A339314 a(n) is the least semiprime k > n-th semiprime s = A001358(n) such that k-s and k+s are both semiprimes.

10, 15, 86, 25, 35, 106, 25, 55, 94, 51, 58, 85, 39, 77, 94, 95, 74, 55, 106, 178, 143, 155, 69, 118, 95, 142, 121, 118, 119, 91, 146, 142, 115, 206, 115, 115, 206, 169, 134, 146, 143, 178, 133, 158, 155, 262, 177, 158, 178, 155, 159, 183, 254, 194, 205, 202, 226, 187, 298, 206, 226, 209

Offset: 1

Author

Zak Seidov, Dec 17 2020

Keywords

• nonn

Comments

Differences k - s: 6, 9, 77, 15, 21, 91, 4, 33, 69, 25, ... with minimal value 4.

What about the maximal value of k - s?

k-s is unbounded, because the gaps between semiprimes are unbounded. In fact, given any n distinct primes, by the Chinese Remainder Theorem there exist n consecutive positive integers that are each divisible by the cube of one of these primes (and thus not semiprimes). - Robert Israel, Dec 27 2020

Examples

			s=4, k=10, 6 and 14 are all semiprimes,
s=6, k=15, 9 and 21 are all semiprimes,
s=9, k=86, 77 and 95 are all semiprimes.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001358.

Programs

Maple

N:= 10^3:
SP:= select(t -> numtheory:-bigomega(t)=2, [$4..N]):
f:= proc(n) local i,s;
  s:= SP[n];
  for i from n+1 do
    if numtheory:-bigomega(SP[i]-s)=2 and numtheory:-bigomega(SP[i]+s)=2 then return SP[i] fi
  od;
end proc:
map(f, [$1..100]); # Robert Israel, Dec 27 2020

PARI

issemip(n) = bigomega(n)==2;
lista(nn) = {my(v = select(issemip, [1..nn])); for (n=1, #v, my(ik=n+1, s=v[n]); while (!(issemip(v[ik]+s) && issemip(v[ik]-s)), ik++; if (ik>#v, return)); print1(v[ik], ", "););} \\ Michel Marcus, Dec 19 2020