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Old 2010-10-25, 00:40   #1530
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Quote:
Originally Posted by cmd View Post
ep

the right mathematics can not cheat

http://www.xkcd.com/747/

http://www.xkcd.com/759/

third element of the first group,
advance of 30 and then tell us where to find last

Last fiddled with by cmd on 2010-10-25 at 00:57 Reason: |o|
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Old 2010-10-25, 02:46   #1531
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Quote:
Originally Posted by 3.14159 View Post
Hopefully it would be accurate..
It should be possible to calculate a reasonable range in which the answer lies. I'm not sure how precise it would be possible to make it.

Last fiddled with by CRGreathouse on 2010-10-25 at 02:47
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Old 2010-10-28, 00:29   #1532
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I think I can sieve by hand for k * b^n ± 1 numbers; I just need to do some modular arithmetic to get rid of each prime p which divides a certain k;

Ex: k * 6^90 + 1;

Step 1; 6^90 mod 7 = 1.

Step 2; Multiply by 6; 6 * 6^90 = 6 mod 7;

6^91 + 1 mod 7 = 0.

7 divides k = {6, 13, 20, 27, 34, 41, 48, 55, 62, 69, 76, 83, 90, 97, 104, 111, ...}

Repeat for 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, and.. as far as you'll be willing to strain your hand..

Although even a computer is approx. 10^4 to 10^8 times faster than you.

Last fiddled with by 3.14159 on 2010-10-28 at 00:40
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Old 2010-10-28, 00:42   #1533
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I'll use that whenever working with no computer assistance; Unless there's an easier method to go by.
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Old 2010-11-01, 13:19   #1534
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Code:
(10:15) gp > for(i=1,#mersenne,print(isprime(floor(prime(i)^(2*Pi)))))
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(10:16) gp > for(i=1,#mersenne,print(isprime(floor(mersenne[i]^(2*Pi)))))
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(10:16) gp >
can we use this fact that all the Mersenne exponents up to #39 work but not all primes up to prime 39 could this help narrow down exponents if used ?

though it may be doubtful these continue and it took me a long time to find a 1 in the prime[i]^(5*Pi) one and they seem to work for Mersenne exponents so far as well.

Last fiddled with by science_man_88 on 2010-11-01 at 13:21
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Old 2010-11-01, 14:02   #1535
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Quote:
Originally Posted by science_man_88 View Post
can we use this fact that all the Mersenne exponents up to #39 work but not all primes up to prime 39 could this help narrow down exponents if used ?
No.

First, we don't know that the pattern will continue -- in fact I doubt it does. Second, there are way too many primes for which this will work -- heuristically, \frac{n}{2\pi\log^2n} up to n.

But if you think you're onto something, try testing
49982629, 49985387, 49985521, 49986509, 49986737, 49987409, 49987759, 49988693, 49989491, ...
for which the relation holds and are toward the small end of the unassigned range. (I don't know which of these may already be taken, possibly all.)

Last fiddled with by CRGreathouse on 2010-11-01 at 14:04
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Old 2010-11-01, 14:07   #1536
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Quote:
Originally Posted by CRGreathouse View Post
No.

First, we don't know that the pattern will continue -- in fact I doubt it does. Second, there are way too many primes for which this will work -- heuristically, \frac{n}{2\pi\log^2n} up to n.

But if you think you're onto something, try testing
49982629, 49985387, 49985521, 49986509, 49986737, 49987409, 49987759, 49988693, 49989491, ...
for which the relation holds and are toward the small end of the unassigned range. (I don't know which of these may already be taken, possibly all.)
okay but how many of those fit for the 5*Pi one ? or the sequence of x*Pi that are :

2 5 8 10 11 12 14 15 18 all Mersenne exponents so far tested by me work to be 0 ( false) for all of these but not every prime will work in these relations.
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Old 2010-11-01, 14:15   #1537
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Quote:
Originally Posted by CRGreathouse View Post
No.

First, we don't know that the pattern will continue -- in fact I doubt it does. Second, there are way too many primes for which this will work -- heuristically, \frac{n}{2\pi\log^2n} up to n.

But if you think you're onto something, try testing
49982629, 49985387, 49985521, 49986509, 49986737, 49987409, 49987759, 49988693, 49989491, ...
for which the relation holds and are toward the small end of the unassigned range. (I don't know which of these may already be taken, possibly all.)
okay but how many of those fit for the 5*Pi one ? or the sequence of x*Pi that are :

2 5 8 10 11 12 14 15 18 all Mersenne exponents so far tested by me work to be 0 ( false) for all of these but not every prime will work in these relations.

f each combo of isprime !isprime etc. knocks out a prime when the range is up to 1,000,000 then this can knock out over 60% of prime exponents in the range (though it might slump lower) without a single LL test being done. if confirmed it would help a lot I think. not quite sure anymore got to figure it out there are 623530 combinations of taking the numbers i've found to work so far off isprime I believe if so then if each one eliminated just one exponent we could eliminate 623530 primes at least from ever needing LL test.

Last fiddled with by science_man_88 on 2010-11-01 at 15:09
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Old 2010-11-01, 15:55   #1538
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Quote:
Originally Posted by science_man_88 View Post
okay but how many of those fit for the 5*Pi one ?
You have Pari/GP, feel free to test. Heuristically, if I understand what you're asking, about \frac{x}{10\pi^2\log^4x} numbers up to x will have this property and the other.

But I emphasize that this is likely to remove as many (as a fraction) of future Mersenne exponents as Mersenne non-exponents.

Edit: I can't decide on what exponent to use in the denominator, but not less than 2 or more than 4.

Last fiddled with by CRGreathouse on 2010-11-01 at 16:10
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Old 2010-11-01, 16:03   #1539
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Quote:
Originally Posted by CRGreathouse View Post
You have Pari/GP, feel free to test. Heuristically, if I understand what you're asking, about \frac{x}{10\pi^2\log^4x} numbers up to x will have this property and the other.

But I emphasize that this is likely to remove as many (as a fraction) of future Mersenne exponents as Mersenne non-exponents.
yeah scrap 18 lol for use for Mersenne exponents yeah well it's based on properties of the rest of them to predict the next lol.
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Old 2010-11-11, 01:47   #1540
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Here's a nice challenge for you all;

Express the number 98874131329600347 as a sum of prime powers, which meet these conditions:

1. No more than 25 primes may be used.

Ex: 230 + 318 + 512 + 196 + 2933 + 5773 + 1913 + 893 + 2232 + 372 + 112 + 731 + 21 = 711

Or, for a harder one; No primes smaller than 250000 may be used, no primes may be used more than once, all exponents must be prime and used once and only once, and no more than 30 primes may be used.

Last fiddled with by 3.14159 on 2010-11-11 at 02:11
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