20150419, 16:59  #1 
"Nathan"
Jul 2008
Maryland, USA
10001011011_{2} Posts 
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...)

20150419, 17:13  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5·7·283 Posts 
Quote:
______________ * 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 sub5strings are not independent (they overlap). Last fiddled with by Batalov on 20150419 at 17:21 

20150419, 17:22  #3 
Sep 2002
Database er0rr
2^{2}×1,063 Posts 
What are the odds of having 16 (or is it 14?) zeroes?
Last fiddled with by paulunderwood on 20150419 at 17:23 
20150419, 17:41  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
26B1_{16} Posts 
You know the answer to that.
Let's see if we hear from someone new. 
20150419, 19:52  #5 
"Daniel Jackson"
May 2011
14285714285714285714
1316_{8} Posts 
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.

20150419, 20:05  #6 
Jun 2014
2^{3}×3×5 Posts 

20150419, 20:07  #7 
Sep 2002
Database er0rr
2^{2}·1,063 Posts 

20150419, 23:55  #8 
Serpentine Vermin Jar
Jul 2014
3·11·101 Posts 
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 (triplecheck confirmed), 1 has been factored (9559841) and 2 of them are still pending a triplecheck if anyone felt like doing it, although I'm 100% sure you will NOT match the weird 0x2 residue. M39847589 M66921341 
20150420, 08:01  #9  
"Nathan"
Jul 2008
Maryland, USA
5·223 Posts 
Quote:
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 16hexit strings containing a quintuplet of A's. Since there are 16^16 possible 16hexit strings overall, the odds of randomly selecting a 16hexit 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 oneinamillion event, but rare enough that it probably doesn't happen every year. What I had first envisioned was a 16hexit residue being constructed by tossing 16 times a 16sided 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... 

20150420, 08:04  #10  
"Nathan"
Jul 2008
Maryland, USA
5×223 Posts 
Quote:


20150420, 09:22  #11  
Jun 2003
5,387 Posts 
Quote:
Quote:


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