Why does this discrete product built from floor and ceiling of squares converge to pi or 1?

Question

I have been experimenting with a structure I call the Discrete Square Residual Structure (DSRS).

For a fixed integer $\mu > 0$, define

$U(n) = \lceil \tfrac{n^2}{\mu} \rceil, \quad L(n) = \lfloor \tfrac{n^2}{\mu} \rfloor.$

Then set

$\Delta_U(n) = U(n+1) - U(n), \quad \Delta_L(n) = L(n+1) - L(n).$

Finally, define the infinite product associated with each $\mu \in \mathbb{N}$:

$$ P(\mu) \;=\; \prod_{n = n_{0}}^{\infty} \frac{\Delta_{L}(n)}{\Delta_{U}(n)}, $$

where the lower bound $n_{0}$ depends on the arithmetic nature of $\mu$ with respect to 4ドル$:

$$ n_{0} = \begin{cases} \mu & \text{if } \mu \not\equiv 0 \pmod{4}, \\[6pt] \dfrac{\mu+2}{2} & \text{if } \mu \equiv 0 \pmod{4}. \end{cases} $$

The product exhibits two distinct behaviors:

• $P(\mu) \to 1$ if $\mu = 1$ or $\mu \equiv 0 \pmod{4}$,
• $P(\mu) \to \dfrac{\pi}{2}$ for most other $\mu \in \mathbb{N}$.

Computational evidence:

• For about 75 percent of integers $\mu$, this product converges rapidly to $\pi$.
• For the remaining $\mu$ (mostly those divisible by 4ドル$ or equal to 1ドル$), the product converges to 1ドル$.
• For example, when $\mu = 3$ or $\mu = 5$, $P(\mu)$ matches $\pi$ to six decimal places at $n = 10^6$.

Question: Why does this binary partition occur? Is there a known framework in number theory or analysis that can explain why these two constants, $\pi$ and 1ドル$, emerge from such a discrete product?

For those interested, I have written a preprint with further details: https://doi.org/10.5281/zenodo.17101750.

• Gerry Myerson: For $\mu=3,ドル you're getting the Wallis product, see en.wikipedia.org/wiki/Wallis_product - in general, your $\Delta$ sequences will have very simple expressions, and probably come down to Wallis or some well-known variant.
• Marco: Are you sure you haven't start the product at $n_0=\lceil \mu/2 \rceil$? Because both theoretical result and python tell me that if you start from $n_0=\mu$ it converges to $\pi/4$ if $\mu \ne 0 \mod 4$
• aadesh tikhe: @GerryMyerson For μ = 2 the construction yields the Wallis product, but this does not hold for μ > 2. I’ve explained the distinction in detail in my preprint: https://doi.org/10.5281/zenodo.17101750 (see the layered mapping in the appendix and Section 8.2 for a full discussion).
• Gerry Myerson: For $\mu=3,ドル the ceiling goes 1,2,3,6,9,12,17,22,27,34,ドル\dotsc$ so $\Delta_U$ goes 1,1,3,3,3,5,5,5,7,ドル\dotsc$. The floor goes 0,1,3,5,8,12,16,21,27,33,ドル\dotsc$ so $\Delta_L$ goes 1,2,2,3,4,4,5,6,6ドル\dotsc$. So the product of the terms $\Delta_L/\Delta_U$ goes $(1/1)(2/1)(2/3)(3/3)(4/3)(4/5)(5/5)(6/5)(6/7)\dotsb$ which, if you drop the terms that give 1ドル,ドル is $(2/1)(2/3)(4/3)(4/5)(6/5)(6/7)\dotsb$ which is Wallis, isn't it? And doesn't Marco's answer end up with a Wallis product?
• aadesh tikhe: @GerryMyerson you’re right, I’m not that deep into math yet, I’m just stepping into it slowly day by day.

Answer 1

Let start by denoting $n=\mu \alpha+\beta$, $\beta<\mu$.

Some calculations shows that $$ \frac{\Delta_{L}(n)}{\Delta_{U}(n)}=\frac{2\alpha + \Delta_{L}(\beta)}{2\alpha +\Delta_{U}(\beta)}, $$ If $\beta^2 \ne 0 \mod \mu$ and $(\beta+1)^2 \ne 0 \mod \mu$ then $\Delta_{L}(n)=\Delta_{U}(n)$ and their quotient is 1ドル$ ( $\alpha \ne 0$ for the choice of $n_0$).

If $\beta^2=0 \mod \mu$ we have $$ \frac{\Delta_{L}(n)}{\Delta_{U}(n)}=\frac{2\alpha + \left\lfloor \frac{2\beta+1}{\mu}\right\rfloor}{2\alpha + \left\lceil \frac{2\beta+1}{\mu}\right\rceil}, $$ while if $(\beta+1)^2 =0 \mod \mu $ $$ \frac{\Delta_{L}(n)}{\Delta_{U}(n)}=\frac{2\alpha + \left\lceil \frac{2\beta+1}{\mu}\right\rceil}{2\alpha + \left\lfloor \frac{2\beta+1}{\mu}\right\rfloor}, $$ So, assuming $\beta^2=0 \mod n$

if $\beta< \frac{\mu-1}{2}$ then

$$ \frac{\Delta_{L}(n)}{\Delta_{U}(n)}=\frac{2\alpha }{2\alpha +1}, $$ If $\beta= \frac{\mu-1}2$ $$ \frac{\Delta_{L}(n)}{\Delta_{U}(n)}=1 $$ otherwise $$ \frac{\Delta_{L}(n)}{\Delta_{U}(n)}=\frac{2\alpha + 1}{2\alpha + 2}, $$ and it is equal to their inverse if $(\beta+1)^2=0 \mod \mu$.

Now we will show the following

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Let $P(\mu):= \prod_{\beta=0}^{\mu-1} \frac{2\alpha + \Delta_{L}(\beta)}{2\alpha +\Delta_{U}(\beta)}$ than $P(\mu)=1 $ if $\mu =0 \mod 4$ and $P(\mu)=\frac{4\alpha(\alpha+1)}{(2\alpha+1)^2}$ otherwise

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Let $\mu= \prod_i p_i^{e_i}$ and let $M= \prod_i p_i^{\lceil e_i/2 \rceil}$.

$\beta^2=0 \mod \mu$ iif $\beta=0 \mod M$ and $(\beta+1)^2=0 \mod \mu$ iif $\beta=-1 \mod M$

The number of $\beta=0 \mod M$ is equal to $T=\mu/M$. and there are also $T$ numbers $\beta$ such that $\beta=-1 \mod M$

Let $a=\left|\left\{ \beta=0 \mod M : \frac{2\beta+1}\mu<1\right\}\right|$ and $b=\left|\left\{ \beta=-1 \mod M : \frac{2\beta+1}\mu<1\right\}\right|$

By counting the $j$-s such that $jM<\mu/2$ we get that $a=\left\lfloor (\mu-1)/2M\right\rfloor+1$ and similarly $b=\left\lfloor \mu/2M\right\rfloor$ and so

$$ P(\mu)=\left(\frac{2\alpha}{2\alpha+1}\right)^a \left(\frac{2\alpha+1}{2\alpha+2}\right)^{T-a} \left(\frac{2\alpha}{2\alpha+1}\right)^{-b} \left(\frac{2\alpha+1}{2\alpha+2}\right)^{-(T-b)} $$ Combining the exponents toghether $$ P(\mu)=\left(\frac{4\alpha(\alpha+1)}{(2\alpha+1)^2}\right)^{a-b} $$ and $a-b=0$ if $T$ is even (i.e. $\mu$ is divisible by 4ドル$) and $=1$ otherwise

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This immediatly prove the result for $\mu$ divisible by 4ドル$

Now i have $$ \prod_{\alpha=1}^\infty \frac{2\alpha(2\alpha+2)}{(2\alpha+1)^2}=\prod_{k=1}^N \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \cdot (N+1)/(2N+1) \to \pi/4 $$ Estimating numerically the case in which $\mu=5$, starting from $n_0=5$ for $N=1000000$ yields a result equal to 0ドル.7853971816527053$ consistent with the theoretical result.

• aadesh tikhe: thank you for showing me proof that this really exists and has a strong link to the Wallis product. I’m still a beginner and very curious to learn more. I want to improve myself and also make progress toward publishing this work. Could you suggest how I should start my journey and what steps I should follow? And Marco, would you be open to mentoring me a bit on this? It would mean a lot.
• Marco: @aadeshtikhe it is quite difficult to suggest anything whiteout knowing at which level of math you are. In general the only theoretical results needed to understand this proof can be founded in any introductory number theory course. Assuming you have already done that then it is all a matter to use the properties of the floor and ceil functions and the Wallis product (that you already know). If there is any step in particular that you find unclear you can ask about it
• aadesh tikhe: Thanks @Marco, honestly I only have high school level math, I never studied number theory formally. Whatever I did here came from experimenting with numbers and following my curiosity. I didn’t fully understand the proof you mentioned, but what I could see is that for any μ and any n, it somehow connects back to the Wallis product. In my preprint I tried to explain a piece of this in section 8.2, where I showed how non-multiples of 4 reduce to μ=2 and how the sequences get more messy for larger μ.
• aadesh tikhe: Except for the Wallis product part, if you look at other parts of the paper the DSRS also generates completely new sequences. That was actually my first aim. When I was exploring μ=2 I checked its sequence in OEIS and found it connects to the Wallis product, so I thought if μ=2 can give π then what about other μ? That’s how DSRS started shaping up. To be honest I am really bad at math compared to you all, this was just the product of some random curiosity. But I want to learn step by step and improve this work, and I’m grateful for your effort and explanations.
• Marco: @aadeshtikhe for finite $N$ the product is the truncated Wallis product times $(N+1)/(2N+1)$ it converges to $\pi/4$. If you start the product at $\mu-1$ (that is what I think happen in your preprint) it converges to $\pi/2$