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#1 |
Jun 2003
1,579 Posts |
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I am interesting in extending the series A113767
https://oeis.org/A113767 I had done some work on this in the past http://www.mersenneforum.org/showthread.php?t=18347 1) How do I submit the two terms found 6071, 218431? 2) Is there an efficient sieve for these forms? 3) Would these sequences be finite? Last fiddled with by Citrix on 2018-04-11 at 23:02 |
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#2 | |
"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
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2) modular remainders are inefficient but give that if a(n-1) is 2 mod 3, k is odd, if a(n-1) is 1 mod 3,k is even.( Oops math for the related sequence) 3) no clue. Last fiddled with by science_man_88 on 2018-04-11 at 23:54 |
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#3 |
Jun 2003
157910 Posts |
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1) I have figured out how to add the terms. Thanks.
2) I am looking for efficient software (not the math). This is basically a fixed k sieve (but the k is very large). Last fiddled with by Citrix on 2018-04-12 at 00:12 |
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#4 |
Nov 2008
91216 Posts |
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It feels like these sequences ought to be finite. The only long-term modular restriction I can think of for members of the sequence is that they cannot be 1 mod p for any p other than 2 and the previous term (correct me if I've missed something!). Since 78557 has covering set {3, 5, 7, 13, 19, 37, 73}, 78557+3*5*7*13*19*37*73*n is a Sierpinski number for every n, and 78557 is not 1 mod any of the primes in its covering set, so eventually the sequence ought to hit a number of this form, if it does not hit another Sierpinski number first.
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#5 | |
"Forget I exist"
Jul 2009
Dumbassville
20C016 Posts |
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Last fiddled with by science_man_88 on 2018-04-12 at 09:08 |
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#6 |
Jun 2003
1,579 Posts |
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Is someone able to modify srsieve to sieve for (((((((((((((((((1*2^1+1)*2^1+1)*2^2+1)*2^1+1)*2^5+1)*2^1+1)*2^1+1)*2^29+1)*2^3+1)*2^37+1)*2^31+1)*2^227+1)*2^835+1)*2^115+1)*2^7615+1)*2^6071+1)*2^218431+1)*2^n+1?
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#7 | |
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5×19×61 Posts |
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Your runtime might be dominated though by calculating k % p as that is a very large k. Whatever you do I don't think it is going to be overly fast. http://www.mersenneforum.org/rogue/mtsieve.html Last fiddled with by henryzz on 2018-04-13 at 06:20 |
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#8 | |
Nov 2008
2×33×43 Posts |
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should tend to 0 (by the prime number theorem), whereas the probability of hitting a Sierpinski number does not. |
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#9 |
Nov 2016
22·691 Posts |
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We can make the list of all primes:
a(m,n+1) is the smallest prime of the form a(m,n)*2^k+1 (k>=0) a(m+1,1) is the smallest prime not in the first m rows Code:
2 3 7 29 59 1889 3779 7559 5 11 23 47 47*2^583+1 13 53 107 857 6857 17 137 1097 140417 For the Riesel analog: a(m,n+1) is the smallest prime of the form a(m,n)*2^k-1 (k>=1) a(m+1,1) is the smallest prime not in the first m rows Code:
2 3 5 19 37 73 9343 7 13 103 823 1685503 11 43 5503 17 67 2143 Last fiddled with by sweety439 on 2018-04-13 at 15:54 |
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#10 | |
Jun 2003
1,579 Posts |
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There is something called modular exponentiation which will calculate k%p very fast. Calculating k%p will take very little time per prime. Addendum: Looks like srsieve is not implemented yet. So this will not help. Last fiddled with by Citrix on 2018-04-15 at 00:13 |
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#11 | |
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5·19·61 Posts |
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