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-   -   The "one billion minus 999,994,000" digits prime number (https://www.mersenneforum.org/showthread.php?t=20568)

a1call 2015-11-09 20:32

[QUOTE=danaj;415567]There is a lot of pushback, and it's rarely gentle. Especially if the subforum has "Math" in the title, which usually means one is expected to have more rigor and background. This forum in general tends to be more focused toward a different audience than many others -- much less beginner level. It's also hurt by starting out wanting to claim the billion digit EFF prize, which immediately sets lots of crank alarm bells ringing. I read some other math forums and they can get quickly overwhelmed by a few people who keep insisting they have rudimentary proofs of Beal's conjecture or FLT, and pretty soon the whole forum becomes useless as it is inundated with daily gibberish. The regulars really don't want that to happen here.

I am assuming you're using the PrimeQ in your function because it isn't your final method. You can't rely on PrimeQ in the long run, or you may as well just run a variant of nextprime. For generating large actual primes, you need to use either a special form that has a very fast proof (e.g. Mersenne) or make sure you have the proof step as you build it. Again a flawed analogy, but given the choice when constructing a semi-prime of multiplying two primes vs. factoring composites until we get two prime factors ... it should be obvious which is the better way to do it.[/QUOTE]
Yes, 1st impressions are key. But I was only being honest not arrogant. I tried to be as humble as I could.
I understand your pointers. The few k digit primes started as a means of putting into practice the concept of generating primes using my theorems and evolved into a hobby of bettering my personal bests till I would reach the limitation of the free WDP account. It was not meant to break any other records. PrimeQ was used because I did not know it was just a probability test. NextPrime would not serve any purpose for me. it would be the same as a random prime generation.

At this point my prime 2Do list is to:
Add to the WDP code that I have written, the codes necessary to generate an automatic mathematical expression of the number.
Try to come up with the math for the sum in Theorem 2 to converge to less than the square of the largest factor in general or in particularly large (Billion+ digits).
Failing that to submit it for publication, knowing well that it may be refused.

a1call 2015-11-09 20:36

[QUOTE=Batalov;415591]If not in error? This "[B]p[/B]" doesn't match the infamous 5121 digit "[B]N[/B]" (which you have surprising trouble understanding not being a variable but "your" number), so there must be an error of some sort.

And why these weird contortions in the number's notation?
In general, (AB)![SUP](B)[/SUP] = A! B[SUP]A[/SUP] because it is simply = {AB}{(A-1)B}{(A-2)B}...B
so why not write it like that? For obfuscation?[/QUOTE]
Sorry for the messy code. It was meant for personal use only. I shared it quickly because I was told I am being secretive so I put it out as is. I will add the code to generate the expression automatically, once I get the chance.

a1call 2015-11-09 22:28

[QUOTE=Batalov;415591]If not in error? This "[B]p[/B]" doesn't match the infamous 5121 digit "[B]N[/B]" (which you have surprising trouble understanding not being a variable but "your" number), so there must be an error of some sort.

And why these weird contortions in the number's notation?
In general, (AB)![SUP](B)[/SUP] = A! B[SUP]A[/SUP] because it is simply = {AB}{(A-1)B}{(A-2)B}...B
so why not write it like that? For obfuscation?[/QUOTE]


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

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

[B](61#-1 +42 x 61#)![SUP](61#)[/SUP] + 61# x 49

[/B]The code to generate the expression as well as the number on the WDP is

[CODE]fctrlcntmns1=42;
primorialBase=61;
bsenmbersbtrctr=1;
cntrbase=1;

prmrlpwr=Prime[primorialBase];
Print ["Largest prime = ",Text[Style[prmrlpwr, Bold,FontSize -> 13,Blue]]]
prmrl=1;
Do[prmrl=prmrl*Prime[n], {n, primorialBase}]
Print["Primorial=",Text[Style[prmrl, Bold,FontSize -> 13,Blue]]];
bsenmbr=prmrl-bsenmbersbtrctr;
Print["Base Number ",Text[Style[bsenmbr, Bold,FontSize -> 13,Blue]]," is Prime= ",Text[Style[PrimeQ[bsenmbr], Bold,FontSize -> 13,Blue]]]
fctrl=bsenmbr;
Do[fctrl=fctrl*(bsenmbr+(prmrl*n)), {n, fctrlcntmns1}]
prmcnddte=fctrl-prmrl*prmrlpwr;
multiplier=1;
Catch[Do[
prmcnddte=fctrl+prmrl*l;
multiplier=l;
; If[PrimeQ[prmcnddte], Throw[prmcnddte]]
,{l,cntrbase,prmrlpwr}]]
Print ["Candidate is prime is ",Text[Style[PrimeQ[prmcnddte], Bold,FontSize -> 23,Blue]],"
",Text[Style[prmcnddte, Bold,FontSize -> 13,Blue]]]
Print["integer length=",Text[Style[IntegerLength[prmcnddte], Bold,FontSize -> 19,Blue]]]
Print["Mathematical Expression = (",primorialBase,"#-1 +",fctrlcntmns1," x ",primorialBase,"#",")!^(",primorialBase,"#) + ",primorialBase,"# x ",multiplier]
Print["End of Run"][/CODE]

R.D. Silverman 2015-11-09 23:12

[QUOTE=R.D. Silverman;415571]It would also be better if he would bother to actually respond to some of the queries. In particular, I
asked about the method used for partitioning his set of primes into two subsets such that the
difference between the two products of the primes in each subset satisfied the sqrt condition.
There was, of course, no response, which in turn makes one suspicious that the OP does not
HAVE a response.[/QUOTE]

He still has not responded....What a shock!!!!

BTW, has anyone bothered to investigate the complexity of the arithmetic involved?

We have a very large set of primes S = {2,3,5, ......R} for suitable R depending on the size
of the final prime we are trying to construct. For billion digit primes, R = 10^500000000

Suppose we somehow (miracle?!) partition it into two sets T,U to satisfy the
sqrt condition. How one does this other than a purely exponential combinatorial
search is beyond me. The OP, as with all cranks ignores this question because
he will find the answer to be (shall we say) inconvenient......

However, to satisfy the sqrt condition, #T will need to be very close to #U.
I now suggest that the readers estimate the product of all the primes in T and U.

Hint: PNT. sum of log(p) p < k ~ k. Exponentiate both sides.
EACH of the products will be near e^(10^500000000 / 2) Now subtract........

The OP has presented one of the worst methods that one might imagine!!!!

a1call 2015-11-09 23:16

The general format of the expression is:

[B](n#-1 +m x n#)![SUP](n#)[/SUP] + n# x k

[/B]for positive integers n, m, and k where k <= m

schickel 2015-11-10 04:58

[QUOTE=a1call;415379][B]A 4581 digit probable prime:
[/B]
[CODE]
111887......381502730280953069
[/CODE]
[/QUOTE] [URL="http://factordb.com/index.php?id=1100000000805487893"]:cool:[/URL]

LaurV 2015-11-10 04:59

[QUOTE=R.D. Silverman;415626]He still has not responded....What a shock!!!!

BTW, has anyone bothered to investigate the complexity of the arithmetic involved?

We have a very large set of primes S = {2,3,5, ......R} for suitable R depending on the size
of the final prime we are trying to construct. For billion digit primes, R = 10^500000000
[/QUOTE]
Of course we did, the "R" was circulated by Batalov in the very beginning (post #43, after the "theorem"), and yourself said about the complexity of the split, in the posts that followed. We understand that. Is the OP who doesn't understand. Or he does, but still tries...

I don't think he has any method to show, he just tries random splits and variables, and checks if they give a prime using Wolfram or whatever. Or others here check for him.

[thinking] I could try this too, but I have to think how to put some of my RL daily work in some attractive form to make the people here do my job for me, now [U]that[/U] would be interesting, not proving primality. :chappy:[/thinking]

a1call 2015-11-10 05:14

[QUOTE=LaurV;415651]Of course we did, the "R" was circulated by Batalov in the very beginning (post #43, after the "theorem"), and yourself said about the complexity of the split, in the posts that followed. We understand that. Is the OP who doesn't understand. Or he does, but still tries...

I don't think he has any method to show, he just tries random splits and variables, and checks if they give a prime using Wolfram or whatever. Or others here check for him.

[thinking] I could try this too, but I have to think how to put some of my RL daily work in some attractive form to make the people here do my job for me, now [U]that[/U] would be interesting, not proving primality. :chappy:[/thinking][/QUOTE]
The largest required known prime for constructing the 5121 digit integer was
[B]P[SUB]61[/SUB]=283[/B]

axn 2015-11-10 05:51

[QUOTE=a1call;415654]The largest required known prime for constructing the 5121 digit integer was
[B]P[SUB]61[/SUB]=283[/B][/QUOTE]

And the largest required prime for constructing the largest known prime is 2. Think about it. *All hail GIMPS*

Dubslow 2015-11-10 06:49

[QUOTE=axn;415660]And the largest required prime for constructing the largest known prime is 2. Think about it. *All hail GIMPS*[/QUOTE]

Debatable. We need to know the prime exponent to construct the Mersenne itself.

schickel 2015-11-10 07:11

[QUOTE=a1call;415593]Thank you [B]schickel[/B], You seem to have a very fast, idle supercomputer. I owe you quite a few CPU clock pulses. I will pay you back if I ever get quantum computer.:smile:

But joking aside, there is a at least one layman member on this board which shall remain anonymous. He might not understand the code. For his sake is the 5121 digit integer a prime or a composite?
Thank you again for your and your computer's time.[/QUOTE]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.


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