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A339314
a(n) is the least semiprime k > n-th semiprime s = A001358(n) such that k-s and k+s are both semiprimes.
1
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
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
KEYWORD
nonn
AUTHOR
Zak Seidov, Dec 17 2020
STATUS
approved