OFFSET
1,1
COMMENTS
When a(n+1) = a(n) + 3 we have that a(n) is a semiprime such that a(n) and a(n)+3 and a(n) + 3 + 3 are all semiprimes, hence at least 3 semiprimes in arithmetic progression with common difference 3. This subsequence begins 115, 155. There cannot be 4 semiprimes in arithmetic progression with common difference 3, starting with k, because modulo 4 we have {k, k+3, k+6, k+9} == {k+0, k+3, k+2, k+1} and one of these must be divisible by 4, hence a nonsemiprime (eliminating k = 4 by inspection).
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
a(1) = 6 because 6 = 2 * 3 is semiprime and 6 + 3 = 9 = 3^2 is semiprime.
a(2) = 22 because 22 = 2 * 11 and 22 + 3 = 25 = 5^2.
a(3) = 35 because 35 = 5 * 7 and 35 + 3 = 38 = 2 * 19.
a(4) = 46 because 46 = 2 * 23 and 46 + 3 = 49 = 7^2.
a(5) = 55 because 55 = 5 * 11 and 55 + 3 = 58 = 2 * 29.
MATHEMATICA
semiprimeQ[n_] := Plus @@ Last /@ FactorInteger[n] == 2; Select[ Range@ 670, semiprimeQ[ # ] && semiprimeQ[ # + 3] &] (* Robert G. Wilson v, Aug 31 2007 *)
SequencePosition[Table[If[PrimeOmega[n]==2, 1, 0], {n, 700}], {1, _, _, 1}][[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 03 2017 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Nov 04 2006
EXTENSIONS
More terms from Robert G. Wilson v, Aug 31 2007
STATUS
approved