20191127, 02:18  #12  
Romulan Interpreter
Jun 2011
Thailand
19·467 Posts 
Quote:
(I say that every time when I change my motto, then I forget about it or don't find anything better, and the motto stays for some weeks...) 

20191127, 08:06  #13 
Jun 2003
2^{2}×3^{2}×131 Posts 
Searched this up to 18million. No hits. Not proceeding any further.

20191127, 16:25  #14 
"Sam"
Nov 2016
2^{2}×79 Posts 

20191127, 17:02  #15 
"Robert Gerbicz"
Oct 2005
Hungary
17·83 Posts 
Not interested in the actual computation, but you can do it in polynomial time per prime if you want all residues for p<N, similarly to the computation of Wilson (prime) residues.
Last fiddled with by R. Gerbicz on 20191127 at 17:03 Reason: typo, grammar 
20200606, 19:25  #16 
Jun 2003
627_{16} Posts 
I have been thinking of a generalized version of the above problem
We can define S(n)=S(n1)+n*(S(n1)S(n2)) For every value of S(0) and S(1) we can get a different sequence (where S(1)>S(0)) Conjecture: For every unique sequence there a prime p such that S(p+x) is always divisible by p (x being a positive natural number) Some thoughts If a prime p divides the sequence for a particular S(0) and S(1) Then prime p also divides all sequences where S'(0)=k*S(0) and S'(1)=k*S(1) (k is a natural number) So we only need to look at the sequence with primitive solutions where S(0)=1 and S(1)>1 S(0)=0 and S(1)>0 is a special case which leads to multiple of the sequence posted in the initial post of this thread. I searched S(0)=1 and S(1)<10000 for all primes p<1,000,000. There are 706 sequences left. Any further thoughts? If the above conjecture is false  what it the density of remaining sequences? 
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