20180411, 22:45  #1 
Jun 2003
2·7·113 Posts 
A113767
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 20180411 at 23:02 
20180411, 23:43  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
2) modular remainders are inefficient but give that if a(n1) is 2 mod 3, k is odd, if a(n1) is 1 mod 3,k is even.( Oops math for the related sequence) 3) no clue. Last fiddled with by science_man_88 on 20180411 at 23:54 

20180412, 00:11  #3 
Jun 2003
1582_{10} Posts 
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 20180412 at 00:12 
20180412, 08:45  #4 
Nov 2008
2322_{10} Posts 
It feels like these sequences ought to be finite. The only longterm 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.

20180412, 09:08  #5  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
Last fiddled with by science_man_88 on 20180412 at 09:08 

20180413, 02:59  #6 
Jun 2003
2·7·113 Posts 
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?

20180413, 06:20  #7  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{5}×5×37 Posts 
Quote:
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 20180413 at 06:20 

20180413, 11:09  #8  
Nov 2008
100100010010_{2} Posts 
Quote:
should tend to 0 (by the prime number theorem), whereas the probability of hitting a Sierpinski number does not. 

20180413, 15:44  #9 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·5·17 Posts 
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^k1 (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 20180413 at 15:54 
20180415, 00:03  #10  
Jun 2003
62E_{16} Posts 
Quote:
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 20180415 at 00:13 

20180415, 10:07  #11  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5920_{10} Posts 
Quote:

