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Old 2010-04-19, 11:52   #1
science_man_88
 
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((2^q-1)*2^(p-1)*((2^p-1)*2^(p-1))/2^(q-1)=((2^q-1)*(2^p-1)))*4^f

2^q-1/((2^(q-1))*2^q-1) = (2^p-1*4^f)/(2^(p-1)*((2^p-1)*2^(p-1)))

1/2^(q-1) = 4^f/2^(2p-2)

f=0 then q=2p-1
f=1 then q = 8p-7
f=2 then q=32p-31
f=3 then q = 128p-127
f=4 then q = 512p-511


the problem is predicting f I made this PARI code but it doesn't allow checking of infinite f values:

allocatemem(932249567);for(f= 0,100000,for(x=25964950,25964952,if(x==25964951 && isprime((2*(4^f))*x-((2*4^f)-1)),print(x))))

I got a list of f values pari gave me back during another session (still open) but all I see is random ( course that's what they say about mersenne exponents)


(09:50) gp > log((2^3-1)*2^(2-1)*((2^2-1)*2^(2-1))/2^(3-1)/((2^3-1)*(2^2-1)))/log(4)
%1 = 0.E-28
(09:50) gp > log((2^5-1)*2^(3-1)*((2^3-1)*2^(3-1))/2^(5-1)/((2^5-1)*(2^3-1)))/log(4)
%2 = 0.E-28
(09:50) gp > log((2^7-1)*2^(5-1)*((2^5-1)*2^(5-1))/2^(7-1)/((2^7-1)*(2^5-1)))/log(4)
%3 = 1.000000000000000000000000000
(09:51) gp > log((2^13-1)*2^(7-1)*((2^7-1)*2^(7-1))/2^(13-1)/((2^13-1)*(2^7-1)))/log(4)
%4 = 0.E-28
(09:51) gp > log((2^17-1)*2^(13-1)*((2^13-1)*2^(13-1))/2^(17-1)/((2^17-1)*(2^13-1)))/log(4)
%5 = 4.000000000000000000000000000
(09:52) gp > log((2^19-1)*2^(17-1)*((2^17-1)*2^(17-1))/2^(19-1)/((2^19-1)*(2^17-1)))/log(4)
%6 = 7.000000000000000000000000000
(09:53) gp > log((2^31-1)*2^(19-1)*((2^19-1)*2^(19-1))/2^(31-1)/((2^31-1)*(2^19-1)))/log(4)
%7 = 3.000000000000000000000000000
(09:53) gp > log((2^61-1)*2^(31-1)*((2^31-1)*2^(31-1))/2^(61-1)/((2^61-1)*(2^31-1)))/log(4)
%8 = 0.E-28
(09:54) gp > log((2^89-1)*2^(61-1)*((2^61-1)*2^(61-1))/2^(89-1)/((2^89-1)*(2^61-1)))/log(4)
%9 = 16.00000000000000000000000000
(09:56) gp > log((2^107-1)*2^(89-1)*((2^89-1)*2^(89-1))/2^(107-1)/((2^107-1)*(2^89-1)))/log(4)
%10 = 35.00000000000000000000000000
(09:56) gp > log((2^127-1)*2^(107-1)*((2^107-1)*2^(107-1))/2^(127-1)/((2^127-1)*(2^107-1)))/log(4)
%11 = 43.00000000000000000000000000
(09:57) gp > log((2^521-1)*2^(127-1)*((2^127-1)*2^(127-1))/2^(521-1)/((2^521-1)*(2^127-1)))/log(4)
%12 = -134.0000000000000000000000000
(09:58) gp > log((2^607-1)*2^(521-1)*((2^521-1)*2^(521-1))/2^(607-1)/((2^607-1)*(2^521-1)))/log(4)
%13 = 217.0000000000000000000000000
(09:59) gp > log((2^1279-1)*2^(607-1)*((2^607-1)*2^(607-1))/2^(1279-1)/((2^1279-1)*(2^607-1)))/log(4)
%14 = -33.00000000000000000000000000
(10:00) gp > log((2^2203-1)*2^(1279-1)*((2^1279-1)*2^(1279-1))/2^(2203-1)/((2^2203-1)*(2^1279-1)))/log(4)
%15 = 177.0000000000000000000000000
(10:01) gp > log((2^2281-1)*2^(2203-1)*((2^2203-1)*2^(2203-1))/2^(2281-1)/((2^2281-1)*(2^2203-1)))/log(4)
%16 = 1062.000000000000000000000000
(10:02) gp > log((2^3217-1)*2^(2281-1)*((2^2281-1)*2^(2281-1))/2^(3217-1)/((2^3217-1)*(2^2281-1)))/log(4)
%17 = 672.0000000000000000000000000
(10:03) gp > log((2^4253-1)*2^(3217-1)*((2^3217-1)*2^(3217-1))/2^(4253-1)/((2^4253-1)*(2^3217-1)))/log(4)
%18 = 1090.000000000000000000000000
(10:04) gp > log((2^4423-1)*2^(4253-1)*((2^4253-1)*2^(4253-1))/2^(4423-1)/((2^4423-1)*(2^4253-1)))/log(4)
%19 = 2041.000000000000000000000000
(10:06) gp > log((2^9689-1)*2^(4423-1)*((2^4423-1)*2^(4423-1))/2^(9689-1)/((2^9689-1)*(2^4423-1)))/log(4)
%20 = -422.0000000000000000000000000
(10:07) gp > log((2^9941-1)*2^(9689-1)*((2^9689-1)*2^(9689-1))/2^(9941-1)/((2^9941-1)*(2^9689-1)))/log(4)
%21 = 4718.000000000000000000000000
(10:08) gp > log((2^11213-1)*2^(9941-1)*((2^9941-1)*2^(9941-1))/2^(11213-1)/((2^11213-1)*(2^9941-1)))/log(4)
%22 = 4334.000000000000000000000000
(10:10) gp > log((2^19937-1)*2^(11213-1)*((2^11213-1)*2^(11213-1))/2^(19937-1)/((2^19937-1)*(2^11213-1)))/log(4)
%23 = 1244.000000000000000000000000
(10:12) gp > log((2^21701-1)*2^(19937-1)*((2^19937-1)*2^(19937-1))/2^(21701-1)/((2^21701-1)*(2^19937-1)))/log(4)
%24 = 9085.999999999999999999999999
(10:15) gp > log((2^23209-1)*2^(21701-1)*((2^21701-1)*2^(21701-1))/2^(23209-1)/((2^23209-1)*(2^21701-1)))/log(4)
%25 = 10096.00000000000000000000000
(10:19) gp > log((2^44497-1)*2^(23209-1)*((2^23209-1)*2^(23209-1))/2^(44497-1)/((2^44497-1)*(2^23209-1)))/log(4)
%26 = 960.0000000000000000000000000
(10:22) gp > log((2^86243-1)*2^(44497-1)*((2^44497-1)*2^(44497-1))/2^(86243-1)/((2^86243-1)*(2^44497-1)))/log(4)
%27 = 1375.000000000000000000000000
(10:23) gp > log((2^110503-1)*2^(86243-1)*((2^86243-1)*2^(86243-1))/2^(110503-1)/((2^110503-1)*(2^86243-1)))/log(4)
%28 = 30991.00000000000000000000000
(10:26) gp > log((2^132049-1)*2^(110503-1)*((2^110503-1)*2^(110503-1))/2^(132049-1)/((2^132049-1)*(2^110503-1)))/log(4)
%29 = 44478.00000000000000000000000
(10:27) gp > log((2^216091-1)*2^(132049-1)*((2^132049-1)*2^(132049-1))/2^(216091-1)/((2^216091-1)*(2^132049-1)))/log(4)
%30 = 24003.00000000000000000000000
(10:29) gp > log((2^756839-1)*2^(216091-1)*((2^216091-1)*2^(216091-1))/2^(756839-1)/((2^756839-1)*(2^216091-1)))/log(4)
%31 = -162329.0000000000000000000000
(10:31) gp > allocatemem(932249567)
(10:31) gp > log((2^859433-1)*2^(756839-1)*((2^756839-1)*2^(756839-1))/2^(859433-1)/((2^859433-1)*(2^756839-1)))/log(4)
%32 = 327122.0000000000000000000000
(10:32) gp > log((2^1257787-1)*2^(859433-1)*((2^859433-1)*2^(859433-1))/2^(1257787-1)/((2^1257787-1)*(2^859433-1)))/log(4)
%33 = 230539.0000000000000000000000
(10:34) gp > log((2^1398269-1)*2^(1257787-1)*((2^1257787-1)*2^(1257787-1))/2^(1398269-1)/((2^1398269-1)*(2^1257787-1)))/log(4)
%34 = 558652.0000000000000000000000
(10:36) gp > log((2^2976221-1)*2^(1398269-1)*((2^1398269-1)*2^(1398269-1))/2^(2976221-1)/((2^2976221-1)*(2^1398269-1)))/log(4)
%35 = -89842.00000000000000000000000
(10:38) gp > log((2^3021377-1)*2^(2976221-1)*((2^2976221-1)*2^(2976221-1))/2^(3021377-1)/((2^3021377-1)*(2^2976221-1)))/log(4)
%36 = 1465532.000000000000000000000
(10:39) gp > log((2^6972593-1)*2^(3021377-1)*((2^3021377-1)*2^(3021377-1))/2^(6972593-1)/((2^6972593-1)*(2^3021377-1)))/log(4)
%37 = -464920.0000000000000000000000
(10:41) gp > log((2^13466917-1)*2^(6972593-1)*((2^6972593-1)*2^(6972593-1))/2^(13466917-1)/((2^13466917-1)*(2^6972593-1)))/log(4)
%38 = 239134.0000000000000000000000
(10:43) gp > log((2^20996011-1)*2^(13466917-1)*((2^13466917-1)*2^(13466917-1))/2^(20996011-1)/((2^20996011-1)*(2^13466917-1)))/log(4)
%39 = 2968911.000000000000000000000
(10:47) gp >

Last fiddled with by science_man_88 on 2010-04-19 at 11:53
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Old 2010-04-19, 13:09   #2
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ones that work aren't always listed like:

x=3 f=0 q=5 but,
x=3 f=1 q=17 works as a relation between two

x=7 f=0 q=13
x=7 f=2 q=193 ( is prime but isn't necessarily a mersenne exponent, that should be checked using the original equation)
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Old 2010-04-19, 13:11   #3
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f=3 also works for 7 to a possible ( disproved) exponent as does
f=5

Last fiddled with by science_man_88 on 2010-04-19 at 13:13
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Old 2010-04-19, 13:53   #4
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if you are solving for f then this what you want.
f = p - q/2 - 1/2
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Old 2010-04-19, 14:04   #5
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if you're sure thanks but I'm looking for integer values of f want to help?

Last fiddled with by science_man_88 on 2010-04-19 at 14:10
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Old 2010-04-19, 14:08   #6
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basically I'm looking for f when both p and q are prime especially if one is a known mersenne exponent then I might look through f values to find one that makes the other prime. then we can generate a possible decrease list of exponents to check

like above 2*p-1 is common but how common does it work where p is prime ? same with the others generate a non repeating list of values of exponents to check. then maybe I'll try testing them my way and gimps can try LL
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Old 2010-04-19, 14:14   #7
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like this x=7 if f = 0 2*p-1 should be prime in this case 2*7-1 = 13 is prime since nothing generates 11 that I know of 11 can't be part of the possible p values.

Last fiddled with by science_man_88 on 2010-04-19 at 14:14
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Old 2010-04-19, 14:25   #8
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f=p-q/2-1/2 is flawed I think as 1 = 7-11/2-1 but (2*4^1)*7-((2*4^1)-1) = 8*7-7 = 7*7 = 49 not 11
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Old 2010-04-19, 14:53   #9
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Code:
After removing unwanted brackets:
(2^q-1)*2^(p-1)*(2^p-1)*2^(p-1)/2^(q-1)=(2^q-1)*(2^p-1)*4^f

Simplifying:
22*(p-1)/2q-1=22*f

2*(p-1)-(q-1)=2*f

f = p-1 - (q-1)/2
q = 2*(p-1-f)+1
p = f+1 + (q-1)/2
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Old 2010-04-19, 14:56   #10
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thanks axn I thought the other one was flawed I haven't tested yours though if it works thanks as we can start making a list lol.
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Old 2010-04-19, 15:01   #11
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all the f values i've tested so far line up with what I originally got so that's a plus
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