Talk:Heegner number

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What does the function denoted Q in this article refer to?

Quadratic Field

No, rational numbers. Charles Matthews 20:57, 12 Mar 2004 (UTC)


I had said that I would either create this page or work on it if someone else created it. Now, someone else has created it and I realize that this topic is way over my head. Nevertheless, I will continue my study of imaginary quadratic fields. PrimeFan 19:38, 13 Mar 2004 (UTC)

--- "Equivalently, its ring of integers has unique factorization"

Should "ring of integers" be "ring of ideals" or something similar? It's been a while since I've taken any number theory classes and I'm not entirely sure on this.

No, "ring of integers" is correct. See this article. An ""ideal" is a subset of a ring that is closed under addition, with the additional property that if any element in the ideal is multiplied by a ring element, the result still lies in the ideal. If we consider the ring of ordinary integers, then if n is any integer the set {..., −3n, −2n, −n, 0, n, 2n, 3n, ...} is an ideal because (a) it's closed under addition, and (b) if we multiply any multiple of n by an integer we obtain another multiple of n. (The concept of an ideal grew out of a desire to understand different sorts of algebraic numbers, where addition and multiplication aren't necessarily tied together the way they are in the natural numbers.) DavidCBryant 19:34, 8 January 2007 (UTC)[reply]

structure of the J(...)=x^3 term[edit]

I read today a recent post on sci.math.research by Titus P. on the additional structure of the integers obtained as exp(pi sqrt n), namely e^(pi sqrt n) - 744 = (12(a(n)^2-1))^3, where the a(n) are consequently extremely close to an integer for Heegner numbers (=3 mod 4). Could somebody sufficiently competent add one or 2 phrases (and/or xrefs) about that? — MFH:Talk 03:53, 16 April 2008 (UTC)[reply]

Mathworld neologism?[edit]

This term appears to be a neologism from mathworld. Google books returns two results, one the print version of mathworld, and the other some pop math book. Google scholar does no better, returning 3 results: all post-date the mathworld article. One result is published in a journal. The other two results are web-published. One references the other, which in turn references mathworld. Does anyone think this isn't a mathworld neologism? RobHar (talk) 03:12, 4 August 2009 (UTC)[reply]

I'll post to Wikipedia talk:WikiProject Mathematics, and if no one replies I'm going to nominate for deletion (and suggest merging content into Stark–Heegner theorem, Ramanujan's constant, and Formula for primes). RobHar (talk) 00:32, 5 August 2009 (UTC)[reply]
There are a little more Google Scholar and Books hits on "Heegner numbers" than "Heegner number". PrimeHunter (talk) 01:24, 5 August 2009 (UTC)[reply]
Definitely not a MathWorld neologism. The term "Heegner numbers" is used by Conway and Guy in The Book of Numbers here (maybe that was what you meant by "some pop math book" ?). This 1996 usage is cited in the MathWorld article, and I have added a ref to our article. Since we have Conway and Guy as a primary source and Weisstein and Sloane/OEIS as secondary sources (plus a few other Google Books hits), I think we can now count this as established terminology. Gandalf61 (talk) 08:43, 5 August 2009 (UTC)[reply]
Alright, thanks. I find it weird that "Heegner numbers" returned more results than "Heegner number". The pop math book I was referring to was "The Kingdom of Infinite Number: A Field Guide"; had the Conway book turned up with the "Heegner number" query, I would've been more accepting. The term is still clearly extremely rare and not used at all in the field, but oh well I guess. Thanks again. RobHar (talk) 14:50, 5 August 2009 (UTC)[reply]

I think it a useful article; if it exists, it needs a name. Is discriminants of complex quadratic fields of class number 1 really an improvement? Septentrionalis PMAnderson 15:46, 5 August 2009 (UTC)[reply]

I think that the entirety of the contents can quite nicely be merged into the three articles I mentioned above, thus rendering this article a mere definition of a rarely used term. RobHar (talk) 16:26, 5 August 2009 (UTC)[reply]
No, Ramanujan's constant should be merged here (there is no corresponding name for the other near-integers); any of this would be disproportionate in Formula for primes; and Stark-Heegner theorem is nicely complementary to this article. Septentrionalis PMAnderson 16:55, 5 August 2009 (UTC)[reply]
This may be a relatively recent neologism. But I don't find it to be of the vexing kind - it's not promoting anything, just giving a label to a short but significant list of discriminants. There is something to be said on both sides, but I think the status quo is OK. Charles Matthews (talk) 07:31, 6 August 2009 (UTC)[reply]
After actually bothering to read the notability guidelines for numbers, I see that this article as it stands violates criterion 4 of Wikipedia:Notability (numbers)#Notability of sequences of numbers, i.e.
Is there at least one commonly accepted name for this sequence?
There is clearly no commonly accepted name for this set of numbers. Do what you will with this. RobHar (talk) 15:21, 7 August 2009 (UTC)[reply]

?[edit]

Does anyone know why is so close to an integer? 4 T C 07:54, 8 January 2010 (UTC)[reply]

Good question. I've just checked it (using SpeedCrunch, if anyone's interested) and you're right, it is very close to one. It comes to approximately;

24591257751.999999822213241.

I don't know the reason for this either, and I'd be interested to know what it is (I'm guessing that with a result that close to being an integer, there's a reason for it).

Meltingpot (talk) 13:15, 9 February 2014 (UTC)[reply]

was this really conjectured by Gauss?[edit]

Can someone give an exact reference to the fact that this was conjectured by Gauss?

Gauss' class number one list, row I.1 in article 303 of Disquisitiones Arithmeticae, is: 1,2,3,4,7 and the completeness of this list was proved by Landau 1902 (see Goldfeld, Bull. AMS, 1985, p.26). Gauss is looking at a slightly different question, with nontrivial overlap, but different. Did he conjecture _elsewhere_ this Heegner list?

129.199.2.17 (talk) 09:20, 26 October 2012 (UTC)[reply]

I don't know off-hand the answer to your question, but the point is that Gauss conjectured something which in modern terms is that those nine numbers give all imaginary quadratic fields of class number 1. It is very common in math to restate conjectures/theorems like this and still attribute them to the old folks. RobHar (talk) 07:16, 27 October 2012 (UTC)[reply]

don't put thousands separators under <math>![edit]

So reads a hidden comment throughout the article. I wonder whether there is any reason not to. I can only suppose that it's because something like <math>e^{\pi \sqrt{163}} = 262,537,412,640,768,743.99999999999925... </math> won't work but <math>e^{\pi \sqrt{163}} = 262{,}537{,}412{,}640{,}768{,}743.99999999999925... </math> does and is used on the article. Better still, though, we could have <math>e^{\pi \sqrt{163}} = 262\,537\,412\,640\,768\,743.999\,999\,999\,999\,25... </math> as in this version. The article is much more readable when numbers are formatted. Can anyone enlighten me as to whether there is some convention not to format numbers within <math> and why this would be so given that all <math> is is a tool for displaying formulae etc.? Jimp 13:10, 19 April 2015 (UTC)[reply]

Okay, I've found the reason and it was just as I'd suspected: "do not fall to such heresy as <math>262,537,412,640,768,743</math>! it is rendered with unreadable spaces after commas". <math>262{,}537{,}412{,}640{,}768{,}743.99999999999925</math>! (as used on the article) isn't nor is <math>262\,537\,412\,640\,768\,743.999\,999\,999\,999\,25</math>!. The latter is the most readable of the two (and better by far than <math>262537412640768743.99999999999925</math>!). Jimp 13:49, 19 April 2015 (UTC)[reply]

Bad formatting of the compound exponentials appears to be a problem throughout this entire wiki article. I fixed one section. Expert Review Needed![edit]

I fixed the formatting in →‎Almost integers and Ramanujan's constant:

with the explanation had bad formatting of the compound exponential, as easily shown by actually computing the number. As previously shown, including on MathWorld (still in error), the final exponentiation was improperly shown as mere multiplication. I have corrected the formatting for both the wiki article itself and the equation from the references. The latter correction may be improper, but it seems best to stop propagating the error. Someone tell me! YodaWhat (talk) 09:20, 25 February 2020 (UTC) undo[reply]

But I am uncertain if it actually IS an error everywhere it seems to be, and I only had the confidence to fix the one area in which I could actually compute the correct numeric answer myself to verify. Certainly the number 262,537,412,640,768,743.999,999,999,999,25... cannot be reached unless the (non-RPN) calculator sequence is e to the pi to the root163

Someone expert in this field (and in Wikipedia protocols) needs to review the entire article now, including my changes!

YodaWhat (talk) 09:20, 25 February 2020 (UTC)[reply]

@YodaWhat: Your changes were wrong and have been reverted. Exponentiation is right-associative (see Exponentiation#Identities and properties), so abc means a(bc) which is not the same as (ab)c. You may also be unaware of the rule (ab)c = ab×c. The expressions could have been written as (ab)c with the parentheses but it's far more common to write ab×c. PrimeHunter (talk) 20:06, 7 March 2020 (UTC)[reply]

My dear Mr. @PrimeHunter:, you are wrong, in every way. I stated my case humbly and simply, in a way that could be verified easily, while the right-associative exponentiation method you first suggest does not pass any 'sanity test' and is disproven easily. Any old scientific calculator would disprove your method in seconds, but you failed to check. Why? We are looking for a numeric result of the order 2.6x10^17, and Pi^root163 is of the order 2,224,255, and e^222 is of the order 2.6x10^96 already. Clearly a drastic mistake! [Though a^(b×c) works.]

Your education has betrayed you, and you should really have checked results numerically, as I did, before insulting me by saying I was wrong. I would not have made bold to make ANY change had I not triple-checked things for myself in a very clear and simple way. You, on the other hand, are merely following DOGMA, a type of mistake that is unfortunately very common. Dogma can be true, in the right time and place, but your element of dogma is incorrectly applied here, resulting in Fogma. Then you try to justify your mistake with an appeal to authority that would get you sharply reprimanded in Debate class. Yours are the kinds of errors which can get one a reputation for being a lazy, lying, pompous ass. Do you want to be known as a pompous ass? No? Do you want to spread lies and claim they are gospel truth? No? Then you will kindly revert your reversion, AND please fix the rest of the article, yes?

I actually took the time to do the computation every which way, and it works properly the way shown in my changes. Please do the computation yourself, and see for yourself. It is easily done to high precision in Windows Calculator in Scientific mode, as shown in these results copied and pasted from that calculator [calc.exe version 6.9.9200.16384]

Here are the calculations:
The square root of 163 = 12.767145334803704661710952009781
e^pi = 2.7182818284590452353602874713527^3.1415926535897932384626433832795 = 23.14069263277926900572908636795
23.14069263277926900572908636795^12.767145334803704661710952009781 = 262,537,412,640,768,743.9999999999994

You cannot trust the typesetter at some website or even of a book, nor even your professors at college to get every detail exactly correct. Nor can you trust that the order of exponentiation used today is the same as used centuries ago. Things change, and most individuals are not entirely consistent, either.

If computed as you suggest, Pi^root163, which equals 2,224,255.524210951847131440033073, then e^2,224,255.524210951847131440033073, the calculator returns an error, invalid input. Of course it cannot handle that calculation, it can compute numbers up to 9.9999999999999999999999999999999x10^9999 "only". It naturally will not accept 2.7182818284590452353602874713527^2,224,255.524210951847131440033073 either.
So there it is. Meanwhile, having already achieved the correct result using what you call the wrong way, I shrug and say 'Equations can confuse, but the numbers don't lie.'

P.S. -- The referenced equation from Mathworld is still wrong. Can that error be fixed? Thank you for your attention. YodaWhat (talk) 21:02, 2 May 2020 (UTC)[reply]

@YodaWhat: As I said, exponentiation is right-associative. This is a mathematical definition. See e.g. the Google search exponentiation is right-associative. Don't trust some old calculator. It's harder to implement right-associative and many calculators don't do it. Modern programming languages usually follow the mathematical definition. See e.g. https://www.wolframalpha.com/input/?i=2%5E2%5E3, or enter 2^2^3 at https://pari.math.u-bordeaux.fr/gp.html. They both compute 2^2^3 = 2^(2^3) = 2^8 = 256, whereas (2^2)^3 = 64. Anyway, the article never says abc so we don't even have to discuss the correct meaning unless you change the article to use that notation incorrectly. It's not an insult to simply say that somebody is wrong. You are however insulting me, and you are still wrong. Please see Wikipedia:No personal attacks and Wikipedia:Civility. By the way, it wasn't me who reverted you, but I agree with the user who did it. The article is correct and MathWorld is correct. only has one exponentiation since π and 163 has the same baseline in the typography. The expression is by syntactic definition, without even having to define exponentiation. PrimeHunter (talk) 23:43, 2 May 2020 (UTC)[reply]

Missing minus sign?[edit]

The article states

Euler's formula, with n taking the values 1,... 40 is equivalent to

n2 + n + 41

Shouldn't this be as follows?

Euler's formula, with n taking the values -1,... -40 is equivalent to

n2 + n + 41 — Preceding unsigned comment added by Mark.sullivan (talkcontribs) 18:47, 11 April 2020 (UTC)[reply]

@Mark.sullivan: Your quote is incomplete. The article says:

Euler's prime-generating polynomial

which gives (distinct) primes for n = 1, ..., 40, is related to the Heegner number 163 = 4 · 41 − 1.

Euler's formula, with taking the values 1,... 40 is equivalent to

with taking the values 0,... 39

This is correct. It would give the same result to take -1,... -40 in but I have never seen a source choose the negative values. The issue is that Euler wrote while modern sources often write so we explain that it merely corresponds to shifting n by 1. PrimeHunter (talk) 23:57, 2 May 2020 (UTC)[reply]