Does anyone know how to reduce this sum of sums into something simpler in order to find a special value? [duplicate]

Question

to clarify the difference between this and the supposed duplicate, these two questions talk about completely different functions with completely different purposes

I was given this from a friend. They asked me to deduce what the equation is of. I played around with trying to compute alpha for some time. Plugging it into f(x), the function appeared to equal $0$ almost everywhere. I could never find the actual value of alpha. It appears to be an infinite irrational decimal number starting with $1.1973...$.

Let $$\alpha = \sum_{m=1}^{\infty} m * (2^{-(\sum_{j=2}^{m} (\lfloor \frac {(j-1)! + 1}{j} \rfloor - \lfloor \frac {(j-1)!}{j} \rfloor))^2}) \\* \left(\left\lfloor \frac {\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)-1}{\sqrt {(\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)-1)^2 + 1}} \right\rfloor + 1\right) * \left(\left\lfloor \frac {-\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)+1}{\sqrt {(\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} (\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)-1)^2 + 1}} \right\rfloor + 1\right)$$

Then let $f(x) = \lfloor 2^{x^2}* \alpha \rfloor - 2^{2x-1}*\lfloor 2^{(x-1)^2}*\alpha \rfloor$.

What is the function f(x)? Is there any way to reduce alpha to something simpler? I think this equation isn't something trivial. It doesn't appear to be $0$. It seems bizarre that alpha consists of so many sums. How do I reduce them into something simpler?

Answer 1

This recurring expression: $(\lfloor \frac {m}{i} \rfloor - \lfloor \frac {m-1}{i} \rfloor)$ effectively gives $1$ if $i$ is a factor of $m$, zero otherwise. So summed over a constant $m$ up to $m$, it would be the divisor function $\sigma_0(m)$. In the case where the sum only runs to include the square root of $m$, it is $\lceil\sigma_0(m)/2\rceil$. In the expression we have some minor complication for $m=1,2$ because the sums run to $2$ in both cases:

$$g(m):=\sum_{i = 1}^{\lfloor \sqrt{m} \rfloor + 1} \left(\left\lfloor \frac {m}{i} \right\rfloor - \left\lfloor \frac {m-1}{i} \right\rfloor\right) = \begin{cases} \sigma_0(m) & m\in \{1,2\} \\ \lceil\sigma_0(m)/2\rceil & m\ge 3\\ \end{cases}$$

Of particular relevance, $g(m)=1$ for primes $>2$ and $g(m)>1$ for composite numbers.

Then if you are aware of Wilson's theorem, you can see that the top expression used as an exponent to $2$, $\sum_{j=2}^{m} (\lfloor \frac {(j-1)! + 1}{j} \rfloor - \lfloor \frac {(j-1)!}{j} \rfloor)$, is counting the number of primes to $m$, that is, $\pi(m)$.

Now, taking $h(m):=g(m)-1$, this simplifies the expression to $$\alpha = \sum_{m=1}^{\infty}\left [ m \left(2^{\large -(\pi(m))^2}\right) \left(\left\lfloor \frac {h(m)}{\sqrt {(h(m))^2 + 1}} \right\rfloor + 1\right) \left(\left\lfloor \frac {-h(m)}{\sqrt {(h(m))^2 + 1}} \right\rfloor + 1\right) \right]$$

And now this expression:

$$\left(\left\lfloor \frac {-h(m)}{\sqrt {(h(m))^2 + 1}} \right\rfloor + 1\right) $$ will be zero for most $m$, but will be $1$ when $h(m)=0$ which is when $m=1$ or $m$ is an odd prime. The similar expression before it, without the minus sign, is always $1$.

So for $m=1$, the expression inside the summation evaluates to $1\cdot2^{-\pi(1)^2}=2^0 = 1$ and the whole expression becomes $$\alpha = 1+\sum_{p \text{ odd prime}}^{\infty} p \cdot 2^{\large -(\pi(p))^2}$$

or, considering $\{p_i\}$ as the primes in sequence, $$\alpha = 1+\sum_{2}^{\infty} p_i \cdot 2^{\large -i^2}$$

Also it turns out that if $2$ were not zeroed out, and $1$ was, the formula result would be unchanged, because for $i=1$, $p_1=2$, and $$ p_1 \cdot 2^{\large -1^2} = 2\cdot2^{-1} = 1$$

and so $$\alpha = \sum_{1}^{\infty} p_i \cdot 2^{\large -i^2}$$

The first few terms of the summation (after $1$, however that is regarded as arising) are $\{ 0.1875, 9.765625\cdot 10^{-3},1.06812\cdot 10^{-4}, 3.27826\cdot 10^{-7}, 1.89175\cdot 10^{-10}, 3.01981\cdot 10^{-14} \}$ after which everything carries on getting vanishingly small and not materially affecting the total, which is $1.197372765$

I don't know what the significance of this value of $\alpha$ has in the definition of $f(x)$ in the question, but clearly we're looking at another "state detection" type formula, that gives zero or non-zero depending on some condition, since without the floor functions

$$ {\Large[}2^{x^2}* \alpha {\Large]} - 2^{2x-1}*{\Large[}2^{(x-1)^2}*\alpha {\Large]} = \alpha (2^{x^2}- 2^{2x-1}*2^{(x^2-2x+1}) = \alpha (2^{x^2}- 2^{x^2}) = 0$$