Smallest integer divisible by all up to $n$ 
What's the smallest positive integer which is divisible by all integers $1, 2, \dots, n$?  Is there a simple way to represent the answer?  Call it $f(n)$ here.

Clearly factorial ($n!$) satisfies the condition of divisibility.  But once you get to $f(4)$ (which is $12$), factorial ($24$) is too large, because it includes as factors both $4$ and $2$, which are redundant.
Multiplying all the prime numbers up to $n$ gives you another estimate, which is this time too low, because you need to include primes multiple times when they're repeated in the factorization of the inputs.
So the answer is somewhere between factorial, and the product of primes.  Is there a simple answer? 
 A: Given $n$, and given a prime number $p$, there's a non-negative integer $r=r(p)$ such that $p^r\le n\lt p^{r+1}$. The number you are looking for is the product of all the numbers $p^r$ over all the primes $p$. 
It is known to be asymptotic to $e^n$, but this is not so easy to prove. 
A: For more, see http://oeis.org/A003418 and references given there.
A: I guess I'll make my comments an answer...
It's the least common multiple of $\{1,\dots,n\}$, that is, $\mathrm{lcm}(1,2,\dots,n)$
Let $p_1,\dots,p_\ell$ be the primes less than or equal to $n$ and $k_i$ be the largest integer such that $p_i^{k_i}<n$, then $\mathrm{lcm}(1,2,\dots,n)=p_1^{k_1}\cdots p_\ell^{k_\ell}$.
A: The answer is
$$f(n) = \operatorname{lcm}(1,2,3,...,n) = \prod\limits_{k=1}^{m-1}p_{k}^\left \lfloor \log_{p_{k}}(n)  \right \rfloor$$
where $p_m$ is the first prime with $p_{m} > n$
This is actually related to one of the Chebyshev functions $\psi(x)$ being 
$$f(n)=e^{\psi(x)}$$
There is a well-known formula that gives explicitly $\psi(x)$ (with a small technical and not very important difference relating the values on the jumps).
$$\psi_{0}(x)=x-\sum_\rho\frac{x^\rho}{\rho} - \log(2\pi) -\log(1-x^{-2})/2$$
where the summation goes over all zeros of Riemann zeta function
So a simple question going very deep immediately: 1 million dollars deep. Strange that we still know so little.
A: Yes: $\text{lcm}(1,2,\ldots,n)$.
A: The least common multiple of a set of numbers can be found by means of prime factorization.
For the positive integers $1$ through $n$, this prime factorization is
$$
\mathrm{lcm} \{1,2,3,\ldots,n\} = 2^{k_2}3^{k_3}5^{k_5}7^{k_7}11^{k_{11}} \cdots
$$
where $k_2$ is the largest power of $2$ found in any prime factorization of $n$ or any smaller positive integer, $k_3$ is the largest power of $3$ found in any of those prime factorizations, and so forth for each prime.
For any prime $p$, clearly
$p^k$ is the smallest positive integer with $p^k$
in its prime factorization.
That is, if $n$ or any smaller positive integer has $p^k$ as a factor,
then $p^k \leq n.$
Conversely, if $p^{k+1}$ is not a divisor of 
$\mathrm{lcm} \{1,2,3,\ldots,n\}$
then $p^{k+1} > n.$
So to find the correct power $k_p$ of any prime $p$ in the factorization
of $\mathrm{lcm} \{1,2,3,\ldots,n\}$ we simply find the largest power of $p$
that is not greater than $n.$
By the way, for $n > 1,$ if $n$ is not itself a power of a prime
then all the powers of primes that can be found in $n$ are already
found in $1,2,3,\ldots, n-1.$
In that case,
$$
\mathrm{lcm} \{1,2,3,\ldots,n\} =
\mathrm{lcm} \{1,2,3,\ldots,n-1\}.
$$
But if $n$ is a power of a prime, for example, $n = p^m$ (with $m>0$),
then the power of $p$ in the factorization of 
$\mathrm{lcm} \{1,2,3,\ldots,n\}$ is $m$
whereas the power of $p$ in the factorization of
$\mathrm{lcm} \{1,2,3,\ldots,n-1\}$ is $m-1$.
All other primes that occur in those factorizations occur with the
same powers, so
$$
\mathrm{lcm} \{1,2,3,\ldots,n\} =
p \times \mathrm{lcm} \{1,2,3,\ldots,n-1\}.
$$
Putting this together,
$$
f(n) = \begin{cases}
p f(n-1) & \text{if $n$ is a power of the prime $p$}\\
f(n-1)   & \text{otherwise.}
\end{cases}
$$
