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Old 2015-04-19, 16:59   #1
NBtarheel_33
 
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Default Fun with LL residues

Check out the LL residue on 72207523. The odds of having a string of five repeating hexits as this residue does is 1 in 16^5, or 1 in 1,048,576. This is an order of magnitude less than the odds of the exponent in question being that of a Mersenne prime! (We really need to start offering prizes for these weird residues...)
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Old 2015-04-19, 17:13   #2
Batalov
 
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Quote:
Originally Posted by NBtarheel_33 View Post
The odds of having a string of five repeating hexits as this residue does is 1 in 16^5, or 1 in 1,048,576.
The odds of having a string of five repeating hexits is ~ 1/16^4 * (number of sub-5-strings in a 14-string = 10), or 1 in ~ 6500.
______________
* This is an estimate of course and that's why I rounded the number, but it is accurate enough. The precise estimate should take into account that sub-5-strings are not independent (they overlap).

Last fiddled with by Batalov on 2015-04-19 at 17:21
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Old 2015-04-19, 17:22   #3
paulunderwood
 
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What are the odds of having 16 (or is it 14?) zeroes?

Last fiddled with by paulunderwood on 2015-04-19 at 17:23
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Old 2015-04-19, 17:41   #4
Batalov
 
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You know the answer to that.

Let's see if we hear from someone new.
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Old 2015-04-19, 19:52   #5
Stargate38
 
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The chance of having 16 zeros is 1 out of 2^64=16^16=18446744073709551616, even though Mersenne primes are a lot more abundant than that.
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Old 2015-04-19, 20:05   #6
legendarymudkip
 
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Quote:
Originally Posted by Stargate38 View Post
The chance of having 16 zeros is 1 out of 2^64=16^16=18446744073709551616, even though Mersenne primes are a lot more abundant than that.
You can have 16 zeros and still not have a Mersenne prime.
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Old 2015-04-19, 20:07   #7
paulunderwood
 
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Quote:
Originally Posted by Stargate38 View Post
The chance of having 16 zeros is 1 out of 2^64=16^16=18446744073709551616, even though Mersenne primes are a lot more abundant than that.
They are more abundant than that at the moment
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Old 2015-04-19, 23:55   #8
Madpoo
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Quote:
Originally Posted by paulunderwood View Post
They are more abundant than that at the moment
If anyone cares, there are 13 exponents where the residue is 0x0000000000000002

Of course it's worth pointing out these were problematic runs. :)

10 are known bad (triple-check confirmed), 1 has been factored (9559841) and 2 of them are still pending a triple-check if anyone felt like doing it, although I'm 100% sure you will NOT match the weird 0x2 residue.

M39847589
M66921341
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Old 2015-04-20, 08:01   #9
NBtarheel_33
 
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Quote:
Originally Posted by Batalov View Post
The odds of having a string of five repeating hexits is ~ 1/16^4 * (number of sub-5-strings in a 14-string = 10), or 1 in ~ 6500.
______________
* This is an estimate of course and that's why I rounded the number, but it is accurate enough. The precise estimate should take into account that sub-5-strings are not independent (they overlap).
I have a 16-hexit string (yes, it's true that I don't know the last two hexits, but it doesn't really matter what they are - it's mildly interesting, certainly, if they are indeed A's as well, but they needn't be). I want five neighboring hexits to be A's (i.e. AAAAA will appear somewhere in the 16-hexit string).

That gives me twelve choices for where the first of the quintuplet of A's can go, and then I have eleven empty slots where I can place any hexit I like. There are, therefore, 12 * 16^11 possible 16-hexit strings containing a quintuplet of A's. Since there are 16^16 possible 16-hexit strings overall, the odds of randomly selecting a 16-hexit string containing a quintuplet of A's ought to be given by 12 * 16^11 / 16^16 = 12 / 16^5 = ~1 / 87,381. Right? Certainly not a one-in-a-million event, but rare enough that it probably doesn't happen every year.

What I had first envisioned was a 16-hexit residue being constructed by tossing 16 times a 16-sided die. The odds of tossing "A" five times in a row (and hence forming "AAAAA" in the resulting residue) would then be given by 1 / 16^5, or 1 / 1,048,576.

It's a pretty neat residue, anyway. But I agree, all zeroes is much more fun...
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Old 2015-04-20, 08:04   #10
NBtarheel_33
 
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Quote:
Originally Posted by Madpoo View Post
If anyone cares, there are 13 exponents where the residue is 0x0000000000000002

Of course it's worth pointing out these were problematic runs. :)

10 are known bad (triple-check confirmed), 1 has been factored (9559841) and 2 of them are still pending a triple-check if anyone felt like doing it, although I'm 100% sure you will NOT match the weird 0x2 residue.

M39847589
M66921341
A residue of 0x2 is LL Hell (HeLL, perhaps?). Square 2, subtract 2, you get 2. Square 2, subtract 2, you get 2. And on and on and on...
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Old 2015-04-20, 09:22   #11
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Quote:
Originally Posted by NBtarheel_33 View Post
I have a 16-hexit string (yes, it's true that I don't know the last two hexits, but it doesn't really matter what they are - it's mildly interesting, certainly, if they are indeed A's as well, but they needn't be). I want five neighboring hexits to be A's (i.e. AAAAA will appear somewhere in the 16-hexit string).
Quote:
Originally Posted by NBtarheel_33 View Post
Check out the LL residue on 72207523. The odds of having a string of five repeating hexits as this residue does is 1 in 16^5, or 1 in 1,048,576.
Your original specification didn't call for 5 repeating 'A's, it called for 5 repeating hexits, including 00000, 11111, ..., FFFFF.
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