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The webpage Rounding Up To $\pi$ defines a certain "rounding up" function by an extremely simple procedure:

Beginning with any positive integer $n$, round up to the nearest multiple of $n-1$, then up to the nearest multiple of $n-2$, and so on, up to the nearest multiple of $1$. Let $f(n)$ denote the result.

E.g., using an obvious notation, the sequence starting with $n=10$ is $$ 10 \xrightarrow{9} 18 \xrightarrow{8} 24 \xrightarrow{7} 28 \xrightarrow{6} 30 \xrightarrow{5} 30 \xrightarrow{4} 32 \xrightarrow{3} 33 \xrightarrow{2} 34 \xrightarrow{1} 34 =:f(10). $$

The interesting thing is that, according to the webpage$^\dagger$, $(f(1),f(2),f(3),\dots)$ is the same sequence that results from a certain sieving method, for which Erdős & Jabotinsky (1958) proved that $$ \lim_{n\to\infty}\frac{n^2}{f(n)} = \pi, $$

so one naturally wonders what happens if, instead of rounding up to the nearest multiple, we round up to the second-nearest multiple, or the third-nearest, etc. Thus, let $f_k(n)$ denote the result of the "rounding up" sequence when using the $k^{\mathrm{th}}$-nearest multiple; e.g., using the second-nearest multiple, the sequence beginning with $10$ is $$ 10 \xrightarrow{9} 27 \xrightarrow{8} 40 \xrightarrow{7} 49 \xrightarrow{6} 60 \xrightarrow{5} 65 \xrightarrow{4} 72 \xrightarrow{3} 75 \xrightarrow{2} 78 \xrightarrow{1} 79 =: f_2(10). $$

(These $f_k$ sequences occur in the OEIS under names like "Generalized Tchoukaillon (or Mancala, or Kalahari) solitaire" or related sieving processes, e.g. at the links $f_1$, $f_2$, $f_3$, $f_4$.)

Now let $g_k(n):={n^2/f_k(n)}$ and, assuming all the limits exist, define the sequence $(G_k)_{k=1,2,3,\dots}$ as $$ G_k := \lim_{n\to\infty} g_k(n) = \lim_{n\to\infty}\frac{n^2}{f_k(n)}. $$

Here's a plot of $(g_k(10^6))_{k=1,\dots,30}$ in which the $g_k$ have apparently converged in at least their first $5$ digits (numerically, it appears that generally the first $d$ digits of $g_k(10^{d+1})$ are those of the limit $G_k$):

Plot of g_k(10^6) and their conjectured limits G_k.

We know that $G_1=\pi,$ and in the plot I've indicated my conjectured values for $G_2,G_3$ as well, based on a simple pattern that I found by searching numerically among the $g_k(n)$; specifically, I found that $g_{k+1}(n) \approx \frac{4}{k^2\,g_k(n)}$, with the absolute error bounded by $5/n$, i.e., $$ \left|g_{k+1}(n) - \frac{4}{k^2\,g_k(n)}\right| < \frac{5}{n}, \quad k=1,2,3,\dots $$ and also that $$ \left|g_{k}(n) - a_k\right| < \frac{5}{n}, \quad k=1,2,3,\dots $$ with the $a_k$ defined recursively by $a_1=\pi,\; a_{k+1}=\frac{4}{k^2\,a_k},\; k=1,2,3,\dots$ This strongly suggests the following ...

Conjecture: $$ \boxed{\quad G_1 = \pi, \quad G_{k+1} = \frac{4}{k^2\,G_k}, \quad k=1,2,3,\dots \quad}\tag{1} $$

i.e., $$ (G_k)_{k=1,2,3,\ldots} = \left(\pi,\, \frac{4}{\pi},\, \frac{\pi}{4},\, \left(\frac{2}{3}\right)^2 \frac{4}{\pi},\, \left(\frac{3}{4} \frac{1}{2}\right)^2 {\pi},\, \left(\frac{4}{5} \frac{2}{3}\right)^2 \frac{4}{\pi},\, \left(\frac{5}{6} \frac{3}{4} \frac{1}{2}\right)^2 {\pi},\,\ldots \right). $$

Question: How to prove the recurrence relation in (1)??

NB: From the conjecture (1) it's straightforward to derive the following formulas that hold for all positive integers $k$:

$$G_k=\begin{cases}\left(\prod_{j=1}^{k-1\over 2}{2j-1\over 2j}\right)^2\pi&\text{if $k$ is odd}, \\ \left(\prod_{j=1}^{k-2\over 2}{2j\over 2j+1}\right)^2 {4\over\pi}&\text{if $k$ is even}\end{cases}\\ $$

$$G_k=\left({(k-2)!!\over(k-1)!!}\right)^2\cdot \begin{cases} \pi&\text{if $k$ is odd},\\[3ex] {4\over\pi}&\text{if $k$ is even}\end{cases}\\ $$

$$G_k\sim{2\over k}$$

where the last line is from the known asymptotic behavior of double factorials.

EDIT: As noted in a comment by Ash Malyshev, the conjectured $(G_k)_{k=1,2,3,\dots}$ can also be written compactly as a ratio of squared gamma functions that I'll call $\rho(k)$:

$$G_k = \left({\Gamma\left({k\over 2}\right) \over \Gamma\left({k+1\over 2}\right)}\right)^2=:\rho(k)$$

This can be established either by deriving it from the above formulas, or by showing that $\rho(k)$ satisfies the defining recursion (1). I've updated the above plot to include the continuous function $\rho$ (blue curve) on the real interval $(0, 30]$: The conjecture is that the blue dots ($G_k$) coincide with this curve at every positive integer $k.$

$^\dagger$ The "rounding up" function $f$ is actually implicit in the proof given by Erdős & Jabotinsky; see this answer by Misha Lavrov for more detail.


For reference, here is the program I used for $f_k(n)$:

# Python
def f(k,n):   
    y = n
    for x in range(n-1,0,-1):  # i.e., for x = n-1, n-2, ..., 1
        y = ((y-1)//x + k)*x   # i.e., y <- k-th multiple of x not less than y
    return y

Also, in view of the possible relevance of sieves for proving conjecture (1), here is a program implementing the following algorithm, which I claim generates the $f_k$:

Starting with seq$_1$ = $(1,2,3,\dots)$, let seq$_{i+1}$ be the result of deleting from seq$_i$ the first $k−1$ elements and then deleting every $(i+1)$th element from those remaining. $f_k(n)$ will be the first element of seq$_n.$

def sieve(k, maxel):  # generates f_k(1), ..., f_k(n) <= maxel 
    seq = list(range(1, maxel+1))   # seq_1 = (1,2,3,...,maxel)
    keep = seq[:1]    # keep the first element of seq_1
    i = 1
    while seq:        # while seq is not empty
        del seq[:k-1]   # delete the first k-1 elements
        del seq[::i+1]  # delete remaining elements with index divisible by i+1
        keep += seq[:1] # keep the first element of seq_(i+1)
        i += 1
    return keep
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    $\begingroup$ This is fascinating ! $\endgroup$ May 13, 2023 at 3:35
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    $\begingroup$ It's often cleaner to think of the the double-factorial as two related but different functions, one for odd numbers and one for even numbers. Both can be expressed in terms of the Gamma function, and the expressions are almost the same, but off by a factor of $\sqrt{\pi/2}$. In your case, taking this point of view reveals that your expressions for $G_k$ simplify to $(\Gamma(k/2) / \Gamma((k+1)/2))^2$ in both the odd and even case. This "explains" why you don't see alternating behavior numerically in the graph of G_k even though "whether pi is in the numerator or denominator" is alternating. $\endgroup$ May 16, 2023 at 3:40
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    $\begingroup$ @AshMalyshev Nice discovery! It's also easy to prove that the form you give must equal the conjectured $G_k$ by showing, using $\Gamma(x+1)=x\Gamma(x)$ and $\Gamma(1/2)=\sqrt{\pi},$ that it satisfies the defining recursion (1). $\endgroup$
    – r.e.s.
    May 16, 2023 at 17:15
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    $\begingroup$ Numerically, it looks like $f_{k+2}(kn+1)=f_k((k+1)n+1)$. If this is true, then $G_{k+2}=\lim_{m\to\infty}\frac{m^2}{f_{k+2}(m)}=\lim_{m\to\infty}\frac{m^2}{f_k(m+(m-1)/k)}=\lim_{m\to\infty}\frac{m^2}{\frac{(m+(m-1)/k)^2}{G_k}}=G_k\frac{1}{(1+1/k)^2}$ if the limits existed. This would then mean your claim is true. $\endgroup$ May 20, 2023 at 4:26
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    $\begingroup$ @VarunVejalla Just an idea: In a comment I made to an answer elsewhere, I describe how to generate any $f_k$ with a corresponding sieve, so a possible route to proving my conjecture would be to (1) show that those sieves indeed generate the $f_k$, and (2) prove that your equation holds as a property of theses sieves. (I just added to my question a program for the sieving algorithm.) $\endgroup$
    – r.e.s.
    May 20, 2023 at 6:33

1 Answer 1

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Let me offer a direct adaptation of that of Rounding up to Pi. Fix $k\geq 1$, and apply the "rounding up" algorithm by selecting the $k$-th nearest multiple, as you describe: this gives us an integer $S_{n,k}(x)$ for every integer $x$ between $1$ and $n$, such that $S_{n,k}(1) = f_k(n)$ and $S_{n,k}(n) = n$.

Most of the trajectory is a sequence of parabola

I start with the observation that the function $$ P_j(x) = (A_j - j x) x $$ coincides with $S_{n,k}$ for small values of $x$ when we choose $j=k$ and $A_k = kn+1$. Rounding up to Pi went on to construct the $A_j$ so that for every $x$, there was one $P_j$ such that $S_{n,1}(x) = P_j(x)$. Keeping with that spirit, we can in fact prove a very similar claim:

Claim 1: for every $n>k$, there exists $k\leq J,X\leq n$ and three $\mathbb{N}$-valued sequences:

  • $(j_x)_{X\leq x \leq n}$ with steps in $\{0,-1\}$ and $j_n = k$,
  • $(x_l)_{k-1 \leq l \leq J}$ decreasing with $x_{k-1} = n$ and $x_J = X-1$,
  • and $(A_l)_{k\leq l \leq J}$ increasing

such that $$ \forall k\leq j \leq J \ , \ \forall x_j < x \leq x_{j-1} \ , \quad S_{n,k}(x) = P_j(x) . \qquad (0)$$

In other words, every $x$ with $X \leq x \leq n$ is "on the parabola $P_{j_x}$" in the sense that $S_{n,k}(x) = P_{j_x}(x)$, and the $x$ that are on the parabola $P_j$ are the integers in $(x_j, x_{j-1}]$.

proof. We start with the following observation: for every $ux < y \leq ux+x$ with $u,x,y$ integers, rounding up $y$ to the $k$-th nearest multiple of $x$ gives $(u+k)x$.

We will prove $(0)$ together with the property $$ \forall k\leq j \leq J \ , \ \forall x_j \leq x < x_{j-1} \ , \quad x(j-k) < A_j - j(x+1) \leq x(j-k+1) . \qquad (0’)$$

We have $(0)$ and $(0)’$ for $x=n$. Let us now proceed by induction. Assume that for some $0< x \leq n$, we have constructed $(j_z)_{x\leq z \leq n}$, $(x_l)_{k\leq l \leq j_x}$ and $(A_l)_{k\leq l \leq j_x}$ so that $(0)$ and $(0’)$ hold when (only there!) pretending that $x_{j_x-1} = x-1$. Write $j=j_x$.

We know that $S_{n,k}(x) = P_j(x) = x (A_j - jx) = (x-1)(A_j-jx) + (A_j-jx)$. We want to round $S_{n,k}(x)$ up to the $k$-th closest multiple of $x-1$. Observe that $A_j-jx > (x-1)(j-k)$ either from $(0’)$ (if $x<n$) or a direct computation (if $x=n$). There are three cases, depending on the value of $A_j - jx$ as it appears in $(0’)$:

  • First case, $(x-1)(j-k) < A_j - jx \leq (x-1)(j-k+1)$: then $$ (x-1)(A_j - jx + j - k) < S_{n,k}(x) \leq (x-1)(A_j - jx + j - k + 1) , $$ so by the observation $S_{n,k}(x-1) = (x-1)(A_j - j(x-1)) = P_{j}(x-1)$. This proves $(0)$ and $(0’)$ for $x-1$, by setting $j_{x-1} = j$.

  • Second case, $(x-1)(j-k+1) < A_j - jx \leq (x-1)(j-k+2)$: then set $j_{x-1} = j+1$, $x_j = x-1$, and observe that similarly $$ S_{n,k}(x-1) = (x-1)(A_j - j(x-1) + 1) = (x-1)(A_{j+1} - (j+1)(x-1)) $$ when we choose $A_{j+1} = A_j + x = A_j + x_j + 1$.

  • Third case, $A_j - jx > (x-1)(j-k+2)$: then set $J = j$, $X=x$, $x_j = x-1$ and stop. Note that by $(0’)$ this can happens only if $j\geq x$. $\square$

Note: We can in fact guess that $J \approx \mathrm{cste} \, n^{2/3}$, as I did in a previous version, by using the approximation $X \approx A_J / 2J$, $A_J \approx n b_J$ and $X \approx J$.

Computing the parameters of the parabola

We already established in the proof that $$ A_k = kn+1 \quad , \quad A_{j+1} = A_j + x_j + 1 . $$ What is the value of $x_j$ for $k\leq j < J$? By $(0’)$ we must have $$\begin{cases} A_j-j(x_j+2) \leq (x_j+1)(j-k+1) &\iff A_j - j + k - 1 \leq (2j-k+1)x_j , \\ A_j-j(x_j+1) > x_j(j-k+1) &\iff A_j - j > (2j-k+1)x_j . \end{cases}$$ This implies $$ \left| x_j - \frac{A_j}{2j-k+1} \right| < 1 . $$ It is tedious to get exact values, so we will work with approximations instead. Define $$ B_k = A_k = kn+1 \quad , \quad \forall j\geq k \ , \ B_{j+1} = B_j + \frac{B_j}{2j-k+1} = B_j \frac{2j-k+2}{2j-k+1} . $$ We can check that $|B_j - A_j|$ is bounded for every $j$ by something independent of $n$. On the other hand, $B_j / n \to b_j > 0$ as $n\to\infty$ with \begin{align} b_{j} &= k \prod_{l=k}^{j-1} \frac{2l-k+2}{2l-k+1} = k \prod_{l=k+1}^{j} \frac{2l-k}{2l-k-1} \\ &= k \frac{\left(1+\frac k 2\right)^{(j-k)}}{\left(\frac{k+1}2\right)^{(j-k)}} = k \frac{\Gamma\left(1+\frac k 2+(j-k)\right)}{\Gamma\left(1+\frac k 2\right)} \frac{\Gamma\left(\frac{k+1}2\right)}{\Gamma\left(\frac{k+1}2 + j-k\right)} \end{align} so $A_j / n \to b_j$ as $n\to\infty$ as well. Using that $\frac{\Gamma(x+1)}{\sqrt{x}\Gamma(x+1/2)} \to 1$ as $x\to\infty$ we find that $$ \frac{b_j}{\sqrt{j}} \underset{j\to\infty}\longrightarrow k \frac{\Gamma\left(\frac{k+1}2\right)}{\Gamma\left(1+\frac k 2\right)} = 2\frac{\Gamma\left(\frac{k+1}2\right)}{\Gamma\left(\frac k 2\right)}, $$ so that as $j\to\infty$ $$ \frac{4 j}{b_j^2} \to G_k. $$

Claim 2: the sequence $(b_j / \sqrt{j})_{j\geq k}$ is monotone increasing.

This is a consequence of the log-convexity of the Gamma function.

The final value

The heuristic used in Rounding up to Pi is $$ f_k(n) := S_{n,k}(1) = \sup_{k\leq l\leq j} \sup_{1\leq x \leq n} P_l(x) . \qquad (*) $$ Let us get a precise version: since $S_{n,k}$ is decreasing and $S_{n,k}(x-1) \leq S_{n,k}(x) + k(x-1)$, we have for every $1\leq x \leq n$ $$ S_{n,k}(x) \leq f_k(n) \leq S_{n,k}(x) + k\frac{x(x-1)}{2} . $$ We have constructed the sequences of Claim 1 for at least all $j\leq J$, where $J = j_X$ and $X$ is the largest integer such that $j_X \geq X$. Since $j_X \leq k + (n-X)$ we must have $J\geq c \sqrt{n}$ for some $c>0$, implying $J\to \infty$ as $n\to\infty$.

We can find functions $r_1, r_2, \dots$ such that for every $j < J$, $$|A_j - n b_j| \leq r_1(j) \quad , \quad \left| x_j - \frac{n b_j}{2j-k+1} \right| \leq r_2(j) \ , $$ which implies (recalling $S_{n,k}(x_j+1) = P_j(x_j+1)$) $$ \left| S_{n,k}(x_j+1) - n^2 b_j^2 \frac{j-k+1}{(2j-k+1)^2} \right| \leq n r_3(j) . $$ Take now any $j$ fixed and arbitrarily large; because $J\to\infty$ as $n\to\infty$ we will have $j<J$ for every $n$ large enough, and the above inequalities will be verified. This gives us, first that $$ \liminf_{n\to\infty} \frac{f_n(k)}{n^2} \geq \liminf_{n\to\infty}\frac{S_{n,k}(x_j+1)}{n^2} \geq b_j^2 \frac{j-k+1}{(2j-k+1)^2} $$ where the right-hand side is larger than $G_k^{-1}-\epsilon$ for every $j$ large enough; then that \begin{align} \limsup_{n\to\infty} \frac{f_n(k)}{n^2} \leq \limsup_{n\to\infty} \left(\frac{S_{n,k}(x_j+1)}{n^2} + k\frac{x_j(x_j-1)}{2n^2}\right) \leq b_j^2 \left( \frac{j-\frac{1}{2}k+1}{(2j-k+1)^2} \right) \end{align} which by the same reasoning is smaller than $G_k^{-1} + \epsilon$ for every $j$ large enough. This concludes the proof of the convergence $$ \frac{n^2}{f_k(n)} \underset{n\to\infty}\longrightarrow G_k , $$ with $G_k$ given by your formulas.

Edits

Edit 1: Some mistakes have slipped in my final computations. I rewrote $b_j$ using Pochhammer symbols and Gamma functions, which makes computation easier afterwards.

Edit 2: Added a proof of the heuristic (*).

Edit 3: Major rewrite, with division into sections. I have tried to made things less informal: many back-of-the-envelope calculations are now almost-rigorous, at the cost of the technical Claim 1.

Edit 4: Added a proof of Claim 2.

Edit 5: Corrected the final proof of the convergence of $f_k(n)$ by bounding the difference between $f_k(n)$ and $S_{n,k}(x_j+1)$ for any fixed $j$.

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  • $\begingroup$ You may have $k$ somehow out-of-place? If I'm not mistaken, WolframAlpha says your $\lim_{j\to\infty}\prod_{l=k+1}^j(...)={k^2\over 2(k+1)}G_k.$ $\endgroup$
    – r.e.s.
    May 24, 2023 at 4:09
  • $\begingroup$ That is because I forgot to take the limit as $c\to\infty$, with $c$ the number of times I have re-done the computations. If you use the formula for $b_j$ and compute $\lim_{j\to\infty} \frac{b_j^2}{4j}$ in WolframAlpha, this gives the correct expression for $G_k^{-1}$ using Gamma functions. $\endgroup$ May 24, 2023 at 8:06
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    $\begingroup$ Could you please clarify the heuristic equation (*) and how it's being used in your final $\lim_{n\to\infty} \frac{n^2}{f_k(n)} \underset{j\to\infty}\longrightarrow...?$ How is $f_k(n)$ (in either place) to be understood as depending on $j$? (BTW, where you wrote $\frac{\Gamma(x+1)}{\sqrt{x}\Gamma(x+1/2)} = 1$, I think you meant to write $1+o_x(1)$ on the RHS.) $\endgroup$
    – r.e.s.
    May 24, 2023 at 15:20
  • $\begingroup$ I have extensively rewritten the proof to make it more rigorous. The derivation of this equation on $A_j$ is now different, and in fact the equation itself is different... Even though it does not change the end result. This should make things solid up to some computations left to the reader in the second section ("computing the parameters of the parabola"), but of course let me know if you find some error! $\endgroup$ May 24, 2023 at 23:14

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