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fact-004

MFT - Fermat Numbers - February 29, 2012
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Math Fact Tuesday! (even though it's Wednesday, but leap days don't count, right?)

Spiked Math Comic - Fermat Numbers


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21 Comments

I avoid using commas in math mode. Notice the extra space after the commas separating your digit triples. If it's a real LaTeX document, I like the siunitx package has a command \num{} that allows for automatic positioning of group separators. But I don't know what limitations the TeX engine is used in your comics - putting the commas in a \mbox{} is one low-tech solution. Please forgive my meddling - I'm actually a LaTeX teacher.

You can also add a negative space after the comma: 123,\!456.

I always avoid using explicit separators as they tend to differ among cultures which can lead to confusion.
(For example in most of Europe we would write 123.456,7 instead of 123,456.7 which is used in the USA/UK.)
Usually I end up using 123\,456 which just adds a little whitespace.

I avoid using commas in math mode too. It simply looks ridiculous.

I assume you mean commas as thousands-separators, rather than commas in general?

I forget why I even put the commas there to begin with! Normally I wouldn't in an ordinary document.

As far as I know, the preferred way is to use braces like in 14{,}328. This *suppresses* the extra space because {,} behaves like \mathord{,}. The negative space solution, on the other hand, in unfortunate circumstances might yield overlapping characters.

Mathematica says its not prime

(lets just forget the overflow error that is print and believe it :P )

Fermat was a fagget

I think you can use these numbers to prove there are infinitely many primes (I believe it was a STEP question) as each Fermat number is co-prime to any other one, so contains at least one unique prime number.

Perhaps off topic, but I just saw an example of a precedence question I've had for a while. The Fermat equation is F(n) = 2^2^n + 1. My question is if it is a general rule that this should be calculated as 2^(2^n) rather than (2^2)^n

It actually does not matter. In fact, 2^(2^n)=(2^2)^n=2^(2n)

2^(2^n)+1 is not the same as (2^2)^n+1.
list out the terms of the former: 3, 5, 17, 257, 65537...
list out terms of the latter: 2,5,17,257, ...

in fact, it should be 2^(2^n)+1

This is not true. (2^2)^n = 2^(2n) is correct, but 2^(2^n) != (2^2)^n. Example: 2^(2^5) = 2^32 != 2^10 = (2^2)^5. The first equation (2^2)^n = 2^(2n) is the reason for the convention that 2^2^n should be read as 2^(2^n) because in the other case it could be written as 2^2n.

I think the factors are of the form k*2^(n+1)+1, not (n+2)+1.

as each Fermat number is co-prime to any other one, so contains at least one unique prime number.

I might be crazy, but doesn't 2^(5+2)=128, not 64? It would still work if m=5 instead of m=10.

I found that F33 is prime. But the proof is too large to fit in this space. Sigh.

That number looks familiar, when I was cheating in Pokemon games

its not prime!!

freaking homework

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