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November 27

Hasse diagram of the "is a divisor of" relation

1. For which positive integer n, the Hasse diagram of all positive divisors of n ordered by the "is a divisor of" relation is a planer graph?
2. If n is a positive integer, and the Hasse diagram of all numbers >=1 and <=n ordered by the "is a divisor of" relation is a planer graph, what is the largest possible value of n?

~2025-36798-45 (talk) 20:21, 27 November 2025 (UTC)

The Hasse diagram based on the divisors of a positive integer is constructed as follows:

1. For a prime power $p^{a}$, the diagram is effectively a line with $a+1$ vertices, e.g. 1-2-4-8 having 4 vertices.

2. For a product of $n$ prime powers $p_{1}^{a_{1}}...p_{n}^{a_{n}}$, it is the Cartesian product of the $n$ line graphs corresponding to the contained factors.

If your number has two prime power factors, then this graph is pretty much just a grid, and obviously planar. If your number has four prime factors, then it must admit the non-planar hypercube graph $Q_{4}$ as a subgraph. If your number has exactly 3 prime factors then it seems to be a bit complicated, and I will have to reflect on this a bit. GalacticShoe (talk) 23:14, 27 November 2025 (UTC)

Doesn't $Q_{3}$ contain the "utility graph" $K_{3,3}$ as a subgraph?  ​‑‑Lambiam 00:22, 28 November 2025 (UTC)

$Q_{3}$ should just be the graph of a 3D cube, which is planar (e.g. File:Cube skeleton.svg) GalacticShoe (talk) 00:57, 28 November 2025 (UTC)

Okay, so when there are three prime factors, if at most one of them has a power above 1, then it is planar; this can be seen as drawing the graph is a series of nested squares. In the article for Hasse diagrams, the example given for factors of 60 is a graph of this form that can be rearranged into a planar graph. Meanwhile, I think the graph corresponding to a number $p_{1}^{2}p_{2}^{2}p_{3}$ (essentially four cubes pasted together in a 2 by 2 grid) is nonplanar, but I can't find a convenient online tool to test planarity, and I'm having an annoying time trying to find a $K_{3,3}$ or $K_{5}$ minor a la Kuratowski's theorem. If it is the case that this is nonplanar, then the criterion would be as follows:

The Hasse diagram of the divisors of a positive integer is planar if and only if the integer either has at most 2 prime factors, or it has exactly 3 prime factors of which only one may have a power greater than or equal to 2.

Or more succinctly,

The Hasse diagram of the divisors of a positive integer is planar if and only if the integer has at most 4 semiprime divisors.

GalacticShoe (talk) 00:51, 28 November 2025 (UTC)

This online SageMath program shows that indeed the graph is nonplanar, confirming the condition mentioned. The list of positive integers with nonplanar divisors Hasse diagram, i.e. the list of positive integers with at least 5 semiprime divisors, starts 180, 210, 252, 300, 330, 360, 390, 396, 420, 450, 462, 468... GalacticShoe (talk) 02:44, 28 November 2025 (UTC)

As for the latter question, the largest value for which the Hasse diagram is still planar is 27. When you add in edges for (4,28) and (14,28), it breaks down, as per this online SageMath program. GalacticShoe (talk) 08:05, 28 November 2025 (UTC)

December 2

I need help here

Hi all, I am stuck on this math problem:

Consider the function

$f(x)={\frac {3.14^{x}\cdot x^{5/2}}{x^{2}+0.5}}$

A space curve $C$ in $\mathbb {R} ^{3}$ is parameterized by

$\mathbf {r} (t)=(x(t),\ y(t),\ z(t))=(t,\ t^{3},\ {\sqrt {t+1}}),\quad t\geq 0.$

The density along this curve at parameter $t$ is given by $\rho (t)=f(t)$.

Compute the exact value of the limit

$\lim _{b\to +\infty }{\frac {1}{b}}\int _{0}^{b}{\frac {3.14^{t}\cdot t^{5/2}}{t^{2}+0.5}}\,dt,$

which represents the long-term average density along the curve as $b\to +\infty$.

Can I just use the average value theorem for integrals? Like, isn’t the limit of ${\frac {1}{b}}\int _{0}^{b}{\frac {3.14^{t}\cdot t^{5/2}}{t^{2}+0.5}}\,dt,$ equal to something in the middle? I would greately appreciate your help here. Thank you. ExtraTerrestrial120 (talk) 14:21, 2 December 2025 (UTC)

The mean value theorem is not going to help. It tells you that some point exists, but not how to find it.

Before we can try to help you, could you tell us where the original problem came from and how it was formulated? This specific problem looks utterly implausible, both as an exercise and as a practical problem.  ​‑‑Lambiam 15:42, 2 December 2025 (UTC)

Is not this limit just zero? Ruslik_Zero 20:42, 2 December 2025 (UTC)

I think it is a bit more. If $t>1,~f(t)>2^{\,t{-}1}.$  ​‑‑Lambiam 23:22, 2 December 2025 (UTC)

The limit is in fact quite a bit more. In the interest of not spoiling the answer to what I presume is a homework question, I'll simply mention that when $t>{\sqrt {1/2}}$, $2t^{2}>t^{2}+0.5$, and when $t>1$, ${\sqrt {t}}>1$. GalacticShoe (talk) 23:53, 2 December 2025 (UTC)

My "a bit more" was meant as an understatement. This may originate from a homework question, but if so, and if this is the original homework question, it is a very strange one, bizarrely convoluted to drive a simple point home. Note that the form of the space curve, and in fact everything before and after the imperative "Compute ... $dt$", is irrelevant to the problem as posed.  ​‑‑Lambiam 06:59, 3 December 2025 (UTC)

It’s also not true that the given integral computes the average density along the curve. ~2025-31850-11 (talk) 11:22, 3 December 2025 (UTC)

I was wondering about that myself. Why is the information about the curve even included in the problem? --RDBury (talk) 17:07, 3 December 2025 (UTC)

I just misread the limit. I thought it b->0. 20:33, 3 December 2025 (UTC)

If b were going to 0, this would be a better question (about the definition of the derivative and the fundamental theorem of calculus—still no meaningful connection to density on a curve, though). ~2025-31850-11 (talk) 02:42, 6 December 2025 (UTC)

December 6

prime divisors of 2^(n+1) choose 2^n, also known as C(2^(n+1),2^n)

when n=13, wolfram alpha says there are 1293 prime distinct prime divisors of N=C(16384,8192). According to the OEIS, the number of primes less than 16384 is 1900, and the number of primes less than 8192 is 1028. So there are 607 primes less than 16384 that don't divide N. But it's known that all primes between 8192 and 16384 will divide N. So the 607 missing primes are from the 1028 primes less than 8192. Now, 1028/(1028-607) is about 2.45. This ratio has been inching upward from smaller values of n. As n->oo, does the ratio go to e~2.71828182846? Rich (talk) 21:46, 6 December 2025 (UTC)

Determining whether a given prime p divides C(m,n) is not too hard: Write out m and n base p, then p|C(m,n) iff each digit in the expression for m is ≥ the corresponding digit in the expression for n. For example does 11 divide C(16384,8192)? 16384 base 11 is 11345₁₁ and 8192 base 11 is 6178₁₁. In the last digit, 8>5 so 11 divides C(16384,8192). The smallest prime that does not divide C(16384,8192) is 61. Note that if p is between 8192 and 16384 then 8192 base p is the single digit 8192 and 16384 base p is 1,16384-p, and 8192>16384-p so p will divide C(16384,8192). On the other hand, if p is between 16384/3 and 8192 then p does not divide C(16384,8192).

I think finding the limit as P→∞ of the proportion of p's < 2P where p divides C(2P,P) would be difficult. You could start with π(2P)-π(P) as first approximation for the number of primes, where π is the Prime-counting function. But you're interested in the relative error of this approximation. You could get additional terms for a better approximation by ignoring primes < √2P (assuming π(√(2P)) is small compared to π(2P)), and looking at the first digit of 2P base p. Then p is included if this digit is odd and excluded if it's even. We then get an alternating series

S = π(2P)-π(2P/2)+π(2P/3)-π(2P/4)+π(2P/5)-π(2P/6)...

and would like the limit of S/π(2P). But I don't know enough about the asymptotics of π to answer this. In any case, I'd be very surprised if the limit is something nice like e. There are many numbers > 2.45, and there are functions that increase very slowly but which don't converge to a limit at all. --RDBury (talk) 10:16, 7 December 2025 (UTC)

It's actually Pi(2^n)/(primes less than 2^n that divide C(2^(n+1),2^n)) that I'm interested in.Rich (talk) 17:22, 7 December 2025 (UTC)

Right, if you know the one you can determine the other. The main point is the alternating series approximating the number of primes dividing C(2P,P); here P is 2ⁿ. RDBury (talk) 20:42, 7 December 2025 (UTC)

thank youRich (talk) 00:01, 8 December 2025 (UTC)

Need for a kind of discrete integral

I need a kind of discrete integral, with a sum that progresses in finite steps like in a computer language , whereas an integral progress in infinitesimal steps.
So I have 2 functions:
$\lambda (i)=i\cdot \Delta \lambda +{\frac {\Delta \lambda }{2}}$
and $B(\lambda )$
I would like to replace the following integral:
$\int _{a}^{b}B(\lambda )\cdot d\lambda$
By the following summation:
$\Sigma _{i=a}^{b}B(\lambda (i))\cdot \Delta \lambda$
I suppose I can also write it theoretically:
$\Sigma _{i=0}^{\infty }B(\lambda (i))\cdot \Delta \lambda$
Therefore, I get a summ, not of infinitesimal areas, but of finite areas.
Does this conform to discrete mathematics, or is there a different notation and if so, what is it?
Malypaet (talk) 23:11, 6 December 2025 (UTC)

Isn't this essentially the question you asked on 20 October with a follow-up on 22 October?

It seems that what you want to do is a crude form of numerical integration; see in particular Numerical integration § Quadrature rules based on step functions. This corresponds to the paragraph in my response of 22 October that starts with "A (not very sophisticated) way to compute an approximation of $\textstyle \int _{a}^{b}\,f(v)\,dv~$", especially the part where I give the formula for when all segments are given equal width.  ​‑‑Lambiam 23:42, 6 December 2025 (UTC)

Yes, but your answers don't satisfy me, because I'm studying a case where the step size is finite and cannot correspond to an integral, where the step size is infinitesimal and applicable neither experimentally nor in computer calculations.

Therefore, I'm not looking for an approximate solution based on an integral, but rather a summation of steps of finite size, regardless of whether the number of steps is finite or infinite.

So I'm looking for a formulation based on summation, respecting mathematical notation in terms of function summation.

My question is whether my proposal is acceptable or needs to be modified to comply with the mathematical rules for summation of functions.
Malypaet (talk) 14:55, 7 December 2025 (UTC)

Have you looked at Summation § Capital-sigma notation?

If you'd write the sum as $v(a)+v(a{+}1)+\cdots +v(b),$ using an ellipsis, the capital-sigma notation is $\textstyle {\sum _{i=a}^{b}v(i)}.$ So, the other way around,

$\Sigma _{i=a}^{b}B\left(\lambda (i))\cdot \Delta \lambda =(B(\lambda (a))+B(\lambda (a{+}1))+\cdots +B(\lambda (i)\right)\cdot \Delta \lambda .$

BTW, it is somewhat unusual to use the notation $\Delta \lambda$ for a variable. A more common use is that it is shorthand for a difference between two changeable quantities, as in $\Delta \lambda =\lambda _{1}-\lambda _{0}.$ For a variable used for a stepwise change, as here, the letter $h$ is conventionally used.

Given the definition $\lambda (i)=i\cdot h+{\frac {h}{2}},$ the above sum comes out as

$\left(B(a\cdot h+{\frac {h}{2}})+B((a{+}1)\cdot h+{\frac {h}{2}})+\cdots +B(b\cdot h+{\frac {h}{2}})\right)\cdot h.$

If that is what you want, it is what you get.  ​‑‑Lambiam 17:55, 7 December 2025 (UTC)

Thanks for the link, it's interesting. But unfortunately, Wikipedia often refers back to integrals.

I found a more relevant explanation elsewhere: Discrete Calculus

So my formula actually corresponds to a discrete integral.

And you are right, it is what I want and get.

However, regarding the unusual use of $\Delta \lambda$, my need is ambiguous.

Actually, I'm using it as a parameter, that is, a constant whose value changes depending on the context:

i.e. the function's domain.

But I don't yet know how to introduce this uncommon usage in my text.

Have you an idea ?
Malypaet (talk) 18:32, 7 December 2025 (UTC)

Is indefinite sum relevant (to the original question)? catslash (talk) 18:44, 7 December 2025 (UTC)

Thanks, but no.

My needs are definitely for discrete integrals.

I use the symbol $\Delta$ before a letter to indicate a parameter with a constant value as a step size value, and linked to a domain of a function. This value may vary depending of the domain.

Whereas in indefinite sum you have the $\Delta$ operator, that I discover here, and it cover another subject.

Thanks again for your help.
Malypaet (talk) 22:12, 7 December 2025 (UTC)

We have an article Discrete calculus that gives a formula for a Riemann sum (essentially the same as a formula I have given before and have referred to above). Does this help? If not, why not? What is it that you need that these formulas do not provide?  ​‑‑Lambiam 04:39, 8 December 2025 (UTC)

If What I write could be a Midpoint rule Rieman summ in this form:

$S=\Sigma _{i=a}^{b}f(x_{i}^{*})\cdot \Delta x$

or

$S=\Sigma _{i={\tilde {0}}}^{\tilde {\infty }}f(x_{i}^{*})\cdot \Delta x$

Where $\Delta x$ is a constant parameter, $x_{i}^{*}\in [x_{i},x_{(i+\Delta x)}]$, $x_{i}^{*}=i\cdot \Delta x+\Delta x$,

${\tilde {0}}>0$, and ${\tilde {\infty }}<\infty$ (These last two values ​​are unknown).

I suppose that $x_{i}$ is a function of i.
Not completely sure of my notations, $x(i)$ may be better ?

Or do you have any corrections to suggest?
Malypaet (talk) 18:36, 8 December 2025 (UTC)

Mathematical notation is a matter of convention, not of rules chiseled in stone. Some of these conventions are more common than others. The advantage of following common conventions, where they exist, is that others (if they are familiar with mathematical notation) will more readily understand the message. Otherwise, if you create your own notation and conventions, you need to explain them if you want to be understood. Writing ${\tilde {0}}$ and ${\tilde {\infty }}$ for the summation bounds instead of using more conventional variable names is unlikely to be helpful.

If you cut up the interval $[a,b]$ into $n$ equal intervals of width $h=(b-a)/n,$ the closed form of the $i$-th interval, counting from $i=0$ up to $i=n{-}1,$ is equal to $[a+ih,a+(i{+}1)h].$ Check that $i=0$ corresponds to $[a,a+h]$ and $i=n{-}1$ corresponds to $[b-h,b].$ If you wish, use $\Delta x$ instead of $h,$ but $h$ is shorter and easier to write.

Using $x_{i}$ to stand for the midpoint of the $i$-th interval, we have $x_{i}=((a+ih)+(a+(i{+}1)h))/2=a+(i{+}{\tfrac {1}{2}})h.$ Summing the function values at these $n$ midpoints, weighted by the length of the interval, can then be written as

$\sum _{i=0}^{n{-}1}f(x_{i})\cdot h.$

 ​‑‑Lambiam 01:29, 9 December 2025 (UTC)

Thank you very much for your advice, I think I now have all the elements to develop my study, both original and as close as possible to the conventions.
Malypaet (talk) 18:38, 9 December 2025 (UTC)

But, $\Delta \lambda$ can also be a shortland for a recurring difference between two variable quantities, $\Delta \lambda =\lambda _{i+1}$-$\lambda _{i}$:

$\Delta \lambda =\lambda _{a+1}-\lambda _{a}=\lambda _{a+2}-\lambda _{a+1}=\lambda _{a+3}-\lambda _{a+2}...=\lambda _{b}-\lambda _{b-1}$ Malypaet (talk) 23:30, 8 December 2025 (UTC)