What about it?

Nice problem!

The sum(in the denominator) is clearly -

$\displaystyle \sum_{j=2}^{m}{\left\lfloor \left\{\frac{(j-1)!}{j}\right\} + \frac{1}{j}\right \rfloor}$

Now, for a non-prime $j$ what happens?

Yes, the fractional part vanishes because the "GIF" part would be an integer(a non-prime would mean it has all factors less than itself and all of them are contained in the numerator factorial term)

Our sum becomes - $\displaystyle \sum_{\text{non-prime j}}{\left\lfloor \left\{\frac{(j-1)!}{j}\right\} + \frac{1}{j} \right \rfloor} = \sum_{\text{non-prime j}}{\left \lfloor \frac{1}{j} \right \rfloor} = 0$

Now, what about primes?

Well, for that, remember Wilson's Theorem from which we see that

$\displaystyle (j-1)! = (j-1) \text{ mod } j$

And hence, $\displaystyle \left\{ \frac { (j-1)! }{ j } \right\} = \frac{j-1}{j}$

And our sum becomes

$\displaystyle \sum_{\text{prime j}}{\left\lfloor \frac{j-1}{j} + \frac{1}{j} \right \rfloor} = \sum_{\text{prime j}}{1} = \pi(m)$

where $\pi(m)$ is the prime counting function .

Hence, our total sum becomes -

$\displaystyle \sum_{m=1}^{{2}^{820}}{\left \lfloor {\left \lfloor \frac{820}{1+\pi(m)} \right \rfloor}^{1/820} \right \rfloor}$

One can now see that the term inside the summation(except the floors) would be zero if the denominator exceeds the numerator.

And for all those which the the term inside floors equals 1, the whole term(with the floors) equals 1.

Also, for those below the 1 value, the term inside the floors $>1$ , hence when raised to the power $1/820$ returns value $>1$ and after taking floor, comes out to be equal to 1.

Hence, our whole term(with the floors) $=1$ for $\pi(m) + 1 \le 820$

As a result,

$\displaystyle \sum_{m=1}^{{2}^{820}}{\left \lfloor {\left \lfloor \frac{820}{1+\pi(m)} \right \rfloor}^{1/820} \right \rfloor} = {\pi}^{-1}(820) -1 = \boxed{6300}$

where " ${\pi}^{-1}(x)$ " is just used for writing purposes, it is not a valid thing. It just means the value of least prime number for which $\pi(p)$ gives the value of $x$ .

Also, we just found a general solution too.

Firstly, $\frac {(j-1)!} j - \left\lfloor \frac {(j-1)!} j \right\rfloor = \large\left\{ \frac{(j-1)!} j \right\} = d\leq 0$ (say)

The only time 'd' is not zero is when $\frac {(j-1)!} j \in N$ This happens whenever the prime factors of 'j' occur at least as many times as needed (the power of the prime factor) in the factorial of (j-1).

But what if 'j' is a perfect power? Let j = $a^n$ where 'a' and 'n' are natural numbers. In the worst case, a is prime and it needs to occur at least n times in the factorial of 'j-1', i.e, $a^n-1\geq a\cdot n \Rightarrow a^n-an-1\nless0$

I checked the plot of the inequality: $a^n-an-1<0$ and good news! For a = 2, n=2 is the only solution for a, n>1; otherwise, for all numbers (inclusive of those that are prime) greater than 2 and for n>1, there are no solutions to the inequality. As for the solution, that is, j=4, if you check, you'll get that the floor function in the summation in the denominator returns the value: $\lfloor 0.75 \rfloor = 0$ .

Now it becomes obvious that the only times $\lfloor d+\frac 1 j \rfloor = 1$ are for when j takes the value of a prime number. It can only take the value of 0 or 1 and it can't go up, for obvious reasons.

I checked for many prime numbers and got that for all of them, the value of the above expression was 1. But I don't know why just yet. Pi Han Goh , a little help on this matter might help. So, I assumed that for all prime numbers, this is the case and proceeded.

Now, $\lfloor 820 \rfloor^ {\frac 1 {820}} = 1$ and the sum becomes 1+1+1+1+......+1+1. But when does it end? It ends when the denominator becomes 820. But when does this happen? The denominator is $1+\sum \lfloor .. \rfloor$ and in the sum, the floor function's value is either a '1' (when j is prime) or a '0' (when j is not prime). So, the denominator becomes 820 when j goes all the way up to the 819th prime (Don't forget about the '1' in the beginning). And this happens as long as j doesn't hit the value of the 820th prime. So, it goes on till 6301 - 1 = $\boxed {6300} (\because \text{820th prime -1})$

Let's start with the summand expression in the denominator $\left \lfloor {\frac {(j-1)!+1}{j}}- \left \lfloor{\frac{(j-1)!}{j}}\right \rfloor \right \rfloor$ This expression equals 1 when $j \mid (j-1)!+1$ , and otherwise it equals 0.

First I show that for all $j>=2$ : $j \mid (j-1)!+1 \Leftrightarrow j \text{ is a prime number}$

If j is composite say $j=ab$ with $1<a<b<j$ , we can see that both a and b occur as distinct factors of $(j-1)!$ , so that j divides $(j-1)!$ . The only composites where we cannot arrange for $a<b$ are squares of primes. if $j=a^2$ and $j>=9$ , then $(j-1)!$ contains the factors $a$ and $2a$ , so also all primes squares $j>=9$ have $j \mid (j-1)!$ too. So, for all composite numbers except 4: $j \mid (j-1)!$ and therefore $j \text { is composite} \Rightarrow j \nmid (j-1)!+1$ .

If j is a prime $j \mid j^2 =(j-1)(j+1)+1=(j-1)((j+1)/2)(2)+1$ . For primes $j>=5$ , the numbers $j-1$ , $(j+1)/2$ and $2$ all occur as distinct factors of $(j-1)!$ , and it is easily checked that $2 \mid (2-1)!+1$ , and $3 \mid (3-1)!+1$ . $j \text { is prime} \Rightarrow j \mid (j-1)!+1$ ■

Corollary: composite values for j do not contribute to the sum, and each prime adds the value 1. So we have that the sum in the denominator is exactly the number of primes less than or equal to m , we denote this with funtion $π(m)$ . We rewrite the expression as $\sum_{m=1}^{2^n} \left \lfloor{\left \lfloor {\frac{n}{1+π(m)}} \right \rfloor} ^{\frac{1}{n}}\right \rfloor$ with $n=820$ . Next notice

• $\left \lfloor{\frac{n}{1+π(m)}}\right \rfloor=0$ when $π(m) \geq n$ , and $\left \lfloor{\frac{n}{1+π(m)}}\right \rfloor >=1$ when $π(m) \lt n$
• Suppose that $\left \lfloor{\frac{n}{1+π(m)}}\right \rfloor ^{\frac1n} >= 2$ , we would need $n \geq 2^n(1+π(m))$ , but it is clear that $n<2^n(1+π(m))$ . From this contradiction our assumption was wrong and hence $\left \lfloor{\frac{n}{1+π(m)}}\right \rfloor ^{\frac1n} \lt 2$ so that $\left \lfloor\left \lfloor{\frac{n}{1+π(m)}}\right \rfloor ^{\frac1n}\right \rfloor \leq 1$ .

So the summand $\left \lfloor{\left \lfloor {\frac{n}{1+π(m)}} \right \rfloor} ^{\frac{1}{n}}\right \rfloor$ takes the value $1$ for every $m$ for which $π(m) < n$ , and takes the value $0$ for every $m$ for which $π(m) >= n$ . This means that the sum just equals the $n^{th}$ prime minus 1, which for n=820 equals $6301-1=\boxed{6300}$ .

I also doublechecked my result for low n using this code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

import math

for n in range(1,11):
    nsum=0;
    for m in range(1,2**n+1):
        msum=0;
        for j in range (2, m+1):
            f=math.factorial(j-1);
            msum += (f+1)//j-(f)//j;
        nsum += math.floor((n//(1+msum))**(1.0/n));
    print ("S({})={}, ".format(n, nsum), end="");

--------------
output::
S(1)=1, S(2)=2, S(3)=4, S(4)=6, S(5)=10, S(6)=12, S(7)=16, S(8)=18, S(9)=22, S(10)=28,