Fun with LL residues
 

Check out the LL residue on [URL=http://www.mersenne.org/report_exponent/?exp_lo=72207523&full=1]72207523[/URL]. 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...)

[QUOTE=NBtarheel_33;400448]The odds of having a string of five repeating hexits as this residue does is 1 in 16^[STRIKE]5[/STRIKE], or 1 in [STRIKE]1,048,576[/STRIKE].[/QUOTE]
The odds of having a string of five repeating hexits is ~ 1/16^[SIZE=3][COLOR=DarkRed]4 * [SIZE=2](number of sub-5-strings in a 14-string = 10)[/SIZE][/COLOR][/SIZE], or 1 in ~[COLOR=DarkRed] 6500[/COLOR].
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* 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).

What are the odds of having 16 (or is it 14?) zeroes? :wink:

You know the answer to that. :smile:

Let's see if we hear from someone new.

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.

[QUOTE=Stargate38;400459]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.[/QUOTE]You can have 16 zeros and still not have a Mersenne prime.

[QUOTE=Stargate38;400459]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.[/QUOTE]

They are more abundant than that at the moment :smile:

[QUOTE=paulunderwood;400462]They are more abundant than that at the moment :smile:[/QUOTE]

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. :smile:

M39847589
M66921341

[QUOTE=Batalov;400450]The odds of having a string of five repeating hexits is ~ 1/16^[SIZE=3][COLOR=darkred]4 * [SIZE=2](number of sub-5-strings in a 14-string = 10)[/SIZE][/COLOR][/SIZE], or 1 in ~[COLOR=darkred] 6500[/COLOR].
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* 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).[/QUOTE]

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...:smile:

[QUOTE=Madpoo;400476]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. :smile:

M39847589
M66921341[/QUOTE]

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...:devil:

[QUOTE=NBtarheel_33;400490]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]

[QUOTE=NBtarheel_33;400448]Check out the LL residue on [URL=http://www.mersenne.org/report_exponent/?exp_lo=72207523&full=1]72207523[/URL]. The odds of having a [B]string of five repeating hexits[/B] as this residue does is 1 in 16^5, or 1 in 1,048,576. [/QUOTE]

Your original specification didn't call for 5 repeating 'A's, it called for 5 repeating hexits, including 00000, 11111, ..., FFFFF.