What is the largest number we can get using $n$ ones, addition, multiplication and brackets? Let's say we have $n$ ones, i.e. $1,1,\dots,1$ $n$ times and are allowed to add them, multiply and insert brackets wherever we want.
What is the largest number we can get for a particular $n$? Is there a closed form or at least an OEIS sequence?
For $n=5$ it appears to be $(1+1)(1+1+1)=6$, for $n=6$ it appears to be $(1+1+1)(1+1+1)=9$, for $n=9$ I found $(1+1+1)(1+1+1)(1+1+1)=27$ to be the largest number.
But I don't see a way to find a general formula. I guess it would make sense to start from the other end - for each number $N$ find a factorization with the least sum of factors or something like that.
 A: Since 
1) it doesn't make sense to create any 1+1+1+... terms with more than 3 1's, since 1+1+1+1+1 is 'worse' than (1+1) times (1+1+1),  and for more 1's it will be even 'better' to multiply several 1+1+.. Terms rather than have one long one, and with 4 1s 1+1+1+1 is just as good as (1+1) times (1+1),
and
2) two 1+1+1 terms is better than three 1+1 terms,
your basic strategy is to get as many 1+1+1 terms as possible. So:
If n mod 3 = 0, the best you can do is $3^{n/3}$
If n mod 3 = 1, get two 1+1 terms, and otherwise 1+1+1 terms, so you get $4*3^{(n-4)/3}$
If n mod 3 = 2, get one 1+1 term in addition to 1+1+1 terms, so $2*3^{(n-2)/3}$
A: Don't know if this could be an "approximate" solution or anything. But try the following.
If the final expression consists of $x_1 \times ... \times x_k$ where $x_1+...+x_k=n$, with $k<n$, Holding n,k fixed, and suppose the $x$'s need not be integers for the moment, the max is $(n/k)^k$. Taking derivative wrt k and get $(n/k)^k ( \ln(n/k)-1)=0)$ to obtain the best $k$ which is about $n/e$, with optimally each term being approx $x \approx 2.7$, which can be as good as $3$ in the case of integers
A: Obviously the problem reduces to finding the maximum of $a_1 \cdot a_2 \cdot ... \cdot a_t$, when $a_1 + a_2 + ... + a_t = n$. Now let's increase $a_1$ by one and decrease $a_2$ by one. Then the value of $a_1 \cdot a_2 \cdot ... \cdot a_t$ will increase if $a_1 \ge a_2+2$. Eventually this leads to concluding that for a "fixed" $t$ the maximum will occur when $a$'s are as close as possible. 
Now let's find the optimal $t$. As previously concluded assume that $a_1 = \cdots = a_t = a$. Then we have $a = \frac nt$. Now $a_1 \cdot a_2 \cdot ... \cdot a_t = \left(\frac nt \right)^t$. This function obtains a maximum at $t=e$, so therefore the maximum is obtained for $t = \frac ne$
As all the variables are integers and as the functions are continuous we have that the we need to choose $t = \left\lfloor \frac ne \right\rfloor$ or $\left\lfloor \frac ne \right\rfloor + 1$. In each case we choose $a_i = \left\lfloor \frac nt \right\rfloor$ or $\left\lfloor \frac nt \right\rfloor + 1$ accordingly, s.t. the sum is $n$. This reduces the problem to few cases, which can be easily checked. 
