Talk:Fermat number

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[edit] old comment

It's worth noticing something. If it makes sense to talk of "probabilities" in this area, then the odds are good that Fermat's little theorem will "make" these numbers prime. That's because there is so much scope for any of the testing sequences to come up with 1 at an intermediate stage, in which case they will stick there. 1 is an attractor for the sequences. PML.

[edit] original proof of 2n + 2k + 1

The article first attributes Euler for proving that factors of Fermat numbers must be of the form 2n + 2k + 1, but then the article later states that Édouard Lucas is the one who proved it. I'm sure both of them aren't the source. Which is it? —Preceding unsigned comment added by 72.235.147.240 (talk) 08:43, 8 October 2007 (UTC)

=== The statements are a little different. Euler proved that any prime factors of a Fermat number 2^{2^{n}}+1 must be of the form 2n + 1k + 1 for some integer k, and Lucas proved that any prime factor of Fermat number 2^{2^{n}}+1 must be of the form k2n + 2 + 1 (for some integer k) if n > 1. For example, Euler's result guarantees only that every prime divisor of the Fermat number 232 + 1 must be of the form 64k + 1, but Lucas's result guarantees that every prime divisor of the Fermat number 232 + 1 must be of the form 128k + 1. Messagetolove 14:59, 8 October 2007 (UTC)

=== Oh, I see. Scrolling up and down, I didn't catch the differences.

[edit] True or false??

True or false: It is likely the F_33 will be prime with current knowledge.

Currently, the primality status of F_33 is unknown. See [1] for details. Maxal 05:45, 13 December 2006 (UTC)

F_34 is more likely to be prime than F_33. It has been proven that F_34 has no prime factors below 2^69.

But, to prove a number prime, you must verify it has no factors less than or equal to its square root. The above info would have proved it to be prime if it were less than 2^138, but it is not, its base 2 logarithm is just over 2^34. (Note that the base 2 logarithm of 2^138 is 138.) Georgia guy 20:54, 17 June 2007 (UTC)

[edit] 2 Squared N?

Is there a way to superscript the n above the 2 in the expression of the function? Right now it doesn't look like "Two to the two-to-the-Nth power" but "Two squared times N." That confused me greatly at first reading. Nevah 03:28, 1 May 2006 (UTC)

[edit] Factually Incorrect

From the article:

If 2n + 1 is prime, it can be shown that n must be a power of 2. (If n = ab where 1 < a, b < n and b is odd, then 2n + 1 ≡ (2a)b + 1 ≡ (−1)b + 1 ≡ 0 (mod 2a + 1).) In other words, every prime of the form 2n + 1 is a Fermat number, and such primes are called Fermat primes. The only known Fermat primes are F0,...,F4.


I don't know much about Fermat numbers but I do know that 2 is a prime number of the form 2n+1 where n is 0. Zero is not a power of two so the first part cannot be correct and the "every prime of the form 2n+1 is a fermat prime" is also incorrect since 2 is of the form 2n+1 and not a fermat prime. I'm gonna try to fix this, but somebody who knows what they're doing really should. --Shadowoftime 18:53, 16 July 2006 (UTC)

Indeed, the statement only holds for nonzero n. -- EJ 19:06, 16 July 2006 (UTC)

I agree with Shadowoftime. I believe that 2 should be a Fermat prime because it is a prime of the form 2n+1. If a prime is of this form, then it is a Fermat prime. --PhiEaglesfan712 14:57, 17 June 2007 (UTC)

Wikipedia content should have reliable sources, and the reliable sources I have seen either define a Fermat prime as a prime of form 2^(2^n)+1 with integer n>=0, or the provably equivalent definition 2^n+1 with integer n>=1. PrimeHunter 21:44, 17 June 2007 (UTC)

[edit] Regarding the phrase...

"The following heuristic argument suggests there are only finitely many Fermat primes: according to the prime number theorem, the "probability" that a number n is prime is at most A/ln(n), where A is a fixed constant. Therefore, the total expected number of Fermat primes is at most

A \sum_{n=0}^{\infty} \frac{1}{\ln F_{n}} = \frac{A}{\ln 2} \sum_{n=0}^{\infty} \frac{1}{\log_{2}(2^{2^{n}}+1)} < \frac{A}{\ln 2} \sum_{n=0}^{\infty} 2^{-n} = \frac{2A}{\ln 2}

It should be stressed that this argument is in no way a rigorous proof. For one thing, the argument assumes that Fermat numbers behave "randomly", yet we have already seen that the factors of Fermat numbers have special properties. Although it is widely believed that there are only finitely many Fermat primes, there are some experts who disagree. [2]"

Name a kind of prime that the same argument suggests is finite, but that in fact is known to be infinite. Georgia guy 20:29, 7 October 2006 (UTC)

[edit] Reciprocal sum

yo, the property "the sum of two fermat number recipricals is irrational" is clearly wrong, a rational + a rational is rational

That quote is false. The article actually says "The sum of the reciprocals of the Fermat numbers is irrational". It's the sum of all reciprocals. A sum of infinitely many rationals can be irrational. PrimeHunter 00:43, 23 December 2006 (UTC)

[edit] F33 status

2^2^33 + 1 is known to have no prime factors below... Georgia guy 23:34, 24 February 2007 (UTC)

[edit] F34 status

2^2^34 + 1 is known to have no prime factors below 2^69 —Preceding unsigned comment added by PhiEaglesfan712 (talk • contribs) ("below 2^69" was added after Georgia guy's below comment)

You mean, it is known to be prime?? Where is your source?? Prove it. Georgia guy 14:57, 17 June 2007 (UTC)

See [3] for details. —Preceding unsigned comment added by PhiEaglesfan712 (talk • contribs)

No, that doesn't say it is known to be prime; it says it has unknown character; if it is prime; I strongly doubt it will be proven before 2030. Georgia guy 18:53, 17 June 2007 (UTC)

However, there is a huge chasm between 2^69 and 2^131072 (the square root of 2^2^34)... If 2^2^34 + 1 was prime... not only do I win $50,000 for the first to find a 10 million digit prime... I would win $400,000 more for finding the first 100 million digit prime and 1 billion digit prime. The number 2^2^34 + 1 has nearly 5.5 billion digits! —Preceding unsigned comment added by PhiEaglesfan712 (talk • contribs)

No, 2^131072 is the square root of 2^262144. The square root of 2^2^34 is 2^(2^33). Georgia guy 20:57, 17 June 2007 (UTC)

If there is a reliable source to known trial factor limits for the smallest Fermat numbers with unknown status, then that may be worth mentioning in the article. The link from PhiEaglesfan712 does not mention a limit of 2^69. PrimeHunter 21:54, 17 June 2007 (UTC)

You are correct. The square root of 2^2^34 is 2^ (2 ^ 33)... I realized the mathematical error after I signed off earlier today... However, my computer tried to factor 2^2^34 to 2^69, and it said that "F34 has no factor to 2^69".

[edit] Luca prime

A prime divisor of Fermat numbers is called Luca prime. The first two Luca prime is 3,5. It is difficult to know whether the given prime is Luca prime. It is not even known if 7 is a Luca prime. We do know that the sum of the the reciprocals of all Luca primes converges.218.133.184.93 16:27, 22 March 2007 (UTC)

The answer is that 7 is definitely not. There are certain primes that no number of the form 2^n+1 is divisible by, and 7 is such a prime. The sequence:

2, 3, 5, 9, 17, 33, 65

Is congruent to 2, 3, 5, 2, 3, 5, 2 mod 7. Georgia guy 16:57, 22 March 2007 (UTC)

I found no mention of "Luca prime" with Google. If the term has been used then it doesn't appear notable enough to mention in Wikipedia. PrimeHunter 01:16, 23 March 2007 (UTC)

Is there any algorithm to know whether the given prime is Luca prime?218.133.184.93 08:10, 23 March 2007 (UTC)

See http://primes.utm.edu/glossary/page.php?sort=FermatDivisor. Your alleged term "Luca prime" is usually called a Fermat divisor. PrimeHunter 15:28, 23 March 2007 (UTC)
The problem is that 3 and 5 are excluded from Fermat divisors.

[edit] Ninth Fermat number??

Recently, there was a discussion about the definition of the "ninth Fermat number". Look at this, as an example:

  • Ninth Floor
  • Eighth Floor
  • Seventh Floor
  • Sixth Floor
  • Fifth Floor
  • Fourth Floor
  • Third Floor
  • Second Floor
  • Ground Floor
  • Basement Floor

As you can see in the above list, the "ninth floor" is technically the tenth floor, but it is designated as the ninth. Does Wikipedia have any article about inconsistent definitions of ordinal numbers?? Georgia guy 13:45, 13 July 2007 (UTC)

See Zeroth. PrimeHunter 16:25, 13 July 2007 (UTC)
So, that means that 220 + 1 = 3, is counted as the zeroth Fermat number. PhiEaglesfan712 17:27, 13 July 2007 (UTC)
Yes. I have also seen it called "the first" but in contexts where "first" appears to mean "the initial" rather than the beginning of a systematic numbering. When numbering gets to "the ninth Fermat number" I have only seen the name about F9 = 229 + 1. PrimeHunter 18:15, 13 July 2007 (UTC)

But keep in mind, that 3 is one Fermat number and 3 and 5 are two. So its nearly unavoidable calling 5 the second. To avoid missunderstandings the first Fermat number should be called Fermat number zero, thats perfect. --Tilman Piesk 13:29, 11 October 2007 (UTC)

[edit] Fermat numbers in hexadecimal representation

Hi, I added the following table of the first nine Fermat numbers in hexadecimal representation, which was now deleted as "redundant". I think, it is more easy to understand these numbers seeing them hexadecimal, because 1.0001hex and 1.0000.0001hex are much more concise than their decimal equivalents. Due to the fact that a normal brain does not see, that 4.294.967.296dec ist simply 1.0000.0001hex I dont concider the second table as redundant. Hexadecimal is the most appropriate representation for Fermat numbers, and the first table is only necessary for people who can´t read them. What are your thoughts? --Tilman Piesk 13:22, 11 October 2007 (UTC)

F0 = 2 1 + 1 = 3 = p(2)
F1 = 2 2 + 1 = 5 = p(3)
F2 = 2 4 + 1 = 11 = p(7)
F3 = 2 8 + 1 = 101 = p(37hex) = p(55dec)
F4 = 210 + 1 = 1.0001 = p(198Fhex) = p(6543dec)
F5 = 220 + 1 = 1.0000.0001
F6 = 240 + 1 = 1.0000.0000.0000.0001
F7 = 280 + 1 = 1.0000.0000.0000.0000.0000.0000.0000.0001
F8 = 2100 + 1 = 1.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0000.0001
I don't think there is any reason to show hexadecimal numbers. They are rarely used in mathematics outside computer science and people who are familiar with them are likely to understand the formula 2n+1 well without listing it in hex. By the way, all your exponents from F4 are wrong. PrimeHunter 22:35, 11 October 2007 (UTC)
I tend to agree that there is not much of an advantage, since from 17 on,they are all of the form 16^m + 1, so they all start with a 1, end with a 1, and have an easily calculated number of zeroes in between. So there does not seem to be a great need to use space to write them down explicitly. However, hexadecimal numbers do have other mathematical interest, see for example Bailey-Borwein-Plouffe formula on the hexadecimal expansion of piMessagetolove 13:56, 12 October 2007 (UTC)

"By the way, all your exponents from F4 are wrong."
This very funny hint shows, that hexadecimal numbers are indeed very strange to a considerable part of the population...
10 = seize, 20 = deux-seize, 40 = quatre-seize, 80=huit-seize, 100=nilā
--Tilman Piesk 13:31, 17 October 2007 (UTC)

OK, I overlooked that you used hexadecimal exponents. I have never seen anybody do that before, and I often see hex numbers. If it's done (which I don't think it should be) then it should be said clearly that exponents are also hex. PrimeHunter 14:06, 17 October 2007 (UTC)