(Reble)Mersenne. p=80,812,807 divides 2^13,468,801-1.

Sun Apr 20 14:34:54 CEST 2003

prime p=80,812,807 divides  Mersenne number, M=2^13,468,801 -1 ..
centiseconds = 15 (k= 6, 10LOGp= 79),

I found p= 6*13.4mill +1 ~= 80.8mill.

n-1 (exponent minus 1) = 13468800 has many divisors , below;
>   .FermatMers.Mersenne.Dec2001.L13468801 // spoolfact

IF a = order of 2 Mod p
i.e. 2^a == 1 mod 80,812,807
THEN (p ~= 80.8million)
does not divide M = 2^(a.b+1)-1.
==2*1-1=1.

see > .Calc.Profile.eisintegsq.Seqfan.2003k143p1

Don Reble says:   20.04.03  23:36

>Testing further terms of A051453 is straightforward but slow,
since the   > numbers get big quickly.
> 
> Once one suspects that an LCM[]+1 is prime (say, by doing strong-
> pseudoprime tests), one should then prove it's prime. It's tempting to
> try the "N-1" method, of Lehmer and Selfridge.
> 
>     If, for each prime factor P_k of N-1, there is a number A_k such that
>         (A_k)^[(N-1)/P_k] /= 1 mod N
>     and
>         (A_k)^[N-1] = 1 mod N
>     then N is prime.

Hello Don R.,

is this usable for Mersenne numbers, do you think? Please.
Perhaps to prove a factor. We only have to do prime factors of
(exponent minus 1.)?
oh dear! I cannot understand my (own old reasoning.)
I am still attempting.

>
> Essentially, one finds a (P_k)-power non-residue, for each P_k factor.
> One must first factor N-1: but that's easy for those LCM[]+1 numbers.
> 
> Each small prime divides the LCM, so one needs to find a quadratic
> non-residue, a cubic non-residue, a pentic non-residue, etc. One way is
> just to check each small number (2,3,...) in turn to see whether it's a
> non-residue; but the least non-residue must be a prime, so one checks
> only those. The process is expected to be quick, since 1/2 of the
> numbers are quadratic non-residues, 2/3rds are cubic non-residues,
> 4/5ths are pentic....
>

CALCULATE ALL PRIME FACTORS OF A (SEQUENCE OF) POSITIVE INTEGERS

ENTER NO./EXPRESSION TO FACTORISE,   = 13468800

step set to default = 1.
13468800 = 2  2  2  2  2  2  2  3  5  5  23  61
13468801 = 13468801
            13468801 prime..1th,cont.
13468802 = 2  6734401

ENTER NO./EXPRESSION TO FACTORISE,
 < 2^31, 0=END?2^3*3*8  (number of divisors)
 = 192

Is the following all a waste? Can we use it?

=1*13468800/=2*6734400/=3*4489600/=4*3367200/=5*2693760/=6*2244800/ 
=8*1683600/=10*1346880/=12*1122400/=15*897920/=16*841800/=20*673440/ 
=23*585600/=24*561200/=25*538752/=30*448960/=32*420900/=40*336720/ 
=46*292800/=48*280600/=50*269376/=60*224480/=61*220800/=64*210450/ 
=69*195200/=75*179584/=80*168360/=92*146400/=96*140300/=100*134688/ 
=115*117120/=120*112240/=122*110400/=128*105225/=138*97600/ 
=150*89792/=160*84180/=183*73600/=184*73200/=192*70150/=200*67344/ 
=230*58560/=240*56120/=244*55200/=276*48800/=300*44896/=305*44160/ 
=320*42090/=345*39040/=366*36800/=368*36600/=384*35075/=400*33672/ 
=460*29280/=480*28060/=488*27600/=552*24400/=575*23424/=600*22448/ 
=610*22080/=640*21045/=690*19520/=732*18400/=736*18300/=800*16836/ 
=915*14720/=920*14640/=960*14030/=976*13800/=1104*12200/=1150*11712/ 
=1200*11224/=1220*11040/=1380*9760/=1403*9600/=1464*9200/=1472*9150/ 
=1525*8832/=1600*8418/=1725*7808/=1830*7360/=1840*7320/=1920*7015/ 
=1952*6900/=2208*6100/=2300*5856/=2400*5612/=2440*5520/=2760*4880/ 
=2806*4800/=2928*4600/=2944*4575/=3050*4416/=3200*4209/=3450*3904/ 
=3660*3680/   progs > IsPrime2kp. powabc1 confirmed prog > SeriCalc4S 

having found factor ~80 million of (2^13.4 million -1)
it may be possible to verify "order of 2 mod 80 million."

I think that is what I did.  But there are 192 possible
indexes that (do not give)
2^(p-1) == 1. mod 80 million
2^ 13.4 million == 2.
Mersenne number = 2^p-1 == 1.

/ don.mcdonald.

Message 4 in thread
From: Rupert ( rupertmccallum at yahoo.com )
Subject: Re: Mersenne. p=80,812,807 divides 2^13,468,801-1.
Newsgroups: sci.math
Date: 2002-08-29 19:06:18 PST
Phil Carmody <thefatphil_demunge at yahoo.co.uk> wrote in message  news:<pan.2002.08.29.17.50.41.547498.31649 at yahoo.co.uk> ...  
> On Thu, 29 Aug 2002 16:45:24 +0300, Mrs Unreliable wrote: 
>

> > don.lotto at paradise.net.nz (Don McDonald) wrote in message 
> >  news:<90961af5.0208281843.3ac8afa2 at posting.google.com> ... 
> >    
> >> p=80,812,807 divides M=2^13,468,801-1   
> > 
> > 80,812,807 divides 2^13,468,801-1 
...> > 80,858,983 divides 2^13,476,497-1
> > 80,865,607 divides 2^13,477,601-1   
>

> And 
> 162-60017709087981968928-60017709087981968928-60017709087981968928-60017
> 7 0 9 08798196892860017709087981968767 divides 10^10^100+10  but
> that's old news. (Patrick Demichel)

Don McD. group -xx- repeats?
>
  http://www.alpertron.com.ar/googol.pl  
>

> Phil  

And 2^15920831-1 is composite, although its first prime factor is 
large. My computer checked that the other month. Now it's checking 
2^14811767-1. 