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Old 2009-07-17, 15:17   #1
spkarra
 
"Sastry Karra"
Jul 2009
Bridgewater, NJ (USA)

338 Posts
Default Peculiar behaviour of prime numbers .....

For the past three months, I am enthusiastically analyzing the known Mersenne’s Prime numbers and trying my best to look for some sort of commonality.

During this process, I found out a peculiar characteristic of prime numbers. I am not sure if this is already proved by other mathematicians or not, but I would like to share with all of you.

I ran my software program to test my conjucture. Due to my laptop technical restrictions, I could run from integer 1 thru integer 16. For some unknown reasons, when the integer value is 17, my Laptop is getting “Hung-Up”. So, I started testing from integer 18 onwards and after letting my Laptop run for more than 9 hrs, it is still unable to tell me if 19^19 minus 2 is a Prime or not.

If this behaviour is not noticed earlier, then I will let my Laptop to run 24X7 for a week and see how far can I verify.



*************************************************************************************************************************************
Conjucture:

If p & q are positive primes, n is a positive integer and q = (p^p – 2^n), then the number of q’s are limited.

( I found only 8 Primes when ran from 2 thru 16).

Example:
2^2 – 2^1 = 2 ;
2^2 – 2^2 = 0 ; - IGNORED SINCE q = ZERO;
3^3 – 2^1 = 25 – Not a Prime;
3^3 – 2^2 = 23 ;
3^3 – 2^3 = 19 ;
3^3 – 2^4 = 11 ;


Here is the Screen Printout of all primes between 3 and 16 tested – found only there are 10 primes with this condition being satisfied.
-----------------------------------------------------------------------------------------------------------------------------------------------
At Tue Jun 23 20:02:05 EDT 2009 Checking if 2 to power of 2 minus 2 2 is a PRIME..
--> 2 to power of 2 minus 2 = 2 is a PRIME.. FIRST ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 3 to power of 3 minus 2 = 25 is a PRIME..
At Tue Jun 23 20:02:05 EDT 2009 Checking if 3 to power of 3 minus 4 = 23 is a PRIME..
--> 3 to power of 3 minus 4 = 23 is a PRIME.. SECOND ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 3 to power of 3 minus 8 = 19 is a PRIME..
--> 3 to power of 3 minus 8 = 19 is a PRIME.. THIRD ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 3 to power of 3 minus 16 = 11 is a PRIME..
--> 3 to power of 3 minus 16 = 11 is a PRIME.. FOURTH ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 2 = 3123 is a PRIME..
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 4 = 3121 is a PRIME..
--> 5 to power of 5 minus 4 = 3121 is a PRIME.. FIFTH ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 8 = 3117 is a PRIME..
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 16 = 3109 is a PRIME..
--> 5 to power of 5 minus 16 = 3109 is a PRIME.. SIXTH ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 32 = 3093 is a PRIME..
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 64 = 3061 is a PRIME..
--> 5 to power of 5 minus 64 = 3061 is a PRIME.. SEVENTH ONE
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 128 = 2997 is a PRIME..
At Tue Jun 23 20:02:05 EDT 2009 Checking if 5 to power of 5 minus 256 = 2869 is a PRIME..
At Tue Jun 23 20:02:05 EDT 2009 Checking if 7 to power of 7 minus 2 = 823541 is a PRIME..
--> 7 to power of 7 minus 2 = 823541 is a PRIME.. EIGTH ONE
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 4 = 823539 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 8 = 823535 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 16 = 823527 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 32 = 823511 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 64 = 823479 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 128 = 823415 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 7 to power of 7 minus 256 = 823287 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 11 to power of 11 minus 2 = 285311670609 is a PRIME..
At Tue Jun 23 20:02:08 EDT 2009 Checking if 11 to power of 11 minus 4 = 285311670607 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 11 to power of 11 minus 8 = 285311670603 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 11 to power of 11 minus 16 = 285311670595 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 11 to power of 11 minus 32 = 285311670579 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 11 to power of 11 minus 64 = 285311670547 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 11 to power of 11 minus 128 = 285311670483 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 11 to power of 11 minus 256 = 285311670355 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 2 = 302875106592251 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 4 = 302875106592249 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 8 = 302875106592245 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 16 = 302875106592237 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 32 = 302875106592221 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 64 = 302875106592189 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 128 = 302875106592125 is a PRIME..
At Tue Jun 23 20:02:10 EDT 2009 Checking if 13 to power of 13 minus 256 = 302875106591997 is a PRIME..
At Thu Jun 25 07:15:28 EDT 2009 Checking if 19 to power of 19 minus 2 = 1978419655660313589123977 is a PRIME.. PROCESS Failed


-----------------------------------------------------------------------------------------------------------------------------------------------

--
Thanks,
Sastry Karra
MS, MBA(MIS)
"Good judgement comes from experience and experience comes from Bad judgement"
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Old 2009-07-17, 15:46   #2
Mr. P-1
 
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pari/gp confirms that 19^19-2 is prime. The result is instant. if your laptop is taking 9 hours then there is something wrong with your algorithm.

Why do you not test past 2^n=256?
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Old 2009-07-17, 16:06   #3
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Incidentally for each p there are about \frac{p.ln(p)}{ln(2)} numbers to test, of which all but one are of the same magnitude as p^p. Ignoring that exception, then by the prime number theorem, you would expect about \frac{2}{ln(2)}p primes. (The constant 2 in the numerator comes from the fact that none of these numbers are even). Your results look to be consistent with this distribution, but in truth, you've tested so few that nothing definite can be said.

The infinite series for all prime p diverges (rather rapidly, in fact), and so I would expect there to be an infinitude of such primes, contrary to your conjecture.

Whether anything can be proven, or whether there are any number-theoretical reasons to doubt my heuristic is beyond by mathematical ability.

Last fiddled with by Mr. P-1 on 2009-07-17 at 16:10
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Old 2009-07-17, 18:07   #4
CRGreathouse
 
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Primes:
Code:
2^2 - 2^1
3^3 - 2^2
3^3 - 2^3
3^3 - 2^4
5^5 - 2^2
5^5 - 2^4
5^5 - 2^6
7^7 - 2^1
7^7 - 2^13
17^17 - 2^6
17^17 - 2^20
17^17 - 2^30
17^17 - 2^40
17^17 - 2^64
19^19 - 2^1
19^19 - 2^5
23^23 - 2^66
23^23 - 2^76
31^31 - 2^77
31^31 - 2^97
41^41 - 2^38
41^41 - 2^214
47^47 - 2^60
47^47 - 2^72
53^53 - 2^30
53^53 - 2^50
59^59 - 2^100
61^61 - 2^91
61^61 - 2^327
67^67 - 2^193
73^73 - 2^177
73^73 - 2^189
73^73 - 2^293
83^83 - 2^22
83^83 - 2^116
83^83 - 2^272
83^83 - 2^274
83^83 - 2^290
83^83 - 2^366
83^83 - 2^436
97^97 - 2^407
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Old 2009-07-17, 18:37   #5
CRGreathouse
 
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Quote:
Originally Posted by Mr. P-1 View Post
Incidentally for each p there are about \frac{p.ln(p)}{ln(2)} numbers to test, of which all but one are of the same magnitude as p^p. Ignoring that exception, then by the prime number theorem, you would expect about \frac{2}{ln(2)}p primes. (The constant 2 in the numerator comes from the fact that none of these numbers are even). Your results look to be consistent with this distribution, but in truth, you've tested so few that nothing definite can be said.
You'd expect 2/log 2 primes per p, so \frac{2p}{\log 2\log p} through p.

You can get a better estimate by considering their behavior mod 3, 5, ..., making the expectation
\frac{2p}{\log 2\log p}\prod_{q\le p}1-\frac1q\approx\frac{2p}{e^\gamma\log2\log^2p},
if I haven't made a mistake. Anyone want to test this against the numbers I've generated so far?
Code:
101^101 - 2^666
103^103 - 2^107
103^103 - 2^159
103^103 - 2^639
103^103 - 2^659
107^107 - 2^6
107^107 - 2^72
107^107 - 2^322
107^107 - 2^352
107^107 - 2^594
109^109 - 2^159
109^109 - 2^679
113^113 - 2^142
113^113 - 2^206
113^113 - 2^488
127^127 - 2^169
127^127 - 2^337
131^131 - 2^70
131^131 - 2^610
131^131 - 2^658
137^137 - 2^134
137^137 - 2^342
137^137 - 2^582
139^139 - 2^227
139^139 - 2^387
139^139 - 2^529
149^149 - 2^304
149^149 - 2^1036
151^151 - 2^181
157^157 - 2^11
157^157 - 2^655
163^163 - 2^263
163^163 - 2^315
163^163 - 2^1011
167^167 - 2^66
167^167 - 2^822
173^173 - 2^8
173^173 - 2^242
173^173 - 2^264
173^173 - 2^452
173^173 - 2^472
173^173 - 2^580
173^173 - 2^710
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Old 2009-07-17, 19:22   #6
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Quote:
Originally Posted by CRGreathouse View Post
You'd expect 2/log 2 primes per p
Why that number, and not p times it?
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Old 2009-07-17, 19:26   #7
CRGreathouse
 
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Quote:
Originally Posted by Mr. P-1 View Post
Why that number, and not p times it?
log(p^p) = p log p, so you expect one in p log p to be prime (or two, because of their parity). There are log_2 p^p = p log p / log 2 per p, so 2 * (p log p / log 2) / (p log p) = 2/log 2.

For other 'small enough' primes there are only q-1 ways of being divisible by q, since q doesn't divide p^p and q doesn't divide 2^n. That's what the correction term is trying to do. It's actually not quite right, since I'm not sure how far to take 'small enough'. Maybe if I do an integral rather than a sum...?
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Old 2009-07-17, 19:41   #8
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Quote:
Originally Posted by CRGreathouse View Post
log(p^p) = p log p, so you expect one in p log p to be prime (or two, because of their parity). There are log_2 p^p = p log p / log 2 per p, so 2 * (p log p / log 2) / (p log p) = 2/log 2.
You're right. I think I forgot to exponentiate p at some point.
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Old 2009-07-20, 22:47   #9
spkarra
 
"Sastry Karra"
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Thanks a lot ....
In my conjecture, I mentioned that "the number of q’s are limited. ".
Now, I found that there are more than I found.....
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