A136799 Last term in a sequence of at least 3 consecutive composite integers.

10, 16, 22, 28, 36, 40, 46, 52, 58, 66, 70, 78, 82, 88, 96, 100, 106, 112, 126, 130, 136, 148, 156, 162, 166, 172, 178, 190, 196, 210, 222, 226, 232, 238, 250, 256, 262, 268, 276, 280, 292, 306, 310, 316, 330, 336, 346, 352, 358, 366, 372, 378, 382, 388, 396

Offset: 1

Author

Enoch Haga, Jan 21 2008

Keywords

• easy
• nonn

Comments

An equivalent definition is "Last term in a sequence of at least 2 consecutive composite integers". - Jon E. Schoenfield, Dec 04 2017

The BASIC program below is useful in testing Grimm's Conjecture, subject of Carlos Rivera's Puzzle 430

Use the program with lines 30 and 70 enabled in the first run and then disabled with lines 31 and 71 enabled in the second run.

Composite numbers m such that m-1 is composite, and m+1 is not. - Martin Michael Musatov, Oct 24 2017

Examples

			a(1)=10 because 10 is the last term in a run of three composites beginning with 8 and ending with 10 (8,9,10).
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Carlos Rivera, Puzzle 430, Grimm's Conjecture, Prime puzzles and problems connection.

Crossrefs

Cf. A136798, A136800, A136801.

Programs

Magma

[p-1: p in PrimesInInterval(4, 420) | not IsPrime(p - 2)]; // Vincenzo Librandi, Apr 11 2019

Mathematica

Select[Prime@ Range@ 78, CompositeQ[# - 2] &] - 1 (* Michael De Vlieger, Oct 23 2015, after PARI *)

PARI

forprime(p=5, 1000, if(isprime(p-2)==0, print1(p-1, ", "))) \\ Altug Alkan, Oct 23 2015

UBASIC

10 'puzzle 430 (gap finder) 20 N=1 30 A=1:S=sqrt(N):print N; 31 'A=1:S=N\2:print N; 40 B=N\A 50 if B*A=N and B=prmdiv(B) then print B; 60 A=A+1 70 if A<=sqrt(N) then 40 71 'if A<=N\2 then 40 80 C=C+1:print C 90 N=N+1: if N=prmdiv(N) then C=0:print:stop:goto 90:else 30

Formula

a(n) = A025584(n+2) - 1. - R. J. Mathar, Jan 24 2008

a(n) ~ n log n. - Charles R Greathouse IV, Oct 27 2015

Extensions

Edited by R. J. Mathar, May 27 2009

a(53) corrected by Bill McEachen, Oct 27 2015