PARI's commands - Page 38

science_man_88 · 2010-08-13, 21:25

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.

science_man_88 · 2010-08-13, 21:34

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

kar_bon · 2010-08-13, 21:35

Example:

$f(x)=\sum_{n=1}^{x}n^{2}$

with:

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

$\infty$

[tex]\infty[/tex]

And:
Here 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!

science_man_88 · 2010-08-13, 21:40

Quote:

Originally Posted by kar_bon

Example:

$f(x)=\sum_{n=1}^{x}n^{2}$

with:

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

$\infty$

[tex]\infty[/tex]

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

kar_bon · 2010-08-13, 21:42

Quote:

Originally Posted by science_man_88

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

The key is using the brackets { and } !

science_man_88 · 2010-08-13, 21:43

Quote:

Originally Posted by kar_bon

The key is using the brackets { and } !

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

3.14159 · 2010-08-13, 21:46

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

In other words: Find a number of the form (k * 1124000727777607680000 + 1) which is divisible by 71⁶.

science_man_88 · 2010-08-13, 21:48

$\pi = 4\sum^{\infty}_{k=0}\frac{(-1)^{k}{2k+1}}$

this is the best I could do.

kar_bon · 2010-08-13, 21:51

Quote:

Originally Posted by science_man_88

$\pi = 4\sum^{\infty}_{k=0}\frac{(-1)^{k}{2k+1}}$

this is the best I could do.

Try this:

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

$\pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}$

science_man_88 · 2010-08-13, 21:54

Quote:

Originally Posted by kar_bon

Try this:

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

$\pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}$

thanks karbon maybe now I can finally express something most people who can help $stand \over will$ lol.

3.14159 · 2010-08-13, 22:00

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)

2010-08-13, 21:34	#409
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	figured it out(except the infinity symbol) lol took a look at flouran's post in the LaTeX thread. Last fiddled with by science_man_88 on 2010-08-13 at 21:35

2010-08-13, 21:35	#410
kar_bon Mar 2006 Germany B5C₁₆ Posts	Example: $f(x)=\sum_{n=1}^{x}n^{2}$ with: [tex]f(x)=\sum_{n=1}^{x}n^{2}[/tex] $\infty$ [tex]\infty[/tex] And: Here 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! Last fiddled with by kar_bon on 2010-08-13 at 21:40

2010-08-13, 21:46	#414
3.14159 May 2010 Prime hunting commission. 2⁴·3·5·7 Posts	Hey: Can anyone solve this? $71^6 \equiv 1(mod 22!)$ In other words: Find a number of the form (k * 1124000727777607680000 + 1) which is divisible by 71⁶. Last fiddled with by 3.14159 on 2010-08-13 at 21:48

2010-08-13, 22:00	#418
3.14159 May 2010 Prime hunting commission. 690₁₆ Posts	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) Last fiddled with by 3.14159 on 2010-08-13 at 22:55

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2010-08-13, 21:25	#408
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	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.

2010-08-13, 21:48	#415
science_man_88 "Forget I exist" Jul 2009 Dumbassville 8384₁₀ Posts	$\pi = 4\sum^{\infty}_{k=0}\frac{(-1)^{k}{2k+1}}$ this is the best I could do.