20160123, 01:02  #1 
Sep 2002
Database er0rr
10602_{8} Posts 
Observation about Mersenne exponents
Code:
? V=[2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457 ,32582657,37156667,42643801,43112609,57885161,74207281] Code:
? f(p)=n=2^p1;x=Mod(3,n);c=0;while(lift(gcd(x1,n))==1,x=x^lift(x);c=c+1;if(c>4,break));if(gcd(x1,n)!=1,print(p" "lift(gcd(x1,n))" "c)) Code:
? for(k=1,#V,p=V[k];f(p)) 2 0 1 3 0 2 5 0 4 7 0 3 13 0 2 31 0 4 61 0 2 127 0 3 607 0 3 1279 0 4 3217 0 4 4423 0 4 23209 0 2 132049 0 4 ^C The real questions are: Why do the resulting exponents have those values? And can the calculation be speeded up? Last fiddled with by paulunderwood on 20160123 at 01:10 
20160123, 01:21  #2  
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
Quote:
Code:
(21:19) gp > for(P=1,1000000,f(P)) 2 0 1 3 0 2 5 0 4 7 0 3 13 0 2 25 0 3 31 0 4 61 0 2 127 0 3 607 0 3 1279 0 4 Last fiddled with by science_man_88 on 20160123 at 01:23 

20160123, 01:27  #3  
Sep 2002
Database er0rr
1182_{16} Posts 
Quote:
The right hand value is the number of iterations to get a 0. My quest is really to understand why such exponents occur, and, with that understanding, see if there is a way to speed up the program using mathematics, so it is faster than LL and therefore predict the next big Mersenne 

20160123, 01:34  #4  
"Forget I exist"
Jul 2009
Dartmouth NS
20342_{8} Posts 
Quote:


20160123, 01:40  #5  
Sep 2002
Database er0rr
2×3^{3}×83 Posts 
Quote:
What I am really after is understanding why those exponents (and later ones presumably) occur 

20160123, 01:45  #6 
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
well we know the LL test shows us Sn =A*Mp maybe something irregular happens starting at these points and working forward the math with the mersennes is probably ordered quite well regardless so I would think would allow for too much regularity however maybe one of them works out the equation to 0 mod the larger Mp's that may help ?
Last fiddled with by science_man_88 on 20160123 at 01:51 
20160123, 14:46  #7 
Sep 2002
Database er0rr
2·3^{3}·83 Posts 
Code:
? f(p)=n=2^p1;x=Mod(3,n);c=0;while(gcd(x1,n)==1,c=c+1;if(c>4,break);x=(1/x)^lift(x1));if(gcd(x1,n)!=1,print(p" "lift(gcd(x1,n))" "c)) Is there a bug in Pari/GP here? I replaced x^lift(x) with (1/x)^lift(x1) which is wrong! Code:
? version() [2, 7, 2] Last fiddled with by paulunderwood on 20160123 at 15:12 
20160123, 15:03  #8  
"Forget I exist"
Jul 2009
Dartmouth NS
20E2_{16} Posts 
Quote:
Quote:
Last fiddled with by science_man_88 on 20160123 at 15:24 

20160123, 17:42  #9 
Sep 2002
Database er0rr
1182_{16} Posts 
I am sure I was taught from an early age that a^(x) = (1/a)^x. However here is more on what Pari/GP gives:
Code:
? x=Mod(3, 170141183460469231731687303715884105727);x^lift(x) Mod(151236607520417094872610936636341427313, 170141183460469231731687303715884105727) ? x=Mod(3, 170141183460469231731687303715884105727);(1/x)^lift(x) Mod(163839658147118519445328514689369879589, 170141183460469231731687303715884105727) ? x=Mod(3, 170141183460469231731687303715884105727);(1/x)^lift(x1) Mod(151236607520417094872610936636341427313, 170141183460469231731687303715884105727) Code:
? n=2^232091;Mod(3,n)^n;gettime() 1273 ? f(23209);gettime() 23209 0 2 1268 
20160123, 17:46  #10 
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
really I thought a^x =1/(a^x) at least in non modular math it does. doh I see that they work out the same. also gettime needs to be before and after the code you do to test the time it's in the code otherwise it times the time between calls of gettime and may time how long it takes you to type/find code I think. the claim is still true in theory of course just pointing something out.
Last fiddled with by science_man_88 on 20160123 at 17:52 
20160123, 18:02  #11  
Aug 2006
5987_{10} Posts 
Quote:
Code:
x=Mod(3, 170141183460469231731687303715884105727); (1/x)^lift(x)==x^lift(x) (1/x)^lift(x)==x^lift(x) Edit: The exponent should really be taken mod phi(P) not mod P, and in this case phi(P) = P1 since P = 170141183460469231731687303715884105727 is prime. So I don't think there is a bug. Last fiddled with by CRGreathouse on 20160123 at 18:30 

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