|
%I #14 Nov 04 2020 05:22:16
%S 101,37,1931,53,73,109,353,389,2393,409,4051,683,8237,3733,691,3331,
%T 5113,3049,216173,1321,22811,1789,165391,3373,22501,15401,180563,5309,
%U 381853,10181,1253621,70067,14011,304597,13523,26759,134507,39569,43133,28111,3475261,45613,4209011,19867,24859
%N a(n) is the first prime p such that q*r mod p = q*r mod s = 12*n, where q,r,s are the next three primes after p.
%C The only common values for q*r mod p and q*r mod s that are not multiples of 12 appear to be 1 for p=2 and 2 for p=3.
%C If we have prime gaps q-p=2, r-q=6*n-2, s-r=2, then q*r == 12*n (mod p) and q*r == 12*n (mod s), so (if p > 12*n) this is a candidate for a(n). Dickson's conjecture implies there are infinitely many such p. Thus a(n) should always exist.
%C It appears that in all cases if p = a(n) and q,r,s are the next three primes, q-p = s-r and n = (q-p)*(r-p)/12.
%H Robert Israel, <a href="/A338615/b338615.txt">Table of n, a(n) for n = 1..106</a>
%e For p=1931 we have q,r,s = 1933, 1949, 1951, and 1933*1949 mod 1931 = 1933*1949 mod 1951 = 36 = 12*3. This is the first time that value appears, so a(3) = 1931.
%p q:= 2: r:= 3: s:= 5:
%p for i from 1 to 10^6 do
%p p:= q; q:= r; r:= s: s:= nextprime(s);
%p v:= q*r mod p; w:= q*r mod s;
%p if v = w and v mod 12 = 0 and not assigned(R[v/12]) then
%p R[v/12]:= p;
%p fi
%p od:
%p for nn from 1 while assigned(R[nn]) do od:
%p seq(R[i],i=1..nn-1);
%K nonn
%O 1,1
%A _J. M. Bergot_ and _Robert Israel_, Nov 03 2020
|
