We all look for patterns in what we see and try to explain them. There's a certain art form in figuring out what patterns are likely to work out and which are probably just coincidence.

Now I see that sm88 has highlighted a second antidiagonal in yellow. It's probably worth pointing out that changes do propagate in that direction, since each cell is just the cell above and to its right times the appropriate number from the top row, mod 9 of course.

[QUOTE=CRGreathouse;225241]Now I see that sm88 has highlighted a second antidiagonal in yellow. It's probably worth pointing out that changes do propagate in that direction, since each cell is just the cell above and to its right times the appropriate number from the top row, mod 9 of course.[/QUOTE]

I even find the longer versions of these down lower but I doubt it helps.

[QUOTE=CRGreathouse]We all look for patterns in what we see and try to explain them. There's a certain art form in figuring out what patterns are likely to work out and which are probably just coincidence.
[/QUOTE]

Yes, but it should not be used to claim that there are patterns where there are not. In maths, most of the time, it turns out to be the law of small numbers, combined with magical thinking. The end.

[QUOTE=3.14159;225245]Yes, but it should not be used to claim that there are patterns where there are not. In maths, most of the time, it turns out to be the law of small numbers, combined with magical thinking. The end.[/QUOTE]

You'd hate Ramsey Theory, methinks. :smile:

A good half of sm88's patterns have had basis in fact -- either they've been entirely correct or a small correction makes them right. (Most of the time the hard part is just getting the idea formalized at all: for example, what prediction is being made in the case of the antidiagonals?)

But what I'd really like to see is sm88 learning enough to prove or disprove his own conjectures.

here's a pattern i know isn't false lol my mother's bad luck (hopefully it's not hereditary). lets see a originally mis-diagnosed rare and terminal breast cancer, now people backed out of our house selling agreement as he's having heart and knee trouble. of course it landed on Friday the 13th of all lucky days of the year lol. what's next for her lol. hopefully not foreclosure.

[QUOTE=CRGreathouse;225249]You'd hate Ramsey Theory, methinks. :smile:

A good half of sm88's patterns have had basis in fact -- either they've been entirely correct or a small correction makes them right. (Most of the time the hard part is just getting the idea formalized at all: for example, what prediction is being made in the case of the antidiagonals?)

But what I'd really like to see is sm88 learning enough to prove or disprove his own conjectures.[/QUOTE]

well if a pattern could be deduced it would figure out what exponents in a range are possible as Mersenne primes exponents when taken along with they must be prime.

anyone want to help me with [TEX] commands ? I've been playing around but I don't know a list to help me save my butt lol.

[QUOTE=science_man_88;225258]well if a pattern could be deduced it would figure out what exponents in a range are possible as Mersenne primes exponents when taken along with they must be prime.[/QUOTE]

Really, this doesn't help for the reasons I've already discussed.

[QUOTE=science_man_88;225261]anyone want to help me with [TEX] commands ? I've been playing around but I don't know a list to help me save my butt lol.[/QUOTE]

Google is your friend. TeX is easiest to learn one step at a time.

you can start a new thread with tex commands if you want.

[QUOTE=CRGreathouse;225263]Really, this doesn't help for the reasons I've already discussed.



Google is your friend. TeX is easiest to learn one step at a time.[/QUOTE]

yeah that's how i figured out what i know lol any good site for it ?

[QUOTE=science_man_88;225264]you can start a new thread with tex commands if you want.[/QUOTE]

Look [url=http://www.mersenneforum.org/showthread.php?t=11884]this thread[/url] and post #29!

found a link to a large set of the math symbols god how do you keep them apart lol seems like millions lol.

I think this forum has this formula.

crank[TEX] \equiv [/TEX]thinker but, [TEX]\therefore[/TEX] mathematician [TEX]\equiv[/TEX] crank + action

[QUOTE=science_man_88;225275]found a link to a large set of the math symbols god how do you keep them apart lol seems like millions lol.[/QUOTE]

You learn them one at a time, forgetting most of what you learn but keeping what you use most often.

Any person who doesn't get frustrated with TeX every now and then either (1) doesn't use it much, or (2) is named Don Knuth.

[QUOTE=CRGreathouse;225276]You learn them one at a time, forgetting most of what you learn but keeping what you use most often.

Any person who doesn't get frustrated with TeX every now and then either (1) doesn't use it much, or (2) is named Don Knuth.[/QUOTE]

bet you can even code that in TeX lol

myself outside of the Greek letters I'm almost lost on most of their meanings lol.

I like I know [TEX]\sum[/TEX] but as I've proved I wouldn't understand all the notation around it anyone care to help in this regard ?

though I've heard of integrals I've never taken calculus so I had no idea what they mean't until I just looked it up.

figured out some of it by looking at Wikipedia wow math is hard to read no wonder people read the worded version more than the equations (though this is a problem that creates misunderstandings between higher and lower mathematicians etc., anyone on this forum knows that,) lol.

[QUOTE=science_man_88;225285] math is hard to read no wonder people read the worded version more than the equations lol.[/QUOTE]

No! Equations are 'easier' to understand because words can interpreted in different ways, especially the person who tries to explain something with his own words but another person even understand something totally different!

[QUOTE=kar_bon;225290]No! Equations are 'easier' to understand because words can interpreted in different ways, especially the person who tries to explain something with his own words but another person even understand something totally different![/QUOTE]

I submit this thread as an example....

[QUOTE=kar_bon;225290]No! Equations are 'easier' to understand because words can interpreted in different ways, especially the person who tries to explain something with his own words but another person even understand something totally different![/QUOTE]

until a second ago by looking on Wikipedia about pi I didn't know what all the symbols around [TEX]\sum[/TEX] mean't.

since I know how to word the formula they talked about I now understand it.

still can't TeX code it though.

[URL="http://upload.wikimedia.org/math/4/9/0/4903aa53b3b3e348a2bf36359d9b2d27.png"]http://upload.wikimedia.org/math/4/9/0/4903aa53b3b3e348a2bf36359d9b2d27.png[/URL]

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)

this is probably already done but I think:

[TEX]M(x)= \sum_{n=1}^{x-1} + 5[/TEX]

not sure though.

[B]Edit:[/B] [TEX]M(x)= \sum_{n=1}^{x-1} M(n)+ 5[/TEX] or [TEX]M(x)= \sum_{n=0}^{x-1} M(n)+ 5[/TEX]

another edit x>3

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

[QUOTE=3.14159;225306]20004 * 901! + 1 is prime! (≈2280-2285 digits)[/QUOTE]

2278 digits!

I should say all that thinking came up to:

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

[QUOTE=Karsten]2278 digits!
[/QUOTE]

You still haven't found me a solution to n * 71[sup]6[/sup] = 1 mod 1124000727777607680000.

[QUOTE=3.14159;225314]You still haven't found me a solution to:

[tex]$71^6 \equiv 1(mod 22!)$[/tex][/QUOTE]

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.

[QUOTE=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.[/QUOTE]

Nope. n * 71[sup]6[/sup] = 1 mod 1124000727777607680000.

[QUOTE=3.14159;225317]Nope. n * 71[sup]6[/sup] = 1 mod 1124000727777607680000.[/QUOTE]

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

[QUOTE=science_man_88]I bow out I checked again to n=100 million and found nothing.
[/QUOTE]

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

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

?

if so who came up with it.

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

[QUOTE=3.14159;225302]Find a number of the form (k * 1124000727777607680000 + 1) which is divisible by 71[sup]6[/sup].[/QUOTE]

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

so 41455616551037280586874880000001, 185440428906759455741024993280001, ...

If you wanted k to be a prime, then
[code]forstep(n=36882197250,1e15,128100283921,if(isprime(n),return(n)))[/code]
takes 0ms to come up with the smallest example, 1702185888223.

[QUOTE=3.14159;225319]And you would have continued checking for another few thousand years, because it's in the 10^20 range.[/QUOTE]

I must be interpreting you the wrong way, then, since I find 17 million instances below 10[SUP]20[/SUP]:
[code]s=0;forstep(n=36882197250,1e20,128100283921,if(ispseudoprime(n),s++));s
time = 24mn, 50,380 ms.
%1 = 17584374[/code]

[QUOTE=CRGreathouse]I must be interpreting you the wrong way, then, since I find 17 million instances below 10[sup]20[/sup]:
[/QUOTE]

No, sm88 was searching from 1 to 10[sup]20[/sup]

Also: I can rig the number to be semiprime: Here's an example:

[code]55762113159666605738616781764882446662982216262889524965004140441546022418664991294967299112901478710275915859674894915215729565500652173245752458035579997432840192000000000000000000000001 = 5846650401949673788584832661 * 9537446114628590934543506256858833034627998057578048449114232547066222757945489740614277575764557950018685132240750168209560234946900002547179019523457586506941[/code]

It's rather cool what you can do with modular exponentiation.

Here's a better example:
[code]289508238708400855287326405336918749133423298889334502223156944093968*150!+1 = 10254288128120939058918287267226262277004901802104088101537671807691 * 1613053686056875801829683444255478970299912076335944456541386114240383845905441917746506601556683535279149882320681534772255292138166095487502436596518188053218548567074676625073642723441629113007713095460989964369203609983928686506798110902140688105800120704029411[/code]

See? I just made a record! :wink: (Imagine someone was in the middle of factoring that one.)

[QUOTE=3.14159;225319]And you would have continued checking for another few thousand years, because it's in the 10^20 range.[/QUOTE]

He would have hit an instance in a few hours, and a prime instance in a few days, if I calculate correctly. Of course they wouldn't be close to 10^20.

[CODE]M(x)= a=0;for(n=0,x-1,a=a+M(n));a=a+x;print(a",")[/CODE]

best i can do in coding my equation.

[QUOTE=CRGreathouse]If you wanted k to be a prime, then[/QUOTE]

Ah: But if you want the cofactor to be prime:

[code]forstep(n=1,a,b,if(isprime((n*22!+1)/b), print(n)))[/code]

Here's an example:
2386628800 * 60! + 1 = 89[sup]4[/sup] * 316518937978282424919538194794587561257209063707380372243938125981964972473329411361

Ex 2:
1974103215468915131950479405951082163*66!+1 = 71[sup]6[/sup] * 89[sup]4[/sup] * 101[sup]8[/sup] *
12347042307394672596544859365776765192013890088140194548358364707496516846616209505096123905841

I know for n>2 MP(n) = 4^k*MP(n-1)+(4^k-1) the hard part is how to define k.

only problem with my codes is deep recursion I think it's called.

I think if i change it to vectors then ask for the sum of the indexes it might work. but it's never a guarantee with me lol.

[QUOTE=science_man_88;225369][CODE]M(x)= a=0;for(n=0,x-1,a=a+M(n));a=a+x;print(a",")[/CODE]

best i can do in coding my equation.[/QUOTE]

Did you mean to return the a, rather than printing it?

[QUOTE=science_man_88;225374]only problem with my codes is deep recursion I think it's called.[/QUOTE]

I'm not sure if the program does what it's intended. If so, then the answer is what you hint at below: memoize the result.

[QUOTE=science_man_88;225375]I think if i change it to vectors then ask for the sum of the indexes it might work. but it's never a guarantee with me lol.[/QUOTE]

[QUOTE=CRGreathouse;225393]Did you mean to return the a, rather than printing it?[/QUOTE]

I've tried both neither give a correct answer for me.

[QUOTE=science_man_88;225395]I've tried both neither give a correct answer for me.[/QUOTE]

I can't correct the program because I don't know what it's trying to do. But certainly the program isn't doing what you intend if it stores the value it returns but doesn't actually return a value.

they both work until M(1) but after that they climb to 35 and 70 for the next ones.

[QUOTE=CRGreathouse;225396]I can't correct the program because I don't know what it's trying to do. But certainly the program isn't doing what you intend if it stores the value it returns but doesn't actually return a value.[/QUOTE]

it's doing the equivalent of the formula I gave above.

[URL="http://www.mersenneforum.org/cgi-bin/mimetex.cgi?M(x)=\sum_{n=0}^{x-1}M(n)%20+%20x"]http://www.mersenneforum.org/cgi-bin/mimetex.cgi?M(x)=\sum_{n=0}^{x-1}M(n)%20+%20x[/URL]

Is x in the sum or out of the sum?

[QUOTE=CRGreathouse;225400]Is x in the sum or out of the sum?[/QUOTE]

out I know I wasn't sure how to make it obvious. sum all M(n) less than M(x) then add x to that sum.

I realize it should of been first now lol.

[QUOTE=science_man_88;225401]out I know I wasn't sure how to make it obvious.[/QUOTE]

[TEX]M(x)=x+\sum_{n=0}^{x-1}M(n)[/TEX] would be the usual way. Of course you could also use parentheses (in which case, in LaTeX, you would want to use \left and \right).

[code]M(x)=x+sum(n=0,x-1,M(n))[/code] is the literal translation into Pari.

Here's a solution with memoization. It stores the values rather than recalculating:
[code]M(x)={
if(x < 1, return(0));
if(Mvec == 'Mvec,Mvec=[]);
if(#Mvec < x,
my(n=#Mvec);
Mvec=vector(x,i,if(i<=n,Mvec[i]));
for(i=n+1,x, Mvec[i]=i+sum(j=1,i-1,Mvec[j]))
);
Mvec[x]
};[/code]

Here's a more efficient version:
[code]M(x)=(1<<x)-1[/code]

[QUOTE=CRGreathouse;225402][TEX]M(x)=x+\sum_{n=0}^{x-1}M(n)[/TEX] would be the usual way. Of course you could also use parentheses (in which case, in LaTeX, you would want to use \left and \right).

[code]M(x)=x+sum(n=0,x-1,M(n))[/code] is the literal translation into Pari.

Here's a solution with memoization. It stores the values rather than recalculating:
[code]M(x)={
if(x < 1, return(0));
if(Mvec == 'Mvec,Mvec=[]);
if(#Mvec < x,
my(n=#Mvec);
Mvec=vector(x,i,if(i<=n,Mvec[i]));
for(i=n+1,x, Mvec[i]=i+sum(j=1,i-1,Mvec[j]))
);
Mvec[x]
};[/code]

Here's a more efficient version:
[code]M(x)=(1<<x)-1[/code][/QUOTE]

memorization* and memorize* for the last few posts but thank's and I have proved it works yeah. is there a formula for the number list that contains the super-perfect numbers and perfect numbers as well like mine ?

[QUOTE=science_man_88;225403]memorization* and memorize*[/QUOTE]

No, memoization and memoize. Different words.

[QUOTE=science_man_88;225403]is there a formula for the number list that contains the super-perfect numbers and perfect numbers as well like mine ?[/QUOTE]

Not sure what you mean.

[TEX]M(x)=x+\sum_{n=0}^{x-1}M(n)[/TEX]

is there a similar formula for the sequences that hold super-perfect and perfect numbers respectively?

[QUOTE=science_man_88;225407][TEX]M(x)=x+\sum_{n=0}^{x-1}M(n)[/TEX]

is there a similar formula for the sequences that hold super-perfect and perfect numbers respectively?[/QUOTE]

Yes, but what sequences? Certainly they're in [url=http://oeis.org/classic/A000027]A000027[/url], but did you have a more specific sequence in mind? I don't know of an explicit formula for the (super-)perfect numbers, of course -- though I wouldn't be surprised if one could be formed with sin and floor. :smile:

[QUOTE=science_man_88;225403]memorization* and memorize* for the last few posts[/QUOTE]

[url]http://en.wikipedia.org/wiki/Memoization[/url]

[QUOTE=axn;225409][url]http://en.wikipedia.org/wiki/Memoization[/url][/QUOTE]

Thanks, axn. I tend to forget that not everyone is a computer programmer. :blush:

[QUOTE=CRGreathouse;225410]Thanks, axn. I tend to forget that not everyone is a computer programmer. :blush:[/QUOTE]

I program a bit (pari only recently) just never heard the term.

A006516 and A019279 is there a similar way to get these if so once we find a way to get one of the three to work only for the correct indexes we can apply it to the other 2.

[QUOTE=science_man_88;225411]I program a bit (pari only recently) just never heard the term.[/QUOTE]

Maybe I should have used the term dynamic programming instead? It just seemed a little highfalutin when all I meant was "don't redo what you already did".

[QUOTE=science_man_88;225412]A006516 and A019279 is there a similar way to get these if so once we find a way to get one of the three to work only for the correct indexes we can apply it to the other 2.[/QUOTE]

[url=http://oeis.org/classic/A006516]A006516[/url] is 2[SUP]2n-1[/SUP] - 2[SUP]n-1[/SUP], so let's look at
[TEX]\sum_{n=1}^{x-1}2^{2n-1}-2^{n-1}=\sum_{n=1}^{x-1}2^{2n-1}-\sum_{n=1}^{x-1}2^{n-1}=1/2\left(\sum_{n=1}^{x-1}4^n-\sum_{n=1}^{x-1}2^n\right)=1/2\left(\frac{4^x-4}{4-1}-\frac{2^x-2}{2-1}\right)=\frac{2^{2x-1}-2}{3}-\frac{2^{x-1}-1}{1}[/TEX]

Can you take this the rest of the way?

[TEX]\frac{2^{2x-1}-2}{3}-\frac{2^{x-1}-1}{1}[/TEX]

3 times the first gives:

[TEX]2^{2x-1}-2[/TEX] add 1 and you get [TEX]2^{p=2x-1}-1[/TEX]


so both can become Mersenne numbers.

the second term's numerator could be turned into a mersenne prime but I doubt that helps the other numerator or fraction.

[QUOTE=science_man_88;225416]the second term's numerator could be turned into a mersenne prime but I doubt that helps the other numerator or fraction.[/QUOTE]

You should be paying attention to the first term, I think, since that's about one-third of the number you want.

[QUOTE=science_man_88;225416]the second term's numerator could be turned into a mersenne prime but I doubt that helps the other numerator or fraction.[/QUOTE]

I'm an idiot the second term's numerator is a Mersenne where p=x-1


times the first by 3 and add 1 to the -2 to get -1 and we get another Mersenne with p=2x-1. so these can be transformed to Mersenne numbers since we have an equation maybe it's time to apply it.

[QUOTE=science_man_88;225420]times the first by 3 and add 1 to the -2 to get -1 and we get another Mersenne with p=2x-1. so these can be transformed to Mersenne numbers since we have an equation maybe it's time to apply it.[/QUOTE]

Write out the full equation first and make sure it's right.

[TEX]\frac{2^{2x-1}-2}{3}-\frac{2^{x-1}-1}{1}[/TEX]

3 times the first gives:

[TEX]2^{2x-1}-2[/TEX] add 1 and you get [TEX]2^{p=2x-1}-1[/TEX]


so both can become Mersenne numbers

I meant with the sum. [TEX]2^{2x-1}-1=2^{2x-1}-1[/TEX] is a boring equation.

[QUOTE=CRGreathouse;225423]I meant with the sum. [TEX]2^{2x-1}-1=2^{2x-1}-1[/TEX] is a boring equation.[/QUOTE]

actually my idea would be [TEX](M(2x-1)-1)/3 -M(x-1)[/TEX] since we know how to calculate any M(x) we can turn this into a sum so if we found one of these using it we could technically link it to a higher Mersenne by the looks of it anyone want to go Mersenne hunting ?

unless we find rules to follow we wouldn't prove primality though.

[QUOTE=science_man_88;225427]actually my idea would be [TEX](M(2x-1)-1)/3 -M(x-1)[/TEX] since we know how to calculate any M(x) we can turn this into a sum so if we found one of these using it we could technically link it to a higher Mersenne by the looks of it anyone want to go Mersenne hunting ?[/QUOTE]

[TEX]((M(2x-1)-1)/3) -M(x-1)[/TEX] sorry.

since we know perfect numbers are a subsequence could we use these rules maybe to predict Mersenne primes ?

[QUOTE=science_man_88;225430]since we know perfect numbers are a subsequence could we use these rules maybe to predict Mersenne primes ?[/QUOTE]

No.

[QUOTE=science_man_88]since we know perfect numbers are a subsequence could we use these rules maybe to predict Mersenne primes ?
[/QUOTE]

No, because perfect numbers are directly divisible by prime Mersenne numbers. You would require knowledge of the next prime Mersenne number to get the next perfect number.

@CRG: Have you run the application?

P.S: Try this: Factor 105813801920266629764526053870673344785675332494211872818049070362444304517536153600000001.

[QUOTE=3.14159;225435]@CRG: Have you run the application?[/QUOTE]

No. You?

[TEX]((M(2x-1)-1)/3) -M(x-1)[/TEX]

[TEX]M(x)=x+\sum_{n=0}^{x-1}M(n)[/TEX]

so if x is replaced by 2x-1 we get

[TEX]M(2x-1)=(2x-1)+\sum_{n=0}^{(2x-1)-1}M(n)[/TEX]

with x-1 instead we get.

[TEX]M(x-1)=(x-1)+\sum_{n=0}^{(x-1)-1}M(n)[/TEX]

so [TEX]((M(2x-1)-1)/3) -M(x-1)[/TEX] becomes:

[TEX]((((2x-1)+\sum_{n=0}^{(2x-1)-1}M(n))-1)/3) -((x-1)+\sum_{n=0}^{(x-1)-1}M(n))[/TEX] if I did the pasting correctly.

[QUOTE=3.14159;225434]No, because perfect numbers are directly divisible by prime Mersenne numbers. You would require knowledge of the next prime Mersenne number to get the next perfect number.[/QUOTE]

I'm not saying that what i'm saying is perfect(x) = mersenne prime(x)*super-perfect (x)

since we have x in these and our formula and since we have M(2x-1) in our equation can we come up with rules for this relation such that M(2x-1) is prime based on the equation.

The number 105813801920266629764526053870673344785675332494211872818049070362444304517536153600000001 is up for grabs! It's a 90-digit number that will only take about an hour of work!

Okay, just kidding about the above: 9841985839884679351627206961366788794469997 * 10751265409411163373749309461564364987789503333

Also: [QUOTE=CRGreathouse]No. You?
[/QUOTE]

Why not?

[QUOTE=science_man_88]since we have x in these and our formula and since we have M(2x-1) in our equation can we come up with rules for this relation such that M(2x-1) is prime based on the equation.
[/QUOTE]

Mersenne numbers do not have any patterns, nor covering divisors. Primes do not follow any patterns besides becoming less common among the larger integers, and are randomly distributed. There is no way to compute the next Mersenne. Simply no way around it.

If it really works as you claim:

I'll offer you a known example: 2[sup]127[/sup] - 1.

I wish for you to find 2[sup]127[/sup] - 1 with all the snippets you posted there.

[QUOTE=3.14159;225440]Why not?[/QUOTE]

Not running Windows at the moment.

[QUOTE=3.14159;225440]Mersenne numbers do not have any patterns, nor covering divisors.[/QUOTE]

To be fair, we haven't proved this. Practically speaking, though, no one doubts it.

[QUOTE=CRGreathouse]Not running Windows at the moment.
[/QUOTE]

HAHAHA. A Mac fainter.

[QUOTE=CRGreathouse]To be fair, we haven't proved this. Practically speaking, though, no one doubts it.
[/QUOTE]

It's unproven? So it's still an open question?

doubt you care but my first name is Roderick now I have to prove it correct as it means famous power.

[QUOTE=3.14159;225444]HAHAHA. A Mac fainter.[/QUOTE]

:yucky: I don't like Macs, especially since I have to maintain some (just a few) at work.

No, this computer is running Linux. Much of the math stuff I use (gp2c, François Morain's ECPP) doesn't work on Windows.

[QUOTE=3.14159;225444]It's unproven? So it's still an open question?[/QUOTE]

Make a formal statement and I'll tell you what I can.

[QUOTE=science_man_88;225445]doubt you care but my first name is Roderick[/QUOTE]

Well met, Roderick. I'm Charles.

[QUOTE=science_man_88;225445]now I have to prove it correct as it means famous power.[/QUOTE]

Good luck with that.

[QUOTE=CRGreathouse;225447]Well met, Roderick. I'm Charles.



Good luck with that.[/QUOTE]



yeah I do have a royal link but I'm not that famous or powerful lol. though I've bugged the OEIS, NASA, the prince of wales(21st cousin ?), obama, harper, my local government(Nova Scotia's Darrell dexter), yeah I found you all over the OEIS lol I figured it out. I only have one thing that got into it so far... [url]http://www.research.att.com/~njas/sequences/A146768[/url]

forgot I also bugged discoverychannel. I'm likely more famous for bugging people than finding anything lol.

[QUOTE=CRGreathouse]No, this computer is running Linux. Much of the math stuff I use (gp2c, François Morain's ECPP) doesn't work on Windows.
[/QUOTE]

Dammit. I wondered why I couldn't get it running. Well, I still have PARI/GP/etc.

[QUOTE=3.14159;225450]Dammit. I wondered why I couldn't get it running. Well, I still have PARI/GP/etc.[/QUOTE]

technically if we could make an gp2c like compile code in Pari it should work for any OS lol.

[QUOTE=3.14159;225450]Dammit. I wondered why I couldn't get it running. Well, I still have PARI/GP/etc.[/QUOTE]

[QUOTE=science_man_88;225451]technically if we could make an gp2c like compile code in Pari it should work for any OS lol.[/QUOTE]

Correct. It's hard to make it compile, though. It can be done -- but I've tried and failed, while it's easy to do in Linux.

Morain's ECPP, though, can be used only on Linux.

[QUOTE=science_man_88;225449]yeah I do have a royal link but I'm not that famous or powerful lol. though I've bugged the OEIS, NASA, the prince of wales(21st cousin ?), obama, harper, my local government(Nova Scotia's Darrell dexter), yeah I found you all over the OEIS lol I figured it out. I only have one thing that got into it so far... [/QUOTE]

I ran into the Prince of Wales when I was over there.

[QUOTE=CRGreathouse;225453]I ran into the Prince of Wales when I was over there.[/QUOTE]

yeah my line according to what we've found is around the time of Robert I Stewart and his daughter Margaret Stewart. which if I found correct Gedcom I got back to Priam of Troy and then back to Adam and Eve (suspect that one of unproven information). we should get back to math and code lol.

[QUOTE=science_man_88;225454](suspect that one of unproven information)[/QUOTE]

So's most of genealogy, frankly.

[QUOTE=science_man_88;225454]we should get back to math and code lol.[/QUOTE]

Do you know where the link on [url]http://oeis.org/classic/A146768[/url] went? It's dead now:
[url]http://www.devalco.de/quadr_Sieb_2x%5E2-1.htm[/url]