

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.


1



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
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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 qp=2, rq=6*n2, sr=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, qp = sr and n = (qp)*(rp)/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..nn1);


CROSSREFS

Sequence in context: A267511 A269503 A067748 * A190757 A212603 A278584
Adjacent sequences: A338612 A338613 A338614 * A338616 A338617 A338618


KEYWORD

nonn


AUTHOR

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


STATUS

approved



