part I have trouble with is the above and below part for sum I use under and over but that creates fractions not just under or over respectively.

figured it out(except the infinity symbol) lol took a look at flouran's post in the LaTeX thread.

Example:

[tex]f(x)=\sum_{n=1}^{x}n^{2}[/tex]

with:
[noparse]
[tex]f(x)=\sum_{n=1}^{x}n^{2}[/tex]
[/noparse]

[tex]\infty[/tex]

[noparse]
[tex]\infty[/tex]
[/noparse]

And:
[url=http://www.mersenneforum.org/showthread.php?t=11183]Here[/url] is an example of the Law of small numbers.

That conjecture was verified upto n=2.3M, so not really small compared to n=1, 2 or 3,
but the first counterexample was higher than that level found the next day!

[QUOTE=kar_bon;225298]Example:

[tex]f(x)=\sum_{n=1}^{x}n^{2}[/tex]

with:
[noparse]
[tex]f(x)=\sum_{n=1}^{x}n^{2}[/tex]
[/noparse]

[tex]\infty[/tex]

[noparse]
[tex]\infty[/tex]
[/noparse][/QUOTE]

so it's [TEX]\pi = \sum^\infty_{k=0}[/TEX] I got into trouble with the fraction god not as easy lol.

[QUOTE=science_man_88;225299]so it's [TEX]\pi = \sum^\infty_{k=0}[/TEX] I got into trouble with the fraction god not as easy lol.[/QUOTE]

The key is using the brackets { and } !

[QUOTE=kar_bon;225300]The key is using the brackets { and } ![/QUOTE]

will that help me I know frac is \frac{} but I have no idea what to do with it.

Hey: Can anyone solve this? [tex]$71^6 \equiv 1(mod 22!)$[/tex]

In other words: Find a number of the form (k * 1124000727777607680000 + 1) which is divisible by 71[sup]6[/sup].

[TEX]\pi = 4\sum^{\infty}_{k=0}\frac{(-1)^{k}{2k+1}}[/TEX]

this is the best [B]I[/B] could do.

[QUOTE=science_man_88;225303][TEX]\pi = 4\sum^{\infty}_{k=0}\frac{(-1)^{k}{2k+1}}[/TEX]

this is the best [B]I[/B] could do.[/QUOTE]

Try this:
[noparse]
[tex]\pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}[/tex]
[/noparse]

[tex]\pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}[/tex]

[QUOTE=kar_bon;225304]Try this:
[noparse]
[tex]\pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}[/tex]
[/noparse]

[tex]\pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}[/tex][/QUOTE]

thanks karbon maybe now I can finally express something most people who can help [TEX] stand \over will[/TEX] lol.

Also: Found a decent factorial-based prime:

20019 * 426! + 1 is an SPRP. (It has 936 digits - Too small. I need somewhat larger primes.)

Sadly, there are no sieve implementations except one I programmed: But that's actually doing the same as pform.

I found a 2293-digit composite pseudoprime. Bummer.

20004 * 901! + 1 is prime! (≈2280-2285 digits)