A202319 Lesser of two semiprimes sandwiched each between semiprimes thus forming a twin semiprime-triple.

214, 143098, 194758, 206134, 273418, 684898, 807658, 1373938, 1391758, 1516534, 1591594, 1610998, 1774798, 1882978, 1891762, 2046454, 2051494, 2163418, 2163958, 2338054, 2359978, 2522518, 2913838, 3108202, 4221754, 4297318, 4334938, 4866118, 4988878, 5108794

Offset: 1

Author

V.J. Pohjola, Dec 16 2011

Keywords

• nonn

Comments

Conjecture: For every natural number k there are infinitely many semiprime pairs sp and sp' both sandwiched between semiprimes such that sp' - sp = 4k.

Note: For the case k=1 the pair of two consecutive semiprime triples will be called twin semiprime-triple with an analogy to twin prime.

The number centered between the triples of a twin semiprime-triple must be divisible by 36. Let m be the middle of such a twin semiprime-triple. It is trivial that m is divisible by 4, and that it is congruent to 0, 4, or 5 (mod 9). If it were congruent to 4, then m-1 and m+2 would both be divisible by 3, hence equal to 3 times a prime. But then those two primes would differ by 1, impossible except for primes 2, 3, which can be checked separately. A similar argument eliminates the case m == 5 (mod 9), so m must be divisible by 9. Conjecture by the author, proved by Franklin T. Adams-Watters, Dec 18 2011.

Members of this sequence must be twice the lesser of a twin prime. - Franklin T. Adams-Watters, Dec 18 2011

A number is in the sequence if and only if it has the form 6k-2, with 2k+-1 being twin primes, 3k+-1 twin primes, and 6k+-1 semiprimes. - Peter Munn, Oct 28 2017

By arguments similar to the above proof that m = a(n)+2 is divisible by 36, it can be shown that (a(n)+2)/36 == {-1, 0, 1} (mod 5) == {-1, 0, 1} (mod 7) and that a(n) == {214, 502, 538, 718, 754, 1042, 1258} (mod 1260). - Jon E. Schoenfield, Feb 26 2022

Examples

			The first twin semiprime-triple is {{213,214,215},{217,218,219}} whereby a(1)=214.
The fifteenth semiprime-triple is {{1891761,1891762,1891763},{1891765,1891766,1891767}} whereby a(15)=1891762.
The separating numbers 216 and 1891764 are divisible by 36.
		

Links

• Donovan Johnson, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A086005, A056809, A001359 Analogs: a(n) and A001359, a(n)+2 and A014574, a(n)+4 and A006512.

Programs

Mathematica

(* Run first to define sp *) nn = 10^8; p = Prime[Range[PrimePi[nn/2]]]; lim = Floor[Sqrt[nn]]; sp = {}; k = 0; While[k++; p[[k]] <= lim, sp = Join[sp, p[[k]]*Take[p, {k, PrimePi[nn/p[[k]]]}]]]; sp = Sort[sp]; lsp = Length[sp]; b = {}; Do[If[sp[[n]] == sp[[n - 1]] + 1 && sp[[n]] == sp[[n + 1]] - 1, AppendTo[b, sp[[n - 1]]]], {n, 2, lsp-1}]; lb = Length[b]; triples = {}; Do[If[b[[i]] == b[[i - 1]] + 4, AppendTo[triples, b[[i - 1]]]], {i, 2, lb}]; triples+1 (* V.J.Pohjola, Dec 18 2011 *)
SequencePosition[PrimeOmega[Range[5200000]],{2,2,2,,2,2,2}][[All,1]]+1 (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale, Jul 06 2017 *)