Mathematics desk
< July 22	<< Jun \| July \| Aug >>	July 24 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

July 23

Prime Pascal

Do we know any interesting facts about the distribution on the primes amongst Pascal's triangle, and more specifically when it comes to the ratio of primes to composites in each subsequent row. — Fly by Night (talk) 01:06, 23 July 2012 (UTC)[reply]

I'm pretty sure the only binomial coefficients that are prime are of the form

{\binom {p}{1}}

and

{\binom {p}{p-1}}

for prime p, although I'm having trouble coming up with a proof. Rckrone (talk) 04:44, 23 July 2012 (UTC)[reply]

{\binom {n}{r}}

is greater than n for 1 < r < n-1, so it can't be prime as no prime factors >n are involved in the calculation of

{\binom {n}{r}}

as n(n-1)...(n-r+1)/r! For prime p,

{\binom {p}{r}}

is divisible by p for 0 < r < p, is prime iff r = 1 or p-1, so Rckrone's conjecture is correct. AndrewWTaylor (talk) 12:48, 23 July 2012 (UTC)[reply]

Thanks Rckrone and AndrewWTaylor. I don't think I thought about it properly before I posted it. But thanks for taking the time to reply. — Fly by Night (talk) 00:19, 24 July 2012 (UTC)[reply]

It's trivial as the answers show. For a variation, if the decimal digits in a row are concatenated then [1] says the result is prime for row 2, 9, 30 and no others in the first 696 rows. I wrote "The search stopped before row 697 which has 104568 digits if anyone feels lucky." PrimeHunter (talk) 13:42, 23 July 2012 (UTC)[reply]

How does that compare with expectations for a series of random numbers of equivalent lengths? 86.160.216.249 (talk) 17:50, 26 July 2012 (UTC)[reply]

Different non-touching edge choices on Regular polyhedra question...

For the Pyramid, there is only one way to choose the maximum number of edges which don't share a vertex, for the Cube there are two ways (all four parallel, and then with two "twisted"), and for the Octahedron there are two that are mirror images. Does anyone have any advice for trying to count the unique ways for the 6 edges on the Icosahedron or the 10 edges on the Dodecahedron?Naraht (talk) 13:49, 23 July 2012 (UTC)[reply]

One way to view this question is as counting the number of maximal matchings of the graph, although you seem to be counting only up to orientation preserving symmetries of the polyhedron. The article mentions the FKT algorithm for planar graphs (which yours are). The icosahedron and dodecahedron are small enough that the FKT algorithm can be done by hand except for computing a large determinant. However once you get the total number of maximal matchings, modding out by the symmetries might be complicated. Rckrone (talk) 17:37, 23 July 2012 (UTC)[reply]

After messing around with it a bit, I think there are only 3 matchings up to rotation for the dodecahedron. Around a given pentagon, there are only 3 arrangements possible: 0 edges from the pentagon in the matching, 1 edge, or 2 edges. If any pentagon has 0 edges, there is only one possible configuration. Otherwise exactly 4 pentagons will have 1 edge in the matching, and there is only one arrangement but with two orientations. I may have missed something though. Rckrone (talk) 04:58, 24 July 2012 (UTC)[reply]

I think so. There is the one that is the zero edges for two opposite pentagons, the other with one image on the top "pentagon" which leads to two choices on the middle/bottom, but the two results there are just the same but inverted, so only two actual matchings.Naraht (talk) 23:55, 25 July 2012 (UTC)[reply]

I think we are in agreement then (as well as with 86.128.4.88). I wasn't sure if you wanted to count mirror images as the same or separate arrangements. If you count them as the same, then I also find only 2 for the dodecahedron (and only 5 for the icosahedron). Rckrone (talk) 15:45, 28 July 2012 (UTC)[reply]

By "pyramid" do you mean square pyramid? I see two fundamentally different ways (excluding mirror images): two parallel edges on the base, and one edge on the base plus one between vertex and base. I'm not sure if I am misunderstanding the question... 86.129.16.51 (talk) 23:35, 24 July 2012 (UTC)[reply]

I'm pretty sure "pyramid" was referring to the tetrahedron. He/she is trying to answer this question for each Platonic solid. Rckrone (talk) 01:42, 25 July 2012 (UTC)[reply]

Ah, yes, of course. Unless otherwise specified, I tend to think of "pyramid" as meaning square pyramid... 86.129.16.51 (talk) 01:50, 25 July 2012 (UTC)[reply]

I just did the icosahedron since I was curious. I broke it up into cases similar to the dodecahedron example. There are always going to be 12 triangles that contain an edge in the matching, and then 8 that don't. You can enumerate the ways that the edges can be arranged around a given triangle that doesn't contain any edges, and then for each arrangement list out the possible ways that the other 3 edges can be chosen. I came up with 8 different matchings up to rotation: 2 that are invariant under flips, and 3 pairs of mirror images. Rckrone (talk) 02:40, 25 July 2012 (UTC)[reply]

I'll have to take a look at this, I was going from the other direction. put a point at the north pole and set the edge from the north pole to the next level down as where things are based from. The line from the south pole then is either 1/10th of the way around, 3/10th or 5/10th of the way around. The ones that are 7/10th and 9/10th will be mirror images and then the three cases (1/10th,3/10th and 5/10th) need to be worked out...Naraht (talk) 23:55, 25 July 2012 (UTC)[reply]

I was playing with this, and for the dodecahedron I got three fundamentally different solutions (excluding rotations) if mirror images are counted as different, and two if mirror images are counted the same. I would not bet my house on this answer, but anyway, that was what I got! 86.128.4.88 (talk) 02:48, 26 July 2012 (UTC)[reply]

Error in article but cannot edit it

http://wiki.riteme.site/wiki/Elementary_algebra

In the article above, there is an example of the commutativity of polynomials (2) and while it states that it may be obvious that (3+5) = (7+5) for the commutative property, obviously that is an incorrect example. The typed example is correct when I went to edit it but the pictures of the numbers, which I cannot edit, are the incorrect numbers. The picture numbers should also show that (3 + 5) = (5 + 3).

Thank you! — Preceding unsigned comment added by Jessicamoreau (talk • contribs) 16:58, 23 July 2012 (UTC)[reply]

The latex code was recently fixed (as you discovered when you tried to edit) but your browser must have the old image cached, so try refreshing that. Rckrone (talk) 17:43, 23 July 2012 (UTC)[reply]

The error was fixed after the message was posted here. The article is being developed pretty actively at the moment -- the error was introduced today and fixed a short time later. Looie496 (talk) 17:47, 23 July 2012 (UTC)[reply]

Primorials

What's the sum of the inverse primorials (1/2+1/6+1/30+1/210+...)? Since it's less than the inverse factorials, it clearly converges, but I can't find the sum. --146.7.96.200 (talk) 21:32, 23 July 2012 (UTC)[reply]

0.70523... Looie496 (talk) 23:04, 23 July 2012 (UTC)[reply]

Or if you want more precision, 0.7052301717918009651474316828882485137435776391091543281922679138139197811480028635861190519840274766...

isc (the new Plouffe's inverter) doesn't say anything about this number other than that it is the sum of inverse primorials. -- Meni Rosenfeld (talk) 09:49, 24 July 2012 (UTC)[reply]

I doubt there is a closed form expression. You'll only be able to approximate it. — Fly by Night (talk) 00:16, 24 July 2012 (UTC)[reply]

This raises a sort of interesting philosophical issue. Even if there was a closed form expression, it's unlikely that it would be easier to evaluate numerically than the series itself, which converges very rapidly. Looie496 (talk) 16:29, 24 July 2012 (UTC)[reply]

The interest in a closed-form expression is presumably driven by mathematical curiosity rather than a practical need to calculate the number to great accuracy... 86.179.3.247 (talk) 17:50, 24 July 2012 (UTC)[reply]

For mathematical curiosity and correctness. These numerical methods never give the correct answer; they give a sequence of incorrect answers that get closer to the correct answer. Numerical approximations are more than adequate for any real-life application. However, this is quite a pure question, with very little real-world application. I would argue that a numerical solution goes against the spirit of such a pure maths type question. — Fly by Night (talk) 17:54, 24 July 2012 (UTC)[reply]

In a way, though, the expressions that are allowed as "closed-form" are somewhat arbitrary. For example, we would, I assume, be happy with 2/e as a closed-form answer, yet e is only calculable as the unreachable limit of a series of approximations. In a different universe, the sum of inverse primorials might be an important number with a designation f, and then we would be happy with, say, 2/f as a "closed form" answer to some problem. 86.129.16.51 (talk) 20:49, 24 July 2012 (UTC)[reply]

I agree totally with the first part. There is something arbitrary about it all. There's nothing stopping you to define a number that is the answer to the OP's sum, and then claim it's a closed form expression. — Fly by Night (talk) 20:54, 24 July 2012 (UTC)[reply]

In reply to talk:Looie496. Yes it is an interesting question, it seems to be the subject of Transcendence theory. Its basically an extension of the idea of transcendental numbers. I guess most numbers will be of this form.--Salix (talk): 20:36, 24 July 2012 (UTC)[reply]

Rate of convergence is a relative notion. No matter how many terms you sum, you'll still have infinitely many incorrect digits. — Fly by Night (talk) 16:38, 24 July 2012 (UTC)[reply]

I doubt it, I'd only have a finite number of digits so only have a finite number wrong. Just to continue the philosophical debate with a bit of reality. ;-) Dmcq (talk) 17:36, 24 July 2012 (UTC)[reply]

Under the convention that 0.5 = 0.5000…, you would have infinitely many incorrect digits. Alternatively, consider the difference between any finite partial sum and the actual limit. All of these differences, i.e. errors, will be infinitely-long, non-trivial chains of digits. — Fly by Night (talk) 17:42, 24 July 2012 (UTC)[reply]

Well a potential infinite then rather than an actual infinite. Dmcq (talk) 18:03, 24 July 2012 (UTC)[reply]

təˈmeɪtoʊ, təˈmɑːtəʊ. — Fly by Night (talk) 20:34, 24 July 2012 (UTC)[reply]

Meni: thanks for linking to the new inverter. I didn't know there was one. We know the old one has been down since at least May. – b_jonas 09:56, 26 July 2012 (UTC)[reply]