[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.