Z}\right)$ are closed sets. By Intersection of Closed Sets is Closed in Topological Space: :$\complement_S\left({Y}\right) \cap X$, $\complement_S\left({Z}\right) \cap X$ are closed sets. By definition of irreducible: :$\complement_S\left({Y}\right) \cap X = X$ or $\complement_S\left({Z}\right) \cap X = X$ By Intersection with Subset is Subset: :$X \subseteq \complement_S\left({Y}\right)$ or $X \subseteq \complement_S\left({Z}\right)$ By Relative Complement is Decreasing and Relative Complement of Relative Complement: :$Y \subseteq \complement_S\left({X}\right)$ or $Z \subseteq \complement_S\left({X}\right)$ Thus by definition of inclusion ordered set: :$Y \preceq \complement_S\left({X}\right)$ or $Z \preceq \complement_S\left({X}\right)$ {{qed}}
{{AimForCont}} that there exists some $i \notin \Fix \sigma$ and some $m \in \Z$ such that $\sigma^m \paren i \in \Fix \sigma$. Then: {{begin-eqn}} {{eqn | l = \paren {\sigma \circ \sigma^m} \paren i       | r = \sigma^{m + 1} \paren i }} {{eqn | r = \sigma^m \paren i       | c = {{Defof|Fixed Element of Permutation}} }} {{eqn | ll= \implies       | l = \sigma^{-m} \sigma^{m + 1} \paren i       | r = \sigma^{-m} \sigma^m \paren i }} {{eqn | ll= \implies       | l = \sigma \paren i       | r = e \paren i }} {{eqn | r = i }} {{end-eqn}} But it was previously established that $i$ was moved by $\sigma$. This is a contradiction. Therefore $\sigma^m \paren i$ is moved by $\sigma$. {{qed}} Category:Symmetric Groups 3oilrb668bi1pz5dlfe43gxusjxsbg0
Let $x \in \complement_S\left({Y}\right)$ By definition of relative complement: :$x \in S \setminus Y$ By definition of difference: :$x \in S$ and $x \notin Y$ By definition of subset: :$x \notin X$ By definition of difference: :$x \in S \setminus X$ Thus by definition of relative complement: :$x \in \complement_S\left({X}\right)$ {{qed}}
{{begin-eqn}} {{eqn | l = 6 \div 1       | r = 6       | c =  }} {{eqn | l = 36 \div 6       | r = 6       | c =  }} {{end-eqn}} Hence the common ratio is $6$. {{qed}}
By definition of topological space: :$X \cup Y, X \cap Y \in \tau$ Thus by Join in Inclusion Ordered Set ans Meet in Inclusion Ordered Set: :the result holds. {{qed}}
By Set is Subset of Union: :$A \subseteq A \cup B$ and $B \subseteq A \cup B$ By definition: :$A \cup B$ is upper bound for $\left\{ {A, B}\right\}$ We will prove that :$\forall C \in X: C$ is upper bound for $\left\{ {A, B}\right\} \implies A \cup B \subseteq C$ Let $C \in X$ such that :$C$ is upper bound for $\left\{ {A, B}\right\}$. By definition of upper bound: :$A \subseteq C$ and $B \subseteq C$ Thus by Union is Smallest Superset: :$A \cup B \subseteq C$ {{qed|lemma}} By definition of supremum: :$\sup \left\{ {A, B}\right\} = A \cup B$ Thus by definition of join: :$A \vee B = A \cup B$ {{qed}}
=== Sufficient Condition === Assume that :$Z$ is prime element in $L$. Let $X, T \in \tau$ such that: :$X \cap Y \subseteq Z$ By Join and Meet in Inclusion Ordered Set of Topology and definition of inclusion ordered set: :$X \wedge Y \preceq Z$ By definition of prime element: :$X \preceq Z$ or $Y \preceq Z$ Thus by definition of inclusion ordered set: :$X \subseteq Z$ or $Y \subseteq Z$ {{qed|lemma}} === Necessary Condition === Assume that: L$\forall X, Y \in \tau: X \cap Y \subseteq Z \implies X \subseteq Z \lor Y \subseteq Z$ Let $X, Y \in \tau$ such that: :$X \wedge Y \preceq Z$ By Join and Meet in Inclusion Ordered Set of Topology and definition of inclusion ordered set: :$X \cap Y \subseteq Z$ By assumption: :$X \subseteq Z$ or $Y \subseteq Z$ Thus by definition of inclusion ordered set: :$X \preceq Z$ or $Y \preceq Z$ {{qed}}
Performing the calculation using long division: <pre>    0.027...   -------- 37)1.00000      74      --      260      259      ---        100         74        ---         ... </pre> This is because $999 = 27 \times 37$. It can be determined by inspection of all smaller integers that this is indeed the $2$nd to have a period of $3$. {{qed}}
{{begin-eqn}} {{eqn | l = C_n       | r = 3 n \paren {n - 1} + 1       | c = Closed Form for Centered Hexagonal Numbers }} {{eqn | r = 6 \paren {\dfrac {\paren {n - 1} n} 2} + 1       | c =  }} {{eqn | r = 6 T_{n - 1} + 1       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{qed}}
By the definition of centered hexagonal number: {{begin-eqn}} {{eqn | l = C_n       | r = 1 + \sum_{k \mathop = 1}^{n - 1} 6 k       | c = {{Defof|Centered Hexagonal Number}} }} {{eqn | r = 1 + 6 \sum_{k \mathop = 1}^{n - 1} k       | c =  }} {{eqn | r = 6 \paren {\dfrac {n \paren {n - 1} } 2} + 1       | c = Closed Form for Triangular Numbers }} {{eqn | r = 3 n \paren {n - 1} + 1       | c =  }} {{end-eqn}} {{qed}}
The smallest composite odd numbers are $9$ and $15$, so trivially $2$ to $16$ and $20$ to $22$ cannot be expressed as the sum of $2$ composite odd numbers. We have: {{begin-eqn}} {{eqn | l = 18       | r = 9 + 9 }} {{eqn | l = 24       | r = 9 + 15 }} {{eqn | l = 30       | r = 21 + 9 }} {{eqn | r = 15 + 15 }} {{eqn | l = 34       | r = 9 + 25 }} {{eqn | l = 36       | r = 9 + 27 }} {{end-eqn}} It remains to investigate $26, 28$ and $32$. This will be done by progressively subtracting smaller composite odd numbers from them, and noting that the difference is not composite. {{begin-eqn}} {{eqn | l = 26 - 9       | r = 17       | c = which is prime }} {{eqn | l = 26 - 15       | r = 11       | c = which is prime }} {{eqn | l = 26 - 21       | r = 5       | c = which is prime }} {{eqn | l = 26 - 25       | r = 1       | c = which is not composite }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 28 - 9       | r = 19       | c = which is prime }} {{eqn | l = 28 - 15       | r = 13       | c = which is prime }} {{eqn | l = 28 - 21       | r = 7       | c = which is prime }} {{eqn | l = 28 - 25       | r = 3       | c = which is prime }} {{eqn | l = 28 - 27       | r = 1       | c = which is not composite }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 32 - 9       | r = 32       | c = which is prime }} {{eqn | l = 32 - 15       | r = 17       | c = which is prime }} {{eqn | l = 32 - 21       | r = 11       | c = which is prime }} {{eqn | l = 32 - 25       | r = 7       | c = which is prime }} {{eqn | l = 32 - 27       | r = 5       | c = which is prime }} {{end-eqn}} It remains to be demonstrated that all even integers greater than $38$ can be expressed as the sum of $2$ composite odd numbers. We note that $9 + 6 k$ is odd and a multiple of $3$. Numbers $18$ and greater of the form $6 n$ can be expressed as: :$\left({9 + 6 k}\right) + 9$ Numbers $34$ and greater of the form $6 n + 4$ can be expressed as: :$\left({9 + 6 k}\right) + 25$ Numbers $44$ and greater of the form $6 n + 2$ can be expressed as: :$\left({9 + 6 k}\right) + 35$ We have that $40$ and $42$ are of the form $6 n + 4$ and $6 n$ respectively. Hence all even integers greater than $38$ are accounted for. {{qed}}
Let $\mathcal F$ be a set of subsets of $S$ such that: :$\mathcal F$ is open cover of $x^\succeq$ By definition of cover: :$x^\succeq \subseteq \bigcup \mathcal F$ By definitions of upper closure of element and reflexivity: :$x \in x^\succeq$ By definition of subset: :$x \in \bigcup \mathcal F$ By definition of union: :$\exists Y \in \mathcal F: x \in Y$ Defime $\mathcal G = \left\{ {Y}\right\}$ By definition of open cover: :$Y$ is open. We will prove that :$x^\succeq \subseteq \bigcup \mathcal G$ Let $y \in x^\succeq$ By definition of upper closure of element: :$x \preceq y$ By Union of Singleton: :$\bigcup \mathcal G = Y$ By assumption: :$Y$ is upper. Thus by definition of upper set: :$x \in \bigcup \mathcal G$ {{qed|lemma}} Then by definition: :$\mathcal G$ is cover of $x^\succeq$ By definitions of singleton and subset: :$\mathcal G \subseteq \mathcal F$ By definition: :$\mathcal G$ is subcover of $\mathcal F$. By Singleton is Finite: :$\mathcal G$ is finite. Thus by definition: :$\mathcal G$ is finite subcover of $\mathcal F$. {{qed}}
Let $a, b \in \R_{> 0}$ We have that the Real Numbers form Ordered Integral Domain. It follows from Positive Elements of Ordered Ring that: :$a \times b \in \R_{> 0}$. {{qed}}
By the power rule for derivatives: :$\map {D_x} {x^a} = a x^{a - 1}$ By power of positive real number is positive, it is seen that: :$x > 0 \implies x^{a - 1} > 0$ By Positive Real Numbers Closed under Multiplication, it follows that $\map {D_x} {x^a} > 0$ for all $x \in \openint 0 {+\infty}$. Hence by Derivative of Monotone Function, $f_a$ is strictly increasing {{qed}} Category:Powers 2cbpp75qizc338uczp8idmqj95sullw
From Tau Function from Prime Decomposition, we have: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where the prime decomposition of $n$ is: :$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ {{begin-eqn}} {{eqn | l = 2 + \map \tau 2       | r = 2 + 2       | rr= = 4       | c = as $2 = 2^1$ }} {{eqn | l = 4 + \map \tau 4       | r = 4 + 7       | rr= = 7       | c = as $4 = 2^2$ }} {{eqn | l = 6 + \map \tau 6       | r = 6 + 4       | rr= = 10       | c = as $6 = 2^1 3^1$ }} {{eqn | l = 8 + \map \tau 8       | r = 8 + 4       | rr= = 12       | c = as $8 = 2^3$ }} {{eqn | l = 10 + \map \tau {10}       | r = 10 + 4       | rr= = 14       | c = as $10 = 2^1 5^1$ }} {{eqn | l = 12 + \map \tau {12}       | r = 12 + 6       | rr= = 18       | c = as $12 = 2^2 3^1$ }} {{eqn | l = 14 + \map \tau {14}       | r = 14 + 4       | rr= = 18       | c = as $14 = 2^1 7^1$ }} {{eqn | l = 16 + \map \tau {16}       | r = 16 + 5       | rr= = 21       | c = as $16 = 2^4$ }} {{eqn | l = 18 + \map \tau {18}       | r = 18 + 6       | rr= = 24       | c = as $18 = 2^1 3^2$ }} {{eqn | l = 20 + \map \tau {20}       | r = 20 + 6       | rr= = 26       | c = as $20 = 2^2 5^1$ }} {{eqn | l = 22 + \map \tau {22}       | r = 22 + 4       | rr= = 26       | c = as $22 = 2^1 11^1$ }} {{eqn | l = 24 + \map \tau {24}       | r = 24 + 8       | rr= = 32       | c = as $24 = 2^3 3^1$ }} {{eqn | l = 26 + \map \tau {26}       | r = 26 + 4       | rr= = 30       | c = as $26 = 2^1 13^1$ }} {{eqn | l = 28 + \map \tau {28}       | r = 28 + 6       | rr= = 34       | c = as $28 = 2^2 7^1$ }} {{eqn | l = 30 + \map \tau {30}       | r = 30 + 8       | rr= = 38       | c = as $30 = 2^1 3^1 5^1$ }} {{eqn | l = 32 + \map \tau {32}       | r = 32 + 6       | rr= = 38       | c = as $32 = 2^5$ }} {{eqn | l = 34 + \map \tau {34}       | r = 34 + 4       | rr= = 38       | c = as $34 = 2^1 17^1$ }} {{end-eqn}} {{qed}}
By Open iff Upper and with Property (S) in Scott Topological Lattice: :$X$ is uooer and has property (S). Let $D$ be a directed subset of $S$ such that :$x \preceq \sup D$ By definition of upper set: :$\sup D \in X$ By definition of property (S): :$\exists y \in D: \forall d \in D: y \preceq d \implies d \in X$ By definitions of infimum and complete lattice: :$\inf X$ is a lower bound for $X$. By definition pf reflexivity: :$y \in X$ Thus by definition of lower bound: :$\exists d \in D: \inf X \preceq d$ Thus by definition pf way below relation: :$\inf X \ll x$ {{qed}}
First we note that $10^5-1 = 41 \times 271 \times 9$ $10$ generates exactly $5$ elements in $\Z_{41}$  Subgroup Generated by One Element is Cyclic $\forall k \in \N: $ $10^{0 + 5 k} \equiv 1 \pmod {41}$ $10^{1 + 5 k} \equiv 10 \pmod {41}$ $10^{2 + 5 k} \equiv 18 \pmod {41}$ $10^{3 + 5 k} \equiv 16 \pmod {41}$ $10^{4 + 5 k} \equiv 37 \pmod {41}$ In $\Z_{41}$: $ \set {10^1, 10^2, 10^3, 10^4, 10^5} \leadsto \set {10, 18, 16, 37, 1}$ Let $n$ be a multiple of $41$ with $5$ digits. Then we have: {{begin-eqn}} {{eqn | l = 41 \times c       | r = a_4 \times 10^4 + a_3 \times 10^3 + a_2 \times 10^2 + a_1 \times 10^1 + a_0 }} {{end-eqn}} Let $m$ be an integer formed by cyclically permuting the digits of $n$. Then we have: {{begin-eqn}} {{eqn | l = 10 \times 41 \times c        | r = 10 \times \paren {a_4 \times 10^4 + a_3 \times 10^3 + a_2 \times 10^2 + a_1 \times 10^1 + a_0}       | c = multiply original number by $10$ }} {{eqn | r =  a_4 \times 10^5 + a_3 \times 10^4 + a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1       | c =  }} {{eqn | r =  a_3 \times 10^4 + a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1 + a_4 \times 10^0       | c = $10^5$ and $10^0 \equiv 1 \pmod {41}$ }} {{end-eqn}} Therefore $m$ is a multiple of $41$. {{qed}}
First we have: {{begin-eqn}} {{eqn | l = 2       | r = 3^1 - 2^0       | c =  }} {{eqn | l = 3       | r = 2^2 - 3^0       | c =  }} {{eqn | l = 5       | r = 3^2 - 2^2       | c =  }} {{eqn | l = 7       | r = 3^2 - 2^1       | c =  }} {{eqn | l = 11       | r = 3^3 - 2^4       | c =  }} {{eqn | l = 13       | r = 2^4 - 3^1       | c =  }} {{eqn | l = 17       | r = 3^4 - 2^6       | c =  }} {{eqn | l = 19       | r = 3^3 - 2^3       | c =  }} {{eqn | l = 23       | r = 3^3 - 2^2       | c =  }} {{eqn | l = 29       | r = 2^5 - 3^1       | c =  }} {{eqn | l = 31       | r = 2^5 - 3^0       | c =  }} {{eqn | l = 37       | r = 2^6 - 3^3       | c =  }} {{end-eqn}} {{AimForCont}} $41 = 2^n - 3^m$. We have that $n > 3$. Thus: :$2^n \equiv 0 \pmod 8$ and as: :$41 \equiv 1 \pmod 8$ we have: :$1 \equiv -3^m \pmod 8$ which is not possible: For any integer $k$,  {{begin-eqn}} {{eqn | l = -3^{2 k}        | r = -9^k }} {{eqn | o = \equiv        | r = -1^k       | rr = \pmod 8       | c = Congruence of Powers }} {{eqn | o = \equiv        | r = -1       | rr = \pmod 8 }} {{eqn | l = -3^{2 k + 1}        | r = 3 \paren {-3^{2 k} } }} {{eqn | o = \equiv        | r = 3 \paren {-1}       | rr = \pmod 8       | c = Modulo Multiplication is Well-Defined }} {{eqn | o = \equiv        | r = -3       | rr = \pmod 8 }} {{end-eqn}} So for any integer $m$, $1 \not \equiv -3^m \pmod 8$. Now suppose $41 = 3^m - 2^n$. We have that $m > 1$ and $n > 2$. Taking $\mod 3$ on both sides: {{begin-eqn}} {{eqn | l = 41        | r = 3^m - 2^n }} {{eqn | ll = \leadsto       | l = -1       | o = \equiv        | r = - 2^n       | rr = \pmod 3       | c =  }} {{eqn | ll = \leadsto       | l = 1       | o = \equiv        | r = 2^n       | rr = \pmod 3       | c =  }} {{eqn | o = \equiv        | r = \paren {-1}^n       | rr = \pmod 3       | c = Congruence of Powers }} {{end-eqn}} which shows that $n$ is even. Taking $\mod 4$ on both sides: {{begin-eqn}} {{eqn | l = 41        | r = 3^m - 2^n }} {{eqn | ll = \leadsto       | l = 1       | o = \equiv        | r = 3^m       | rr = \pmod 4       | c =  }} {{eqn | o = \equiv        | r = \paren {-1}^m       | rr = \pmod 4       | c = Congruence of Powers }} {{end-eqn}} which shows that $m$ is even as well. Now we take $\mod 5$ on both sides: {{begin-eqn}} {{eqn | l = 41        | r = 3^m - 2^n }} {{eqn | ll = \leadsto       | l = 1       | o = \equiv        | r = 3^m - 2^n       | rr = \pmod 5       | c =  }} {{eqn | o = \equiv        | r = 9^{m / 2} - 4^{n / 2}       | rr = \pmod 5       | c =  }} {{eqn | o = \equiv        | r = \paren {-1}^{m / 2} - \paren {-1}^{n / 2}       | rr = \pmod 5       | c = Congruence of Powers }} {{end-eqn}} But $\paren {-1}^{m / 2} - \paren {-1}^{n / 2}$ can only be $0$ or $\pm 2$, not $1$, which is a contradiction. Therefore $41$ is not the difference between a power of $2$ and a power of $3$. {{qed}}
This follows by Open iff Upper and with Property (S) in Scott Topological Lattice and definition Scott sigma. {{qed}}
Let $p$ be an Euler lucky number. Let $f_p: \Z \to \Z$ be the mapping defined as: :$\forall n \in \Z: f_p \left({n}\right) = n^2 + n + p$ Let $m = n - 1$. Then: {{begin-eqn}} {{eqn | l = f_p \left({m}\right)       | r = f_p \left({n - 1}\right)       | c =  }} {{eqn | r = \left({n - 1}\right)^2 + \left({n - 1}\right) + p       | c =  }} {{eqn | r = n^2 - 2 n + 1 + n - 1 + p       | c =  }} {{eqn | r = n^2 - n + p       | c =  }} {{end-eqn}} We have that $f_p \left({n}\right)$ is prime for $0 \le n < p - 1$. Thus $f_p \left({m}\right)$ is prime for $0 \le \left({n - 1}\right) < p - 1$. and so $f_p \left({m}\right)$ is prime for $1 \le n < p$. {{qed}} Category:Euler Lucky Numbers 9u1fq9eu5bstdob86swq2xd0iz5qbem
By Scott Topology equals to Scott Sigma: :$\tau = \map \sigma L$ By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice: :$x^\preceq = \set x^-$ where $\set x^-$ denotes the topological closure of $\set x$. By Topological Closure of Singleton is Irreducible: :$x^\preceq$ is topologically irreducible Thus by Complement of Irreducible Topological Subset is Prime Element: :$\relcomp S {x^\preceq}$ is a prime element in $D$. By definitions of reflexivity and lower closure of element: :$x \in x^\preceq$ By definitions of relative complement and difference: :$x \notin \relcomp S {x^\preceq}$ Hence by definition set equality: :$\relcomp S {x^\preceq} \ne S$ {{qed}}