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Aaron Meyerowitz
Aaron Meyerowitz
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Here are a few computed examples, but first comments which move things along, although not very far.

I'll write examples as $\frac{n!}{a!b!}$ with $a \ge b.$ We might define the virtue of such an example to be $v=a+b-n.$ Then we know that $\frac{(k!)!}{(k!-1)!k!}$ has virtue $k-1.$

One can look the denominators (after reduction to lowest terms) of $\frac{n!}{a!(b+1)!},\frac{n!}{(a+1)!b!}$ and $\frac{(n-1)!}{a!b!}$ to see what primes it was which denied a slightly more virtuous example. My observation from small cases is that, perhaps unsurprisingly, is is frequently just $2$ or some few small primes.

An trivial yet helpful equivalent condition which keeps the numbers smaller is that $b!$ divide $(a+1)(a+2)\cdots(a+b-v).$ So we are looking for the product of relatively few consecutive integers to be a multiple of $b!$

I was excited by the fact that $7!=71^2-1=70\cdot 72$ giving virtue $4$ to $\frac{72!}{69!7!}$ which is ahead of $\frac{120!}{119!5!}.$ However the first examples with virtue $4$ are actually $\frac{36!}{23!17!}$ and $\frac{36!}{29!11!}.$ Observe that $\frac{35!}{23!17!},\frac{36!}{24!17!}$ and $\frac{36!}{23!18!}$ have denominators $4,8$ and $2$. The best examples seem to have $b$ much larger than $v.$

There are three more values of $n$ giving virtue $4$ before

  • $\frac{56!}{47!14!}$ , the first of $5$ values of $n$ giving virtue $5$ before:
  • $\frac{100!}{59!47!}$ , the first of $10$ values of $n$ giving virtue $6$ before:
  • $\frac{162!}{107!62!}$ and $\frac{162!}{110!59!}$, the first of $2$ values of $n$ giving virtue $7$ before:
  • $\frac{256!}{142!122!},\frac{256!}{202!62!},\frac{256!}{239!25!}$ and $\frac{256!}{241!23!}$, the only value of $n$ giving virtue $8$ before:
  • $\frac{261!}{223!47!}$ and $\frac{261!}{239!31!}$ of virtue $9.$

Another example with virtue $9$ is $\frac{273!}{223!59!}.$ That is the only one for $n=273.$ I do not know if there are any other examples before the first of virtue $10.$

Note that $224=2^57$ and $240=2^43\ 5$ are just a bit too far for the denominator in more than one case.

Later

One might add that $225=15^2$ and $243=3^5$ along with $242=2\ 11^2$ are also potential obstacles in the denominator.

Here are a few computed examples, but first comments which move things along, although not very far.

I'll write examples as $\frac{n!}{a!b!}$ with $a \ge b.$ We might define the virtue of such an example to be $v=a+b-n.$ Then we know that $\frac{(k!)!}{(k!-1)!k!}$ has virtue $k-1.$

One can look the denominators (after reduction to lowest terms) of $\frac{n!}{a!(b+1)!},\frac{n!}{(a+1)!b!}$ and $\frac{(n-1)!}{a!b!}$ to see what primes it was which denied a slightly more virtuous example. My observation from small cases is that, perhaps unsurprisingly, is is frequently just $2$ or some few small primes.

An trivial yet helpful equivalent condition which keeps the numbers smaller is that $b!$ divide $(a+1)(a+2)\cdots(a+b-v).$ So we are looking for the product of relatively few consecutive integers to be a multiple of $b!$

I was excited by the fact that $7!=71^2-1=70\cdot 72$ giving virtue $4$ to $\frac{72!}{69!7!}$ which is ahead of $\frac{120!}{119!5!}.$ However the first examples with virtue $4$ are actually $\frac{36!}{23!17!}$ and $\frac{36!}{29!11!}.$ Observe that $\frac{35!}{23!17!},\frac{36!}{24!17!}$ and $\frac{36!}{23!18!}$ have denominators $4,8$ and $2$. The best examples seem to have $b$ much larger than $v.$

There are three more values of $n$ giving virtue $4$ before

  • $\frac{56!}{47!14!}$ , the first of $5$ values of $n$ giving virtue $5$ before:
  • $\frac{100!}{59!47!}$ , the first of $10$ values of $n$ giving virtue $6$ before:
  • $\frac{162!}{107!62!}$ and $\frac{162!}{110!59!}$, the first of $2$ values of $n$ giving virtue $7$ before:
  • $\frac{256!}{142!122!},\frac{256!}{202!62!},\frac{256!}{239!25!}$ and $\frac{256!}{241!23!}$, the only value of $n$ giving virtue $8$ before:
  • $\frac{261!}{223!47!}$ and $\frac{261!}{239!31!}$ of virtue $9.$

Another example with virtue $9$ is $\frac{273!}{223!59!}.$ That is the only one for $n=273.$ I do not know if there are any other examples before the first of virtue $10.$

Note that $224=2^57$ and $240=2^43\ 5$ are just a bit too far for the denominator in more than one case.

Here are a few computed examples, but first comments which move things along, although not very far.

I'll write examples as $\frac{n!}{a!b!}$ with $a \ge b.$ We might define the virtue of such an example to be $v=a+b-n.$ Then we know that $\frac{(k!)!}{(k!-1)!k!}$ has virtue $k-1.$

One can look the denominators (after reduction to lowest terms) of $\frac{n!}{a!(b+1)!},\frac{n!}{(a+1)!b!}$ and $\frac{(n-1)!}{a!b!}$ to see what primes it was which denied a slightly more virtuous example. My observation from small cases is that, perhaps unsurprisingly, is is frequently just $2$ or some few small primes.

An trivial yet helpful equivalent condition which keeps the numbers smaller is that $b!$ divide $(a+1)(a+2)\cdots(a+b-v).$ So we are looking for the product of relatively few consecutive integers to be a multiple of $b!$

I was excited by the fact that $7!=71^2-1=70\cdot 72$ giving virtue $4$ to $\frac{72!}{69!7!}$ which is ahead of $\frac{120!}{119!5!}.$ However the first examples with virtue $4$ are actually $\frac{36!}{23!17!}$ and $\frac{36!}{29!11!}.$ Observe that $\frac{35!}{23!17!},\frac{36!}{24!17!}$ and $\frac{36!}{23!18!}$ have denominators $4,8$ and $2$. The best examples seem to have $b$ much larger than $v.$

There are three more values of $n$ giving virtue $4$ before

  • $\frac{56!}{47!14!}$ , the first of $5$ values of $n$ giving virtue $5$ before:
  • $\frac{100!}{59!47!}$ , the first of $10$ values of $n$ giving virtue $6$ before:
  • $\frac{162!}{107!62!}$ and $\frac{162!}{110!59!}$, the first of $2$ values of $n$ giving virtue $7$ before:
  • $\frac{256!}{142!122!},\frac{256!}{202!62!},\frac{256!}{239!25!}$ and $\frac{256!}{241!23!}$, the only value of $n$ giving virtue $8$ before:
  • $\frac{261!}{223!47!}$ and $\frac{261!}{239!31!}$ of virtue $9.$

Another example with virtue $9$ is $\frac{273!}{223!59!}.$ That is the only one for $n=273.$ I do not know if there are any other examples before the first of virtue $10.$

Note that $224=2^57$ and $240=2^43\ 5$ are just a bit too far for the denominator in more than one case.

Later

One might add that $225=15^2$ and $243=3^5$ along with $242=2\ 11^2$ are also potential obstacles in the denominator.

Source Link
Aaron Meyerowitz
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

Here are a few computed examples, but first comments which move things along, although not very far.

I'll write examples as $\frac{n!}{a!b!}$ with $a \ge b.$ We might define the virtue of such an example to be $v=a+b-n.$ Then we know that $\frac{(k!)!}{(k!-1)!k!}$ has virtue $k-1.$

One can look the denominators (after reduction to lowest terms) of $\frac{n!}{a!(b+1)!},\frac{n!}{(a+1)!b!}$ and $\frac{(n-1)!}{a!b!}$ to see what primes it was which denied a slightly more virtuous example. My observation from small cases is that, perhaps unsurprisingly, is is frequently just $2$ or some few small primes.

An trivial yet helpful equivalent condition which keeps the numbers smaller is that $b!$ divide $(a+1)(a+2)\cdots(a+b-v).$ So we are looking for the product of relatively few consecutive integers to be a multiple of $b!$

I was excited by the fact that $7!=71^2-1=70\cdot 72$ giving virtue $4$ to $\frac{72!}{69!7!}$ which is ahead of $\frac{120!}{119!5!}.$ However the first examples with virtue $4$ are actually $\frac{36!}{23!17!}$ and $\frac{36!}{29!11!}.$ Observe that $\frac{35!}{23!17!},\frac{36!}{24!17!}$ and $\frac{36!}{23!18!}$ have denominators $4,8$ and $2$. The best examples seem to have $b$ much larger than $v.$

There are three more values of $n$ giving virtue $4$ before

  • $\frac{56!}{47!14!}$ , the first of $5$ values of $n$ giving virtue $5$ before:
  • $\frac{100!}{59!47!}$ , the first of $10$ values of $n$ giving virtue $6$ before:
  • $\frac{162!}{107!62!}$ and $\frac{162!}{110!59!}$, the first of $2$ values of $n$ giving virtue $7$ before:
  • $\frac{256!}{142!122!},\frac{256!}{202!62!},\frac{256!}{239!25!}$ and $\frac{256!}{241!23!}$, the only value of $n$ giving virtue $8$ before:
  • $\frac{261!}{223!47!}$ and $\frac{261!}{239!31!}$ of virtue $9.$

Another example with virtue $9$ is $\frac{273!}{223!59!}.$ That is the only one for $n=273.$ I do not know if there are any other examples before the first of virtue $10.$

Note that $224=2^57$ and $240=2^43\ 5$ are just a bit too far for the denominator in more than one case.