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#1530 | |
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"(^r'°:.:)^n;e'e"
Nov 2008
;t:.:;^
33×37 Posts |
Quote:
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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#1531 |
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Aug 2006
3×1,993 Posts |
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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#1532 |
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May 2010
Prime hunting commission.
110100100002 Posts |
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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#1533 |
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May 2010
Prime hunting commission.
32208 Posts |
I'll use that whenever working with no computer assistance; Unless there's an easier method to go by.
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#1534 |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
Code:
(10:15) gp > for(i=1,#mersenne,print(isprime(floor(prime(i)^(2*Pi))))) 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (10:16) gp > for(i=1,#mersenne,print(isprime(floor(mersenne[i]^(2*Pi))))) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (10:16) gp > 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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#1535 | |
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Aug 2006
3×1,993 Posts |
Quote:
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, 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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#1536 | |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
Quote:
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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#1537 | |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
Quote:
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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#1538 |
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Aug 2006
3·1,993 Posts |
You have Pari/GP, feel free to test. Heuristically, if I understand what you're asking, about
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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#1539 | |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
Quote:
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#1540 |
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May 2010
Prime hunting commission.
69016 Posts |
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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