hcju/proofwikiqa huggingface dataset:
Note that there cannot be quadruplets of such numbers, since one of the quadruplets must be divisible by $4$, making it non-sphenic. We have: {{begin-eqn}} {{eqn | l = 1309       | r = 7 \times 11 \times 17 }} {{eqn | l = 1310       | r = 2 \times 5 \times 131 }} {{eqn | l = 1311       | r = 3 \times 19 \times 23 }} {{eqn | l = 1885       | r = 5 \times 13 \times 29 }} {{eqn | l = 1886       | r = 2 \times 23 \times 41 }} {{eqn | l = 1887       | r = 3 \times 17 \times 37 }} {{eqn | l = 2013       | r = 3 \times 11 \times 61 }} {{eqn | l = 2014       | r = 2 \times 19 \times 53 }} {{eqn | l = 2015       | r = 5 \times 13 \times 31 }} {{end-eqn}} hence each number above is sphenic. {{qed}}
We will show that: {{begin-eqn}} {{eqn | l = 1375       | r = 11 \times 5^3       | c =  }} {{eqn | l = 1376       | r = 172 \times 2^3       | c =  }} {{eqn | l = 1377       | r = 51 \times 3^3       | c =  }} {{end-eqn}} is the smallest such triplet. Each number in such triplets of consecutive integers is divisible by a cube of some prime number. Only $2, 3, 5, 7, 11$ are less than $\sqrt [3] {1377}$. Since the numbers involved are small, we can check the result by brute force. For general results one is encouraged to use the Chinese Remainder Theorem. === Case $1$: a number is divisible by $11^3$ === The only multiple of $11^3$ less than $1377$ is $1331$, and: {{begin-eqn}} {{eqn | l = 1330       | r = 2 \times 5 \times 7 \times 19 }} {{eqn | l = 1332       | r = 2^2 \times 3^2 \times 37 }} {{end-eqn}} Since neither $1330$ nor $1332$ are divisible by a cube of some prime number, $1331$ is not in a triplet. {{qed|lemma}} === Case $2$: a number is divisible by $7^3$ === The only multiples of $7^3$ less than $1377$ are $343, 686, 1029, 1372$, and: {{begin-eqn}} {{eqn | l = 342       | r = 2 \times 3^2 \times 19 }} {{eqn | l = 344       | r = 2^3 \times 43 }} {{eqn | l = 345       | r = 3 \times 5 \times 23 }} {{eqn | l = 685       | r = 5 \times 137 }} {{eqn | l = 687       | r = 3 \times 229 }} {{eqn | l = 1028       | r = 2^2 \times 257 }} {{eqn | l = 1030       | r = 2 \times 5 \times 103 }} {{eqn | l = 1371       | r = 3 \times 457 }} {{eqn | l = 1373       | o = \text {is}       | r = \text {prime} }} {{end-eqn}} Hence none of these numbers is in a triplet. {{qed|lemma}} === Case $3$: the numbers are divisible by $2^3, 3^3, 5^3$ respectively === Let $n = k \times 5^3$. We show that $k$ cannot be divisible by $3$ or $4$. Suppose $3 \divides k$. Then none of $n \pm 1, n \pm 2$ are divisible by $3$, and consequently $3^3$. Suppose $4 \divides k$. Then none of $n \pm 1, n \pm 2$ are divisible by $4$, and consequently $2^3$. The only multiples of $5^3$ less than $1377$ are $125, 250, 375, 500, 625, 750, 875, 1000, 1125, 1250, 1375$, and we eliminate $375, 500, 750, 1000, 1125$ due to the reasons above. Now: {{begin-eqn}} {{eqn | l = 124       | r = 2^2 \times 31 }} {{eqn | l = 126       | r = 2 \times 3^2 \times 7 }} {{eqn | l = 249       | r = 3 \times 83 }} {{eqn | l = 251       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 623       | r = 7 \times 89 }} {{eqn | l = 624       | r = 2^4 \times 3 \times 13 }} {{eqn | l = 626       | r = 2 \times 313 }} {{eqn | l = 874       | r = 2 \times 19 \times 23 }} {{eqn | l = 876       | r = 2^2 \times 3 \times 73 }} {{eqn | l = 1030       | r = 2 \times 5 \times 103 }} {{eqn | l = 1249       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 1251       | r = 3^2 \times 139 }} {{end-eqn}} Hence none of these numbers is in a triplet. {{qed}}
{{begin-eqn}} {{eqn | l = \int_{\closedint 0 1^n} \frac 1 {1 - \prod_{i \mathop = 1}^n x_i} \prod_{i \mathop = 1}^n \rd x_i        | r = \int_{\closedint 0 1^n} \sum_{j \mathop = 1}^\infty \paren {\prod_{i \mathop = 1}^n x_i}^{j - 1} \prod_{i \mathop = 1}^n \rd x_i        | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \sum_{j \mathop = 1}^\infty \prod_{i \mathop = 1}^n \int_0^1 x^{j - 1}_i \rd x_i        | c = Fubini's Theorem }} {{eqn | r = \sum_{j \mathop = 1}^\infty \frac 1 {j^n} }} {{eqn | r = \map \zeta n       | c = {{Defof|Riemann Zeta Function}} }} {{end-eqn}} {{qed}} Category:Riemann Zeta Function Category:Analytic Number Theory jwrs1ween35w33fz057fglkz65ihfz8
{{begin-eqn}} {{eqn | l = S_n       | r = \sum_{k \mathop = 1}^n H_k       | c = {{Defof|Hexagonal Pyramidal Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n k \paren {2 k - 1}       | c = Closed Form for Hexagonal Numbers }} {{eqn | r = 3 \sum_{k \mathop = 1}^n 2 k^2 - \sum_{k \mathop = 1}^n k       | c =  }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 6 - \sum_{k \mathop = 1}^n k       | c = Sum of Sequence of Squares }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} } 6 - \dfrac {n \paren {n + 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {2 n \paren {n + 1} \paren {2 n + 1} - 3 n \paren {n + 1} } 6       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {4 n + 2 - 3} } 6       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {4 n - 1} } 6       | c =  }} {{end-eqn}} {{qed}} Category:Closed Forms Category:Pyramidal Numbers 1b9e5x8004owfzmnt6dfbmrdwpbcibr
{{begin-eqn}} {{eqn | l = 1       | r = \dfrac {1 \paren {1 + 1} \paren {1 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \dfrac {1 \times \paren {1 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 10       | r = \dfrac {3 \paren {3 + 1} \paren {3 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \dfrac {4 \times \paren {4 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 120       | r = \dfrac {8 \paren {8 + 1} \paren {8 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \dfrac {15 \times \paren {15 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1540       | r = \dfrac {20 \paren {20 + 1} \paren {20 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \dfrac {55 \times \paren {55 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 7140       | r = \dfrac {34 \paren {34 + 1} \paren {34 + 2} } 6       | c = Closed Form for Tetrahedral Numbers }} {{eqn | r = \dfrac {119 \times \paren {119 + 1} } 2       | c = Closed Form for Triangular Numbers }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the only such instances.}}
By definitions of composition of mappings and restriction of mapping: :$\left({g \circ f}\right) \restriction S: S \to Z$ and $g \circ \left({f \restriction S}\right):S \to Z$ Let $s \in S$. By definition of restriction of mapping: :$\left({\left({g \circ f}\right) \restriction S}\right)\left({s}\right) = \left({g \circ f}\right)\left({s}\right)$ Thus {{begin-eqn}} {{eqn | l = \left({g \circ \left({f \restriction S}\right)}\right)\left({s}\right)       | r = g\left({\left({f \restriction S}\right)\left({s}\right)}\right)       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = g\left({f\left({s}\right)}\right)       | c = {{Defof|Restriction of Mapping}} }} {{eqn | r = \left({\left({g \circ f}\right) \restriction S}\right)\left({s}\right)       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} {{qed}}
The cases $1$ and $5$ are trivial. Now we show that $1549 - a^b$ is never prime for $a \ge 1$ and $b \ge 2$. It suffices to show the result for prime values of $b$. We first prove: {{begin-eqn}} {{eqn | n = 1       | l = 2       | o = \divides       | r = a }} {{eqn | n = 2       | l = 3       | o = \divides       | r = a       | rr = \text {if } b = 2 }} {{eqn | n = 3       | l = a^b       | o = \not \equiv       | r = 4       | rr = \pmod {10} }} {{end-eqn}} For $(1)$: Suppose $a$ is odd. Then so is $a^b$. Therefore $1549 - a^b$ is even. Hence $1549 - a^b$ is prime only if it is equal to $2$. But $1547$ is not a perfect power. {{qed|lemma}} For $(2)$: Suppose $b = 2$ and $a$ is not divisible by $3$. By Corollary to Square Modulo $3$: :$3 \divides \paren {a^2 - 1}$ Hence: :$3 \divides \paren {1548 - a^2 + 1} = 1549 - a^b$ So $1549 - a^b$ is prime only if it is equal to $3$. But $1547$ is not a perfect square. {{qed|lemma}} For $(3)$: Suppose $a^b \equiv 4 \pmod {10}$. Then $1549 - a^b \equiv 5 \pmod {10}$ So $1549 - a^b$ is prime only if it is equal to $5$. But $1544$ is not a perfect power. {{qed|lemma}} Hence we only need to check the cases: :For $b = 2: \: 6^2, 12^2, 18^2, 24^2, 30^2, 36^2$ since $42^2 > 1549$ :For $b \ne 2: \: 2^3, 2^5, 2^7, 4^3, 4^5, 6^3, 8^3, 10^3$ since $2^{11}, 4^7, 6^5 > 1549$ of which $12^2 \equiv 18^2 \equiv 4^3 \equiv 4^5 \equiv 4 \pmod {10}$, so they are rejected. We have: {{begin-eqn}} {{eqn | l = 1549 - 6^2       | r = 1513 }} {{eqn | r = 17 \times 89 }} {{eqn | l = 1549 - 24^2       | r = 973 }} {{eqn | r = 7 \times 139 }} {{eqn | l = 1549 - 30^2       | r = 649 }} {{eqn | r = 11 \times 59 }} {{eqn | l = 1549 - 36^2       | r = 253 }} {{eqn | r = 11 \times 23 }} {{eqn | l = 1549 - 2^3       | r = 1541 }} {{eqn | r = 23 \times 67 }} {{eqn | l = 1549 - 2^5       | r = 1517 }} {{eqn | r = 37 \times 41 }} {{eqn | l = 1549 - 2^7       | r = 1421 }} {{eqn | r = 7^2 \times 29 }} {{eqn | l = 1549 - 6^3       | r = 1333 }} {{eqn | r = 31 \times 43 }} {{eqn | l = 1549 - 8^3       | r = 1037 }} {{eqn | r = 17 \times 61 }} {{eqn | l = 1549 - 10^3       | r = 549 }} {{eqn | r = 3^2 \times 61 }} {{end-eqn}} and none of the above are prime. {{ProofWanted|For $1771561 {{=}} 11^6$}} {{ProofWanted|In addition to the above, don't we also need to demonstrate that for $5 < n < 1549$, an odd number *does* have such an expression?}}
By definition, a harshad number is divisible by the sum of its digits base $10$. Let $n$ be a $1$-digit positive integer. The sum of the digits of $n$ is trivially $n$. The result follows from Integer Divides Itself. {{qed}} Category:Harshad Numbers kxykvvanwxcz0ibl71hfbpqgniu0v7p
{{ProofWanted|We have the list of Poulet numbers and Fermat pseudoprimes base $3$, but not of base $5$. Once we get that list, we can find the numbers on the list for both.}}
The number of $2$-digit integers that can be formed from the digits of $1782$ equals the number of $2$-permutations of $\set {1, 7, 8, 2}$. That is: :$\set {17, 18, 12, 71, 78, 72, 81, 87, 82, 21, 27, 28}$ Hence: :$17 + 18 + 12 + 71 + 78 + 72 + 81 + 87 + 82 + 21 + 27 + 28 = 594 = \dfrac {1782} 3$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = 1880       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 8^2 + 8^2 + 0^2       | r = 1 + 64 + 64 + 0       | c =  }} {{eqn | r = 129       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 2^2 + 9^2       | r = 1 + 4 + 81       | c =  }} {{eqn | r = 86       | c =  }} {{eqn | ll= \leadsto       | l = 8^2 + 6^2       | r = 64 + 36       | c =  }} {{eqn | r = 100       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 0^2 + 0^2       | r = 1 + 0 + 0       | c =  }} {{eqn | r = 1       | c = and so $1880$ is happy }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = 1881       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 8^2 + 8^2 + 1^2       | r = 1 + 64 + 64 + 1       | c =  }} {{eqn | r = 130       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 3^2 + 0^2       | r = 1 + 9 + 0       | c =  }} {{eqn | r = 10       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 0^2       | r = 1 + 0       | c =  }} {{eqn | r = 1       | c = and so $1881$ is happy }} {{end-eqn}} {{begin-eqn}} {{eqn | o =        | r = 1882       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 8^2 + 8^2 + 2^2       | r = 1 + 64 + 64 + 4       | c =  }} {{eqn | r = 133       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 3^2 + 3^2       | r = 1 + 9 + 9       | c =  }} {{eqn | r = 19       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 9^2       | r = 1 + 81       | c =  }} {{eqn | r = 82       | c =  }} {{eqn | ll= \leadsto       | l = 8^2 + 2^2       | r = 64 + 4       | c =  }} {{eqn | r = 68       | c =  }} {{eqn | ll= \leadsto       | l = 6^2 + 8^2       | r = 36 + 64       | c =  }} {{eqn | r = 100       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 0^2 + 0^2       | r = 1 + 0 + 0       | c =  }} {{eqn | r = 1       | c = and so $1882$ is happy }} {{end-eqn}} {{ProofWanted|It remains to be shown that this is the smallest such triple}}
Established by inspecting the sequence of Mersenne numbers: :$3, 7, 31, 127, 2047, 8191, 131 \, 071, 524 \, 287, 8 \, 388 \, 607, 536 \, 870 \, 911, 2 \, 147 \, 483 \, 647, \ldots$ and removing from it the sequence of Mersenne primes: :$3, 7, 31, 127, 8191, 131 \, 071, 524 \, 287, 2 \, 147 \, 483 \, 647, \ldots$ {{qed}}
{{begin-eqn}} {{eqn	| l = \int \frac {\d x} {x^2 + a^2} 	| r = \frac 1 a \int \frac {\d t} {t^2 + 1} 	| c = Substitution of $x \to a t$}} {{eqn	| r = \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} } 	| c = Factoring}} {{eqn	| r = \frac 1 {2 a} \paren {\int \frac {\d t} {1 + i t} + \int \frac {\d t} {1 - i t} } 	| c = {{Defof|Partial Fractions Expansion}} }} {{eqn	| r = \frac 1 {2 a} \paren {i \map \ln {1 - i t} - i \map \ln {1 + i t} } + C 	| c = Primitive of Reciprocal }} {{eqn	| r = \frac i {2 a} \map \ln {\frac {1 - i t} {1 + i t} } + C 	| c = Sum of Logarithms }} {{eqn	| r = \frac 1 a \arctan \frac x a + C 	| c = Arctangent Logarithmic Formulation and substituting back $t \to \dfrac x a$ }} {{end-eqn}} {{qed}}
Let k represent the number of $9$s in the middle of the number. For $k > 0$ We can rewrite the number as follows: {{begin-eqn}} {{eqn | l = \sqbrk {21 (9) 78}_{10}       | r = 21 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 78       | c = {{Defof|Geometric Series}} }} {{end-eqn}} Taking numbers of this form and multiplying by $4$ produces: {{begin-eqn}} {{eqn | l = 4 \times \paren {21 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 78 }       | r = 84 \times 10^{k + 2 } + 3600 \sum_{i \mathop = 0}^{k - 1} 10^i + 312 }} {{end-eqn}} The first part is composed of $k + 4$ digits. The first two digits will be $84$ followed by $k +2$ digits of $0$ {{begin-eqn}} {{eqn | l = 84 \times 10^{k + 2 }        | r = 84 \times 10^{k + 2 }  }} {{end-eqn}} The sum in the middle is composed of $k + 3$ digits. The first digit will be $3$ followed by $k - 1$ digits of $9$ and then the remaining three digits at the end are $600$ {{begin-eqn}} {{eqn | l = 3600 \sum_{i \mathop = 0}^{k - 1} 10^i       | r = 399 \cdots 99600 }} {{end-eqn}} Summing the three pieces, the final answer will have $k + 4$ digits. The first digit is $8$  followed by $7$ which is the sum of the $4$ from the first part and the $3$ of the middle part followed by $k$ digits of $9$ where the last $9$ is the sum of the $6$ from the middle part and the $3$ of the last part and then ending in $12$: {{begin-eqn}} {{eqn | l = \sqbrk {21 (9) 78}_{10} \times 4       | r = \sqbrk {87 (9) 12}_{10}       | c = }} {{end-eqn}} {{qed}}
We obtain the prime decomposition of all $17$ of these integers: {{begin-eqn}} {{eqn | l = 2184       | r = 2^3 \times 3 \times 7 \times 13 }} {{eqn | l = 2185       | r = 5 \times 19 \times 23 }} {{eqn | l = 2186       | r = 2 \times 1093 }} {{eqn | l = 2187       | r = 3^7 }} {{eqn | l = 2188       | r = 2^2 \times 547 }} {{eqn | l = 2189       | r = 11 \times 199 }} {{eqn | l = 2190       | r = 2 \times 3 \times 5 \times 73 }} {{eqn | l = 2191       | r = 7 \times 313 }} {{eqn | l = 2192       | r = 2^4 \times 137 }} {{eqn | l = 2193       | r = 3 \times 17 \times 43 }} {{eqn | l = 2194       | r = 2 \times 1097 }} {{eqn | l = 2195       | r = 5 \times 439 }} {{eqn | l = 2196       | r = 2^2 \times 3^2 \times 61 }} {{eqn | l = 2197       | r = 13^3 }} {{eqn | l = 2198       | r = 2 \times 7 \times 157 }} {{eqn | l = 2199       | r = 3 \times 733 }} {{eqn | l = 2200       | r = 2^3 \times 5^2 \times 11 }} {{end-eqn}} It can be seen by inspection that each of the integers in this sequence shares at least one prime factor with at least one other. It is then worth noting that: {{begin-eqn}} {{eqn | l = 2183       | r = 37 \times 59 }} {{eqn | l = 2201       | r = 31 \times 71 }} {{end-eqn}} and it can be seen that the sequence can be extended neither upwards nor downwards. {{qed}}
Define $B := \left\{ {\complement_S\left({x^\succeq}\right): x \in S}\right\}$ where $x^\succeq$ denotes the upper closure of $x$. By definition of generated topology: :$\tau\left({B}\right)$ is a topology on $S$ where $B$ is a sub-basis of $\tau \left({B}\right)$. Thus by definition of lower topology: :$T := \left({S, \preceq, \tau\left({B}\right)}\right)$ has a lower topology. Thus by definition: :$T$ is a topological space. {{qed}}
Take the examples: :$(1): \quad \sequence {x_n} = \dfrac 1 n$ :$(2): \quad \sequence {y_n} = -\dfrac 1 n$ Then : :$\forall n \in \N_{>0}: \dfrac 1 n > 0, -\dfrac 1 n < 0$ From Sequence of Reciprocals is Null Sequence, we have :$x_n \to 0$ :$y_n \to 0$ as $n \to \infty$. However, it is clearly '''false''' that $0 > 0$ and $0 < 0$. {{qed}}
By Zsigmondy's Theorem, there exists a prime divisor $p$ of $a^{2 n} - b^{2 n}$ which does not divide $a^k - b^k$ for all $k < 2 n$ unless: :$n = 1$ and $a + b$ is a power of $2$ :$n = 3$, $a = 2$, $b = 1$ In particular, $p$ does not divide $a^{2 k} - b^{2 k} = \paren {a^k - b^k} \paren {a^k + b^k}$ for $k < n$. It remains to check the case $n = 1$ and $a + b$ a power of $2$. We have to show that $a^2 + b^2$ has an odd prime divisor. Since $a$ and $b$ are coprime, both $a$ and $b$ are odd. By Square Modulo 4, $a^2 + b^2 \equiv 2 \pmod 4$. Because $a > b > 0$, $a^2 + b^2 > 2$. But $4 \divides 2^k$ for $k > 1$. Thus $a^2 + b^2$ is not a power of $2$. Hence $a^2 + b^2$ has an odd prime divisor. {{qed}} {{Namedfor|Karl Zsigmondy|cat = Zsigmondy}} Category:Number Theory 3spcigpduvdts8wuzbojjnubqlu3z3v
By Product of Cyclotomic Polynomials, $p\mid \Phi_n(a) \mid a^n-1$. Thus $a$ is coprime to $p$. By Fermat's Little Theorem, $1\equiv a^n\equiv a^q\pmod p$. Thus $\operatorname{ord}_p(a) \leq q$. Suppose $\operatorname{ord}_p(a) = k < q$. By Product of Cyclotomic Polynomials, $p\mid \Phi_d(a)$ for some $d\mid k$. Then $a$ is a double root of $x^q-1$ modulo $p$. By Double Root of Polynomial is Root of Derivative, $q\equiv0\pmod p$. This is a contradiction, thus $k=q$. {{qed}} Category:Cyclotomic Polynomials 888kvf5vm9j2n8ofx0xc0w4sdd4dxze
Define: : $B := \left\{ {\complement_S \left({x^\succeq}\right): x \in S}\right\}$ where $x^\succeq$ denotes the upper closure of $x$. Thus: {{begin-eqn}} {{eqn | l = \tau_1       | r = \tau \left({B}\right)       | c = {{Defof|Topology Generated by Synthetic Sub-Basis}} }} {{eqn | r = \tau_2       | c = {{Defof|Topology Generated by Synthetic Sub-Basis}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\Phi_{2^n} } x       | r = \prod_{\zeta} \paren {x - \zeta}       | c = where the product runs over all primitive complex $2^n$th roots of unity }} {{eqn | r = \prod_{\substack {1 \mathop \le k \mathop \le 2^n \\ \gcd \set {k, 2^n} = 1} } \paren {x - \map \exp {\frac {2 \pi i k} {2^n} } }       | c = Condition for Complex Root of Unity to be Primitive }} {{eqn | r = \prod_{\substack {1 \mathop \le k \mathop \le 2^n \\ k \text { odd} } } \paren {x - \map \exp {\frac {k i \pi} {2^{n - 1} } } }       | c =  }} {{eqn | r = \prod_{k \mathop = 0}^{2^{n - 1} - 1} \paren {x - \map \exp {\frac {\paren {2 k + 1} i \pi} {2^{n - 1} } } }       | c = by writing each odd number as $2 k + 1$ }} {{eqn | r = x^{2^{n - 1} } + 1       | c = Factorisation of $z^n + 1$ }} {{end-eqn}} {{qed}} Category:Cyclotomic Polynomials 5f6f3syoexhclaxk8bbaw0tlv0k0h4y
We apply 1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways: {{:1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways}} Find the maximum powers of the primes in each equation, and choose the largest that appears: {{begin-eqn}} {{eqn | l = 42       | r = 2^1 \times 3^1 \times 7^1 }} {{eqn | l = 24       | r = 2^3 \times 3^1 }} {{eqn | l = 18       | r = 2^1 \times 3^2 }} {{eqn | l = 30       | r = 2^1 \times 3^1 \times 5^1 }} {{eqn | l = 20       | r = 2^2 \times 5^1 }} {{eqn | l = 12       | r = 2^2 \times 3^1 }} {{end-eqn}} Therefore the smallest number would be: :$2^3 \times 3^2 \times 5^1 \times 7^1 = 2520$ {{qed}}
We have the following: : $1$ is a Special Highly Composite Number : $2$ is a Special Highly Composite Number : $6$ is a Special Highly Composite Number : $12$ is a Special Highly Composite Number : $60$ is a Special Highly Composite Number : $2520$ is a Special Highly Composite Number By inspection of the sequence of highly composite numbers: :$1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, \ldots$ it can be seen that there are no more special highly composite numbers less than $2520$. Let $n > 18$. From Ratio between Consecutive Highly Composite Numbers Greater than 2520 is Less than 2, the $n$th highly composite number does not divide the $n+1$th. Hence the $n$th highly composite number is not a special highly composite number. The result follows. {{qed}}
This follows from the Lifting The Exponent Lemma for p=2 with $y$ replaced by $-y$. {{qed}}
Let $n = \displaystyle \prod_{k \mathop \in \N} {p_k}^{r_k}$ be highly composite. By definition of $\tau$ function: :$\tau \left({n}\right) = \displaystyle \prod_{k \mathop \in \N} \left({r_k + 1}\right)$ {{AimForCont}} $r_{l + 1} > r_l$ for some $l \in \N$. Consider $m \in \Z$ whose prime decomposition of $n$ is expressed as: :$m = \displaystyle \prod_{k \mathop \in \N} {p_k}^{s_k}$ where: :$\forall j < l: s_j = r_j$ :$\forall j > l + 1: s_j = r_j$ :$s_l = r_{l + 1}$ :$s_{l + 1} = r_l$ We have that: {{begin-eqn}} {{eqn | l = \tau \left({m}\right)       | r = \prod_{k \mathop \in \N} \left({s_k + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | r = \prod_{k \mathop \in \N} \left({r_k + 1}\right)       | c = as all the $s$s are the same as all the $r$s, but in a different order }} {{eqn | r = \tau \left({n}\right)       | c = {{Defof|Tau Function}} }} {{end-eqn}} Thus: : $\tau \left({m}\right) = \tau \left({n}\right)$ Now we have that: :$r_l < r_{l + 1}$ and so: :$s_l > s_{l + 1}$ and so: {{begin-eqn}} {{eqn | l = \frac { {p_l}^{r_l} {p_{l + 1} }^{r_{l + 1} } } { {p_l}^{s_l} {p_{l + 1} }^{s_{l + 1} } }       | r = \frac { {p_l}^{r_l} {p_{l + 1} }^{r_{l + 1} } } { {p_l}^{r_{l + 1} } {p_{l + 1} }^{r_l} }       | c =  }} {{eqn | r = \frac { {p_l}^c} { {p_{l + 1}^c} }       | c = where $c = r_{l + 1} - r_l$ }} {{end-eqn}} Then: :$p_{l + 1} > p_l$ and so: :${p_{l + 1} }^c > {p_l}^c$ from which it follows that: :$m < n$ while: :$\tau \left({m}\right) = \tau \left({n}\right)$ But $n$ is highly composite. This means that if $\tau \left({m}\right) = \tau \left({n}\right)$, then $n \le m$. This is contradicted by $n > m$. Thus by Proof by Contradiction: :$\forall k \in \N: r_k \ge r_{k + 1}$ {{qed}}
{{begin-eqn}} {{eqn 	| l = \map {\frac \d {\d x} } {\sinh x} 	| r = -i \map {\frac \d {\d x} } {\sin i x} 	| c = Hyperbolic Sine in terms of Sine  }} {{eqn 	| r = \cos i x  	| c = Derivative of Sine Function  }} {{eqn 	| r = \cosh x  	| c = Hyperbolic Cosine in terms of Cosine  }} {{end-eqn}} {{qed}}
Define $B := \left\{ {\complement_S\left({y^\succeq}\right): y \in S}\right\}$ By definition of lower topology: :$B$ is sub-basis of $T$. By definition of sub-basis: :$B \subseteq \tau$ By definition of $B$: :$\complement_S\left({x^\succeq}\right) \in B$ Thus by definition of subset: :$\complement_S\left({x^\succeq}\right) \in \tau$ Thus by definition: :$x^\succeq$ is closed. {{qed}}