Coincidences in $\{\sum_k\left\lfloor\frac{k}2\sqrt p\right\rfloor2^{-k}\}$ and $\sum_k(\lfloor\frac{k}2\sqrt p\rfloor\bmod2)2^{-k}$

Question

Let us consider the following sums:

$$f(p) = \Big\{\sum_{k=1}^\infty \Big\lfloor{\frac{k}{2}\sqrt{p}\Big\rfloor}\cdot2^{-k}\Big\}$$ $$g(p) = \sum_{k=1}^\infty \Big(\Big\lfloor{\frac{k}{2}\sqrt{p}\Big\rfloor \mod{2}}\Big)\cdot{2^{-k}}$$ Here, $\{\cdot\}$ represents the fractional part function, and $p$ is a positive integer, typically a prime.

I have empirically obtained interesting identities for a few values of $p$. They are exact up to 15 decimals, which is the maximum precision I can get on my machine without using exact arithmetic. Are these identities correct? Are there more to be found, maybe a generalization?

• If $p = 23$, then $f(p) = 9/31$ and $g(p) = 7/31$
• If $p = 41$, then $f(p) = 2/31$ and $g(p) = 10/31$
• If $p = 97$, then $f(p) = 3g(p)$
• If $p = 143$, then $f(p) = g(p) = 1$ (correct up to 15 decimals)

Typically, the binary digits of $g(p)$ are 50/50 zero's and one's. This not the case for $f(p)$. Again, this is based on empirical evidence only. I have found in the past numbers that agreed up to the first 10 million digits but were actually different, so please take this with a grain of salt.

Update

More details can be found in my new question, here.

Context

Of course this is related to my study of normal numbers, but more specifically, with the following operator denoted as $*$, associated with a group on $[0, 1]$, and defined as follows: if $x, y \in [0,1]$ then $x * y$ is the number whose $k$-th binary digit is the sum of the $k$-th binary digits of $x$ and $y$, modulo $2$.

Answer 1

I can at least disprove your last identity and explain why $f(143)$ gets so close to an integer. We’re looking at the sum $$\sum_{k=1}^\infty \frac{\lfloor k\sqrt{143}/2\rfloor}{2^k}\approx 10.99999...$$ Notice that this sum is almost equal to $$\sum_{k=1}^\infty \frac{\lfloor k\sqrt{144}/2\rfloor}{2^k}=\sum_{k=1}^\infty\frac{6k}{2^k}=12$$ But when we change $144$ to $143$, we are basically lowering the value of the term $\lfloor \cdot \rfloor$ inside of the floor just a little bit, but since the floor rounds down, each floor expression in the sum is decreased by $1$, and the sum as a whole is decreased by approximately $1/2+1/4+...=1$. However, it’s also decreased a little bit more, because $\sqrt{143}$ is not quite $12$, but it’s so close that the floor expressions $\lfloor \cdot \rfloor$ actually agree for the first couple terms. That is, $$\lfloor \sqrt{143}/2\rfloor = 6-1$$ $$\lfloor 2\cdot \sqrt{143}/2\rfloor=12-1$$ $$\lfloor 3\cdot \sqrt{143}/2\rfloor=18-1$$ $$...$$ $$\lfloor 47\cdot \sqrt{143}/2\rfloor=6\cdot 47-1$$ $$\lfloor 48\cdot \sqrt{143}/2\rfloor=6\cdot 48-2\color{red}{\ne 6\cdot 48-1}$$ So the reason your conjecture is so close (but actually incorrect) is that this pattern holds for a long time, but fails after $k=48$, at which point the terms are being divided by $2^{48}$ and are thus incredibly small.