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A243937 Even numbers n>=6 for which lpf(n-1) > lpf(n-3), where lpf = least prime factor.

%I A243937 #30 Sep 03 2014 10:29:22
%S A243937 6,8,12,14,18,20,24,30,32,36,38,42,44,48,54,60,62,66,68,72,74,78,80,
%T A243937 84,90,96,98,102,104,108,110,114,120,122,126,128,132,138,140,144,150,
%U A243937 152,156,158,162,164,168,174,180,182,186,188,192,194,198,200,204,210
%N A243937 Even numbers n>=6 for which lpf(n-1) > lpf(n-3), where lpf = least prime factor.
%C A243937 Complement of A245024 over even n >= 6.
%C A243937 Conjecture: All differences are 2, 4 or 6 such that there are no two consecutive terms 2 (..., 2, 2, ...), no two consecutive terms 4, while consecutive terms 6 occur 1, 2, 3 or 4 times; also consecutive pairs of terms 2, 4 appear 1, 2, 3 or 4 times. The conjecture is verified up to n = 2.5*10^7. - _Vladimir Shevelev_ and _Peter J. C. Moses_, Jul 11 2014
%C A243937 Divisibility by 3 means 6m is in the sequence for all m > 0, and 6m + 4 never is, while 6m + 2 is undetermined. Divisibility by 5 means 30m + 8 is always in the sequence, and 30m + 26 never is. This proves the above conjecture. - _Jens Kruse Andersen_, Aug 19 2014
%C A243937 Note that,
%C A243937 1) Since numbers of the form 6*k evidently are in the sequence, then the counting function of the terms not exceeding x is not less than x/6.
%C A243937 2) Sequence {a(n)-1} contains all primes greater than 3 in the natural order. The subsequence of other terms of {a(n)-1} is 35, 65, 77, 95, ... - _Vladimir Shevelev_, Jul 15 2014
%H A243937 Jens Kruse Andersen, <a href="/A243937/b243937.txt">Table of n, a(n) for n = 1..10000</a>
%o A243937 (PARI) select(n->factor(n-1)[1,1]>factor(n-3)[1,1], vector(200, x, 2*x+4)) \\ _Jens Kruse Andersen_, Aug 19 2014
%Y A243937 Cf. A242719, A242720, A245024.
%K A243937 nonn
%O A243937 1,1
%A A243937 _Vladimir Shevelev_, Jul 10 2014
%E A243937 More terms from _Peter J. C. Moses_, Jul 10 2014