PARI's commands - Page 39

science_man_88 · 2010-08-13, 23:00

this is probably already done but I think:

$M(x)= \sum_{n=1}^{x-1} + 5$

not sure though.

Edit: $M(x)= \sum_{n=1}^{x-1} M(n)+ 5$ or $M(x)= \sum_{n=0}^{x-1} M(n)+ 5$

another edit x>3

I have learn to think over it now lol technically replace 5 with $\sum _{p=0}^{2}M(p)$ all add (x-4)

kar_bon · 2010-08-13, 23:13

Quote:

Originally Posted by 3.14159

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

2278 digits!

science_man_88 · 2010-08-13, 23:23

I should say all that thinking came up to:

for x>3, $M(x)=\sum_{n=0}^{x-1}M(n)$ all plus x.

3.14159 · 2010-08-13, 23:29

Quote:

Originally Posted by Karsten

2278 digits!

You still haven't found me a solution to n * 71⁶ = 1 mod 1124000727777607680000.

science_man_88 · 2010-08-13, 23:31

Quote:

Originally Posted by 3.14159

You still haven't found me a solution to:

$71^6 \equiv 1(mod 22!)$

outside of proving it i have no idea lol

71^6 = 128100283921
22! = 1124000727777607680000 ?

so if pari did what I think you are wrong.

now i saw the n and altered my pari code so far i've checked over 1 million. going toward 1 billion.

I gave up.

3.14159 · 2010-08-13, 23:41

Quote:

Originally Posted by science_m

outside of proving it i have no idea lol

71^6 = 128100283921
22! = 1124000727777607680000 ?

so if pari did what I think you are wrong.

Nope. n * 71⁶ = 1 mod 1124000727777607680000.

science_man_88 · 2010-08-13, 23:50

Quote:

Originally Posted by 3.14159

Nope. n * 71⁶ = 1 mod 1124000727777607680000.

I bow out I checked again to n=100 million and found nothing.

3.14159 · 2010-08-13, 23:53

Quote:

Originally Posted by science_man_88

I bow out I checked again to n=100 million and found nothing.

And you would have continued checking for another few thousand years, because it's in the 10^20 range.

science_man_88 · 2010-08-14, 00:40

So Pi have you ever heard of:
for x>3 $M(x)=\sum_{n=0}^{x-1}M(n) + x$

?

if so who came up with it.

I realized this works for all of them above M(0) = 0

CRGreathouse · 2010-08-14, 01:09

Quote:

Originally Posted by 3.14159

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

Code:

> Mod(-1,71^6)/1124000727777607680000
%1 = Mod(36882197250, 128100283921)

so 41455616551037280586874880000001, 185440428906759455741024993280001, ...

If you wanted k to be a prime, then

Code:

forstep(n=36882197250,1e15,128100283921,if(isprime(n),return(n)))

takes 0ms to come up with the smallest example, 1702185888223.

CRGreathouse · 2010-08-14, 02:19

Quote:

Originally Posted by 3.14159

And you would have continued checking for another few thousand years, because it's in the 10^20 range.

I must be interpreting you the wrong way, then, since I find 17 million instances below 10²⁰:

Code:

s=0;forstep(n=36882197250,1e20,128100283921,if(ispseudoprime(n),s++));s
time = 24mn, 50,380 ms.
%1 = 17584374

2010-08-13, 23:00	#419
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	this is probably already done but I think: $M(x)= \sum_{n=1}^{x-1} + 5$ not sure though. Edit: $M(x)= \sum_{n=1}^{x-1} M(n)+ 5$ or $M(x)= \sum_{n=0}^{x-1} M(n)+ 5$ another edit x>3 I have learn to think over it now lol technically replace 5 with $\sum _{p=0}^{2}M(p)$ all add (x-4) Last fiddled with by science_man_88 on 2010-08-13 at 23:15

2010-08-13, 23:23	#421
science_man_88 "Forget I exist" Jul 2009 Dumbassville 10000011000000₂ Posts	I should say all that thinking came up to: for x>3, $M(x)=\sum_{n=0}^{x-1}M(n)$ all plus x. Last fiddled with by science_man_88 on 2010-08-13 at 23:23

2010-08-14, 00:40	#427
science_man_88 "Forget I exist" Jul 2009 Dumbassville 20300₈ Posts	So Pi have you ever heard of: for x>3 $M(x)=\sum_{n=0}^{x-1}M(n) + x$ ? if so who came up with it. I realized this works for all of them above M(0) = 0 Last fiddled with by science_man_88 on 2010-08-14 at 00:44

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