The "one billion minus 999,994,000" digits prime number

[Batalov]

Originally Posted by a1call

Thank you for editing the title. It bothered me too.

The correct and auto generated mathematical expression for the 5121 digit prime(?) number is:

(61#-1 +42 x 61#)!^(61#) + 61# x 49

This is still wrong, because you'll need to look up what 61# really means.
I'd already given up on getting the right answer from you, when I saw from your later post that you think that "61#" means p₆₁# .
So, for everyone's benefit, what N really was is, after all

N=(283#-1)*(2*283#-1)*(3*283#-1)*(4*283#-1)*(5*283#-1)*(6*283#-1)*(7*283#-1)*(8*283#-1)*(9*283#-1)*(10*283#-1)*(11*283#-1)*(12*283#-1)*(13*283#-1)*(14*283#-1)*(15*283#-1)*(16*283#-1)*(17*283#-1)*(18*283#-1)*(19*283#-1)*(20*283#-1)*(21*283#-1)*(22*283#-1)*(23*283#-1)*(24*283#-1)*(25*283#-1)*(26*283#-1)*(27*283#-1)*(28*283#-1)*(29*283#-1)*(30*283#-1)*(31*283#-1)*(32*283#-1)*(33*283#-1)*(34*283#-1)*(35*283#-1)*(36*283#-1)*(37*283#-1)*(38*283#-1)*(39*283#-1)*(40*283#-1)*(41*283#-1)*(42*283#-1)*(43*283#-1)+49*283#
and 
1088*N^3+1 is prime (because N was proven by Schickel)

Well, (283#-1)*(2*283#-1)*(3*283#-1)*(4*283#-1)*(5*283#-1)*(6*283#-1)*(7*283#-1)*(8*283#-1)*(9*283#-1)*(10*283#-1)*(11*283#-1)*(12*283#-1)*(13*283#-1)*(14*283#-1)*(15*283#-1)*(16*283#-1)*(17*283#-1)*(18*283#-1)*(19*283#-1)*(20*283#-1)*(21*283#-1)*(22*283#-1)*(23*283#-1)*(24*283#-1)*(25*283#-1)*(26*283#-1)*(27*283#-1)*(28*283#-1)*(29*283#-1)*(30*283#-1)*(31*283#-1)*(32*283#-1)*(33*283#-1)*(34*283#-1)*(35*283#-1)*(36*283#-1)*(37*283#-1)*(38*283#-1)*(39*283#-1)*(40*283#-1)*(41*283#-1)*(42*283#-1)*(43*283#-1)+2310*283# is a PRP, too, and (gasp!) look at that - 2310 is 11#. Isn't that a miracle?
...No, it isn't - this is just a coincidence. (and now Frank has another prime to prove, as he seems to enjoy doing them... ;-)

[R.D. Silverman]

Originally Posted by LaurV

Of course we did, the "R" was circulated by Batalov in the very beginning

This is true. But this does not answer my question about the complexity of the arithmetic involved.

(post #43, after the "theorem"), and yourself said about the complexity of the split, in the posts that followed.

I suggested a method. But I do not know whether there might be better. I suspect that
the problem is equivalent to the cutting stock problem in O.R. (or the subset sum problem; take logs
to turn it into a summation problem). These are NPC.

However, I have seen no DISCUSSION.

[a1call]

Originally Posted by schickel

No supercomputer needed. Just an AMD X6 and a couple of hours. The other one will be done by tomorrow.

ETA: Ooops! Looks like I messed up the links earlier. I thought I had done the 5121 but I didn't. Maybe start tomorrow unless someone else beats me to it.

It's very kind of you regardless schickel, Thank you.

I can not do the runs myself anytime soon.

[a1call]

Originally Posted by Batalov

This is still wrong, because you'll need to look up what 61# really means.
I'd already given up on getting the right answer from you, when I saw from your later post that you think that "61#" means p₆₁# .
So, for everyone's benefit, what N really was is, after all

N=(283#-1)*(2*283#-1)*(3*283#-1)*(4*283#-1)*(5*283#-1)*(6*283#-1)*(7*283#-1)*(8*283#-1)*(9*283#-1)*(10*283#-1)*(11*283#-1)*(12*283#-1)*(13*283#-1)*(14*283#-1)*(15*283#-1)*(16*283#-1)*(17*283#-1)*(18*283#-1)*(19*283#-1)*(20*283#-1)*(21*283#-1)*(22*283#-1)*(23*283#-1)*(24*283#-1)*(25*283#-1)*(26*283#-1)*(27*283#-1)*(28*283#-1)*(29*283#-1)*(30*283#-1)*(31*283#-1)*(32*283#-1)*(33*283#-1)*(34*283#-1)*(35*283#-1)*(36*283#-1)*(37*283#-1)*(38*283#-1)*(39*283#-1)*(40*283#-1)*(41*283#-1)*(42*283#-1)*(43*283#-1)+49*283#
and 
1088*N^3+1 is prime (because N was proven by Schickel)

Well, (283#-1)*(2*283#-1)*(3*283#-1)*(4*283#-1)*(5*283#-1)*(6*283#-1)*(7*283#-1)*(8*283#-1)*(9*283#-1)*(10*283#-1)*(11*283#-1)*(12*283#-1)*(13*283#-1)*(14*283#-1)*(15*283#-1)*(16*283#-1)*(17*283#-1)*(18*283#-1)*(19*283#-1)*(20*283#-1)*(21*283#-1)*(22*283#-1)*(23*283#-1)*(24*283#-1)*(25*283#-1)*(26*283#-1)*(27*283#-1)*(28*283#-1)*(29*283#-1)*(30*283#-1)*(31*283#-1)*(32*283#-1)*(33*283#-1)*(34*283#-1)*(35*283#-1)*(36*283#-1)*(37*283#-1)*(38*283#-1)*(39*283#-1)*(40*283#-1)*(41*283#-1)*(42*283#-1)*(43*283#-1)+2310*283# is a PRP, too, and (gasp!) look at that - 2310 is 11#. Isn't that a miracle?
...No, it isn't - this is just a coincidence. (and now Frank has another prime to prove, as he seems to enjoy doing them... ;-)

Thank you for the correction? Batalov,
I think I was absent the day the teacher talked about that.
I have to cut down on my prime fighting hours so here is another try without fixing the code itself.

Would this compute?
(P₆₁#-1 +42 x P₆₁#)!^{([B][B]P[/B][SUB]61[/SUB][B]#[/B])[/B]} + P₆₁# x 49

And The general format of the expression is:

(P_n#-1 +m x P_n#)!^{([B][B]P[/B][SUB]n[/SUB][B]#[/B])[/B]} + P_n# x k

for positive integers n, m, and k where k <= P_n

I would beg to differ on the coincidence claim though. I wouldn't expect that from a mathematician that you seem to be. Think about the statistics of it.

Thank you for the reply.

[VBCurtis]

"think about the statistics of it"? Are you saying it *has* to be prime because of some property Serge didn't notice, or that it's merely *likely* to be prime due to its form? If the latter, this method of yours is utterly worthless; we have tons of forms that are "likely" to be prime but still need to be tested for primality.

If the former, you best get about proving how you KNOW it is prime without asking wolfram.

[a1call]

Originally Posted by VBCurtis

"think about the statistics of it"? Are you saying it *has* to be prime because of some property Serge didn't notice, or that it's merely *likely* to be prime due to its form? If the latter, this method of yours is utterly worthless; we have tons of forms that are "likely" to be prime but still need to be tested for primality.

If the former, you best get about proving how you KNOW it is prime without asking wolfram.

I think I will just quote from my previous posts sometimes rather than repeating things over and over.

Originally Posted by a1call

A theorem such as theorem 1 is of value for finding primes even when it does not converge to less than the square since it can be used to construct routines which yield primes with astronomically higher probability than random trials. For example odd numbers are twice as likely to be primes than integers in general. Factor out divisible by 3s and you improve your odds substantially more.
....
My Theorem 1 and to a greater degree Theorem 2 (undisclosed so far) has a much more probable solution.

[VBCurtis]

Well, then your method is utterly useless for finding large primes, since the form of your numbers doesn't allow for any of the special primality tests that run quickly. You should look into what the largest proven-by-ECPP prime is on the top 5000 website (primes.utm.edu), and also how long the proofs of those records took.

You might also enjoy quantifying just how much more likely your "astronomically more likely" claim is. If your method produces a list of numbers of, say, 10k digits of which 1% are expected to be prime, is that useful? Compare to what a sieve to 1e12 leaves for such a list; specifically, if I sieve a large pool of 10k digit numbers to 1e12, what percentage of remaining candidates should be prime? How do the candidates-per-expected-prime change at 100k digits for your method vs a sieve?

It sounds like all you've done is think up an expression for which numbers are merely more likely to be prime than a sieved list of candidate numbers. While possibly interesting as a curiosity if it's vastly more likely, the lack of provability for candidates even half the size of 5000th place on the top 5000 renders it useless as a way of finding large primes.

[science_man_88]

Originally Posted by a1call

And The general format of the expression is:

(P_n#-1 +m x P_n#)!^{([B][B]P[/B][SUB]n[/SUB][B]#[/B])[/B]} + P_n# x k

for positive integers n, m, and k where k <= P_n

a rearrangement on each side turns this into (((m+1)*{{P_n}#}-1)!) ^{{P_n}#} +{{P_n}#} * k = a*{{P_n}#} +k*{{P_n}#} doesn't it ? edit: my math even with edits doesn't show up so I am removing the Tex tags

[a1call]

Originally Posted by VBCurtis

Well, then your method is utterly useless for finding large primes, since the form of your numbers doesn't allow for any of the special primality tests that run quickly. You should look into what the largest proven-by-ECPP prime is on the top 5000 website (primes.utm.edu), and also how long the proofs of those records took.

You might also enjoy quantifying just how much more likely your "astronomically more likely" claim is. If your method produces a list of numbers of, say, 10k digits of which 1% are expected to be prime, is that useful? Compare to what a sieve to 1e12 leaves for such a list; specifically, if I sieve a large pool of 10k digit numbers to 1e12, what percentage of remaining candidates should be prime? How do the candidates-per-expected-prime change at 100k digits for your method vs a sieve?

It sounds like all you've done is think up an expression for which numbers are merely more likely to be prime than a sieved list of candidate numbers. While possibly interesting as a curiosity if it's vastly more likely, the lack of provability for candidates even half the size of 5000th place on the top 5000 renders it useless as a way of finding large primes.

All very good points. Thank you.
But please try to go through my numerous previous posts in this thread explaining reportedly this is not an attempt at finding a record breaking prime. I have already pointed out why I went through the exercise that I did not that long ago.
My ultimate goal is to have a sum converge to less than the square of the largest factor of all the addends. I do not have that as of yet. But I think it is feasible to be formulated which would be the proof of primality for these types of primes. Through all the posts and insults, I have yet to see any valid argument as to why such a formulation can not exist.

[CRGreathouse]

So as far as I can tell, the idea is to construct a big number B and a small number s and check B + s, B + 2s, B + 3s, ... until a prime is found.

Originally Posted by a1call

My ultimate goal is to have a sum converge to less than the square of the largest factor of all the addends. [...] I have yet to see any valid argument as to why such a formulation can not exist.

Do you mean that you expect $(a\cdot p\#+b)^c+d$ to be less than $p^2$ for judicious choices of a, b, c, and d?

[a1call]

Originally Posted by CRGreathouse

So as far as I can tell, the idea is to construct a big number B and a small number s and check B + s, B + 2s, B + 3s, ... until a prime is found.



Do you mean that you expect $(a\cdot p\#+b)^c+d$ to be less than $p^2$ for judicious choices of a, b, c, and d?

Not quite even though I am lost in the 2nd half of your post.

The idea is to find a sum made of addends similar to my post 39 which are co-prime (for Theorem 1, not so for Theorem 2). The truncated sum (no need to evaluate all digits) can be determined to be smaller than the (largest-factor)^2 (no need to determine the largest prime factor).

please let me know if that makes sense.