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 2012-07-16, 20:54 #12 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100101011001102 Posts For 62, the PRP is 3490-digit. For 81, the PRP is 4834-digit. For 84, the PRP is 3057-digit. (these would be easy to prove prime) Last fiddled with by Batalov on 2012-07-16 at 21:08
2012-07-16, 21:22   #13
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

5,923 Posts

Quote:
 Originally Posted by Batalov For 62, the PRP is 3490-digit. For 81, the PRP is 4834-digit. For 84, the PRP is 3057-digit. (these would be easy to prove prime)
How far are you testing?

 2012-07-16, 21:25 #14 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·4,787 Posts I have 50000 Pi digits dumped from gp and then prefiltered by a simple perl script that takes care of small factors of 2,5 (easy!) and 3 (simple sum of digits). 7 and 11 could be easily added but is not of significant help. Then the candidate file goes to pfgw -f. All very easy, very 'umble. For 20, 80, 96 and 98, the PRPs would be larger than 30,000 digits now. Last fiddled with by Batalov on 2012-07-17 at 02:43
2012-07-17, 07:17   #15
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Dec 2008

B316 Posts

Quote:
 Originally Posted by Batalov Would you care to expand on this? How would you move on from the first instance to the next one? By proof?
By using a not-completely-insane method. In the outer loop, you take the first N digits of pi; in the inner loop, you take the last M digits of those N digits and test if that number begins with the specified digit(s) and is prime. Optimizations are possible: the outer loop can skip those N where the last digit is even, and the inner loop can keep a table of the positions of the starting digit(s).

 2012-07-17, 07:57 #16 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·4,787 Posts That's not what I meant. The inner loop is already done to death. You simply cannot use the outer loop (as the problem is stated). Suppose we are searching for a(20). Code: pi = 3. 14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679 82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196... Suppose we've checked the substrings starting from the red 20 up to a length of million and didn't find a prime. That doesn't give us the right to move on to the blue 20, find 2089 and say that we are done. We can only move on to the next instance of the starting point after we have proven that there are no primes formed by the first instance. Can we prove that?
 2012-07-17, 08:21 #17 axn     Jun 2003 3·17·101 Posts Do you (the general "you") define the "first prime instance" using the starting position or the ending position? Explicit clarification is called for.
 2012-07-17, 08:29 #18 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·4,787 Posts The problem is stated somewhat loosely and is open to interpretations. Clearly we can all agree that the first, say, '3' after the prefix is not what OP intended. (Even though he wrote, "the first prime constructed from the subsequent digits".) But I can see how you can interpret that as the string becomes longer it absorbes more instances of the prefix and you can hop onto that prefix. I think that this variant is too easy. It probably can be solved up to 5-6-digit values. Too easy. Indeed, I am interested in the hard variant (For a given number n, find the leftmost instance of a prime substring in decimal expansion of pi that starts with n.) It can still be not the first useable prefix. P.S. For the easy interpretation, 1 --> 41, isn't it? Maybe not (that's yet another interpretation). But surely, for the easy interpretation, 9 --> 97 Last fiddled with by Batalov on 2012-07-17 at 08:44
2012-07-17, 15:14   #19
bsquared

"Ben"
Feb 2007

2×1,789 Posts

Quote:
 Originally Posted by Batalov T We can only move on to the next instance of the starting point after we have proven that there are no primes formed by the first instance. Can we prove that?
IANANT, but the infinite sum of 1/ln(n) diverges, so even accounting for the fact that on average we only sum 4 of every 10 terms, I would think that the probability that there exists a prime would be 1.

edit: here is where it would be nice for RDS to come around, slap me upside the head, and give a correct answer. I just put my $0.02 worth of thought into it. Last fiddled with by bsquared on 2012-07-17 at 15:15  2012-07-17, 18:26 #20 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 225468 Posts 80 -> 8034825342...6216977871<41938> Formula: floor(Pi*10^42021)%(10^41938). (Submitted to Lifchitz&Lifchitz...) edit: I just put my$0.25 in the swear jar. B-> Last fiddled with by Batalov on 2012-07-18 at 01:01 Reason: Formula
 2012-07-17, 20:42 #21 davar55     May 2004 New York City 5×7×112 Posts Indeed, the problem being solved by Batalov and henryzz et.al. is the one I intended in the OP. Only if no prime exists in pi at a found point should the next occurrence of the integer be used. I see why this wasn't all clear in the OP.
 2012-07-18, 03:00 #22 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40

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