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

1,1

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

Links

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

Example

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.

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

Prog

(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

Crossrefs

Cf. A001358.

Sequence in context: A020139 A056522 A056511 * A166626 A238759 A278349

Adjacent sequences: A339311 A339312 A339313 * A339315 A339316 A339317

Keyword

nonn

Author

Zak Seidov, Dec 17 2020

Status

approved