A338615 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.

101, 37, 1931, 53, 73, 109, 353, 389, 2393, 409, 4051, 683, 8237, 3733, 691, 3331, 5113, 3049, 216173, 1321, 22811, 1789, 165391, 3373, 22501, 15401, 180563, 5309, 381853, 10181, 1253621, 70067, 14011, 304597, 13523, 26759, 134507, 39569, 43133, 28111, 3475261, 45613, 4209011, 19867, 24859

Offset

1,1

Comments

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.

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.

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.

Links

Robert Israel, Table of n, a(n) for n = 1..106

Example

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.

Maple

q:= 2: r:= 3: s:= 5:
for i from 1 to 10^6 do
  p:= q; q:= r; r:= s: s:= nextprime(s);
  v:= q*r mod p; w:= q*r mod s;
  if v = w and v mod 12 = 0 and not assigned(R[v/12]) then
       R[v/12]:= p;
  fi
od:
for nn from 1 while assigned(R[nn]) do od:
seq(R[i], i=1..nn-1);

Crossrefs

Sequence in context: A375784 A269503 A067748 * A395183 A190757 A212603

Adjacent sequences: A338612 A338613 A338614 * A338616 A338617 A338618

Keyword

nonn

Author

J. M. Bergot and Robert Israel, Nov 03 2020

Status

approved