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A136799
Last term in a sequence of at least 3 consecutive composite integers.
7
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
1,1
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
LINKS
Carlos Rivera, Puzzle 430, Grimm's Conjecture, Prime puzzles and problems connection.
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
EXAMPLE
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).
MATHEMATICA
Select[Prime@ Range@ 78, CompositeQ[# - 2] &] - 1 (* Michael De Vlieger, Oct 23 2015, after PARI *)
PROG
(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
(PARI) forprime(p=5, 1000, if(isprime(p-2)==0, print1(p-1, ", "))) \\ Altug Alkan, Oct 23 2015
(Magma) [p-1: p in PrimesInInterval(4, 420) | not IsPrime(p - 2)]; // Vincenzo Librandi, Apr 11 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Enoch Haga, Jan 21 2008
EXTENSIONS
Edited by R. J. Mathar, May 27 2009
a(53) corrected by Bill McEachen, Oct 27 2015
STATUS
approved