Special Highly Composite Number/Examples/2520
== Example of Special Highly Composite Number ==  $2520$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.  == Proof == By inspection of the sequence of highly composite numbers, $2520$ is highly composite. For reference, the prime decomposition of $2520$ is: :$2520 = 2^3 \times 3^2 \times 5 \times 7$  $n > 2520$ is a highly composite number which is not divisible by $2520$. We have that $60$ is a special highly composite number. Therefore $60$ is a divisor of $n$. It follows that $3$, $4$ and $5$ are all divisors of $n$. But as $2520$ is not a divisor of $n$, it follows that at least one of $7$, $8$ and $9$ is not a divisor of $n$. These will be investigated on a case-by-case basis. :$(1): \quad 7$  is not a divisor of $n$. By Prime Decomposition of Highly Composite Number we have that: :$n = 2^a \times 3^b \times 5^c$ where $a \ge b \ge c \ge 1$. Suppose that $a < 3$. Then: :$n \le 2^2 \times 3^2 \times 5^2 = 900$ which is too small. So we have that $a \ge 3$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 3} \times 3^b \times 5^c \times 7       | o = <       | r = 2^a \times 3^b \times 5^c       | c = because $7 < 2^3 = 8$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 3^b \times 5^c \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3^b \times 5^c}\right)       | c = as $2^a \times 3^b \times 5^c$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 7}\right) \times \tau \left({3^b \times 5^c}\right)       | o = <       | r = \tau \left({2^a}\right) \times \tau \left({3^b \times 5^c}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 7}\right)       | o = <       | r = \tau \left({2^a}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({a - 3}\right) + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 2 \left({a - 2}\right)       | o = <       | r = a + 1       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 5       | c =  }} {{end-eqn}} Now suppose that $b < 2$. Then: :$n \le 2^5 \times 3 \times 5 = 480$ which is too small. So we have that $b \ge 2$. Then: {{begin-eqn}} {{eqn | l = 2^a \times 3^{b - 2} \times 5^c \times 7       | o = <       | r = 2^a \times 3^b \times 5^c       | c = because $7 < 3^2 = 9$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 3^{b - 2} \times 5^c \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3^b \times 5^c}\right)       | c = as $2^a \times 3^b \times 5^c$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 5^c}\right) \times \tau \left({3^{b - 2} \times 7}\right)       | o = <       | r = \tau \left({2^a \times 5^c}\right) \times \tau \left({3^b}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({3^{b - 2} \times 7}\right)       | o = <       | r = \tau \left({3^b}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({b - 2}\right) + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({b + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 2 \left({b - 1}\right)       | o = <       | r = b + 1       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = <       | r = 3       | c =  }} {{end-eqn}} But $c \le b$ and so $c < 3$ as well. Thus we have upper bounds on $a$, $b$ and $c$. Since $2^a \times 3^b \times 5^c > 2520$, it must be the case that: :$n = 2^4 \times 3^2 \times 5^2$ which gives that $n = 3600$. But: : from {{TauLink|3600}} we have that $\tau \left({3600}\right) = 45$ : from {{TauLink|2520}} we have that $\tau \left({2520}\right) = 48$ This contradicts our hypothesis that $3600$ is highly composite. By Proof by Contradiction it follows that $7$ must be a divisor of $n$. {{qed|lemma}} :$(2): \quad 9$ is not a divisor of $n$, but $7$ is. By Prime Decomposition of Highly Composite Number we have that: :$n = 2^a \times 3^1 \times 5^1 \times 7^1 \times 11^e \times r$ where: :$e$ is either $0$ or $1$ :$r$ is a possibly vacuous square-free product of prime numbers strictly greater than $11$. Suppose $e = 1$. Then: {{begin-eqn}} {{eqn | l = 2^a \times 3^3 \times 5 \times 7 \times r       | o = <       | r = 2^a \times 3 \times 5 \times 7 \times 11 \times r       | c = because $9 = 3^2 < 11$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 3^3 \times 5 \times 7 \times r}\right)       | o = <       | r = \tau \left({2^a \times 3 \times 5 \times 7 \times 11 \times r}\right)       | c = as $2^a \times 3 \times 5 \times 7 \times 11 \times r$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 5 \times 7 \times r}\right) \times \tau \left({3^3}\right)       | o = <       | r = \tau \left({2^a \times 5 \times 7 \times r}\right) \times \tau \left({3 \times 11}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({3^3}\right)       | o = <       | r = \tau \left({3 \times 11}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = 3 + 1       | o = <       | r = \left({1 + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 4       | o = <       | r = 4       | c = which is an absurdity }} {{end-eqn}} So $e = 0$ and so by Prime Decomposition of Highly Composite Number $r = 1$. Thus: :$n = 2^a \times 3 \times 5 \times 7$ We have that $n > 2520$, so: :$(2 \text a): a \ge 5$ Then: {{begin-eqn}} {{eqn | l = 2^{a - 2} \times 3^2 \times 5 \times 7       | o = <       | r = 2^a \times 3 \times 5 \times 7       | c = because $3 < 2^2 = 4$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 2} \times 3^2 \times 5 \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3 \times 5 \times 7}\right)       | c = as $2^a \times 3 \times 5 \times 7$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 2} \times 3^2}\right) \times \tau \left({5 \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3}\right) \times \tau \left({5 \times 7}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 2} \times 3^2}\right)       | o = <       | r = \tau \left({2^a \times 3}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({a - 2}\right) + 1}\right) \left({2 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 3 \left({a - 1}\right)       | o = <       | r = 2 \left({a + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 5       | c = which is a contradiction of $(2 \text a)$ }} {{end-eqn}} It follows by Proof by Contradiction that $9$ is a divisor of $n$. {{qed|lemma}} :$(3): \quad 8$ is not a divisor of $n$, but $7$ and $9$ both are. By Prime Decomposition of Highly Composite Number we have that: :$n = 2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r$ where $r$ is a possibly vacuous product of prime numbers strictly greater than $11$. Suppose: :$(3 \text a): \quad e > 0$ Then: {{begin-eqn}} {{eqn | l = 2^5 \times 3^2 \times 5^c \times 7^d \times 11^{e - 1} \times r       | o = <       | r = 2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r       | c = because $8 = 2^3 < 11$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 3^2 \times 5^c \times 7^d \times 11^{e - 1} \times r}\right)       | o = <       | r = \tau \left({2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r}\right)       | c = as $2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 11^{e - 1} }\right) \times \tau \left({3^2 \times 5^c \times 7^d \times r}\right)       | o = <       | r = \tau \left({2^2 \times 11^e}\right) \times \tau \left({3^2 \times 5^c \times 7^d \times r}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 11^{e - 1} }\right)       | o = <       | r = \tau \left({2^2 \times 11^e}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({5 + 1}\right) \left({\left({e - 1}\right) + 1}\right)       | o = <       | r = \left({2 + 1}\right) \left({e + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 6 e       | o = <       | r = 3 \left({e + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = e       | o = <       | r = 1       | c = which is a contradiction of $(3 \text a)$ }} {{end-eqn}} So $e = 0$ and so Prime Decomposition of Highly Composite Number $r = 1$. Thus: :$n = 2^2 \times 3^2 \times 5^c \times 7^d$ Suppose $c = 2$. Then: {{begin-eqn}} {{eqn | l = 2^4 \times 3^2 \times 5^1 \times 7^d       | o = <       | r = 2^2 \times 3^2 \times 5^2 \times 7^d       | c = because $4 = 2^2 < 5$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^4 \times 3^2 \times 5^1 \times 7^d}\right)       | o = <       | r = \tau \left({2^2 \times 3^2 \times 5^2 \times 7^d}\right)       | c = as $2^2 \times 3^2 \times 5^2 \times 7^d$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^4 \times 5}\right) \times \tau \left({3^2 \times 7^d}\right)       | o = <       | r = \tau \left({2^2 \times 5^2}\right) \times \tau \left({3^2 \times 7^d}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^4 \times 5}\right)       | o = <       | r = \tau \left({2^2 \times 5^2}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({4 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({2 + 1}\right) \left({2 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 10       | o = <       | r = 9       | c = which is an absurdity }} {{end-eqn}} The remaining possibility is that $c = 1$ and $d = 1$: Thus: :$n = 2^2 \times 3^2 \times 5 \times 7 = 1260$ But this is a contradiction of our supposition that $n > 2520$. It follows by Proof by Contradiction that $8$ is a divisor of $n$. {{qed|lemma}} By Proof by Cases it is seen that the existence of a highly composite $n$ not divisible by $2520$ leads to a contradiction. The result then follows by Proof by Contradiction.  == Sources == * {{Citation|title = An Interesting Subset of the Highly Composite Numbers|author = Steven Ratering|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 64|issue = 5|date = Dec. 1991|startpage = 343|endpage = 346|jstor = 2690653}}