A123017 Semiprimes k such that k+3 is also a semiprime.

6, 22, 35, 46, 55, 62, 74, 82, 91, 115, 118, 119, 142, 143, 155, 158, 166, 202, 203, 206, 214, 215, 218, 259, 262, 295, 298, 299, 302, 323, 326, 355, 358, 362, 391, 395, 451, 466, 478, 482, 502, 511, 514, 526, 535, 542, 551, 559, 562, 583, 586, 611, 623, 626

Offset: 1

Author

Jonathan Vos Post, Nov 04 2006

Keywords

• easy
• nonn

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).

Examples

			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.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001358, A056809, A070552, A092125, A092126, A092127, A092128, A092129.

Programs

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 *)

Formula

{a(n)} = {k such that k is in A001358 and k+3 is in A001358}.

Extensions

More terms from Robert G. Wilson v, Aug 31 2007