%I
%S 10,16,22,26,28,34,40,46,50,52,56,58,64,70,76,82,86,88,92,94,100,106,
%T 112,116,118,124,130,134,136,142,146,148,154,160,166,170,172,176,178,
%U 184,190,196,202,206,208,214,220,226,232,236,238,244,250,254,256,260
%N Even numbers n for which lpf(n1) < lpf(n3), where lpf = least prime factor.
%C By the definition, either a(n)==1 (mod 3) or, for every pair of primes (p,q), p>q>=3, a(n)==1 (mod p) and a(n) not==3 (mod q).
%C Conjecture: All differences are 2,4 or 6 such that 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 4,2 appear 1,2,3 or 4 times.
%C Conjecture is verified up to n = 2.5*10^7.  _Vladimir Shevelev_ and _Peter J. C. Moses_, Jul 11 2014
%C The first comment is wrong as stated. This would fix it: for every pair of primes (p,q), p>q>=3, if a(n)==1 (mod p) then a(n) not==3 (mod q). Divisibility by 3 means 6m+4 is in the sequence for all m>0, and 6m never is, while 6m+2 is undetermined. Divisibility by 5 means 30m+26 is always in the sequence, and 30m+8 never is. This proves the above conjecture.  _Jens Kruse Andersen_, Jul 13 2014
%C Note that the sequence {a(n)3} contains all odd primes, except for lesser primes in twin primes pairs (A001359). Other terms of {a(n)3} are 25,49,55,85,91,...  _Vladimir Shevelev_, Jul 15 2014
%H Amiram Eldar, <a href="/A245024/b245024.txt">Table of n, a(n) for n = 1..10000</a>
%p lpf:= n > min(numtheory:factorset(n)):
%p select(n > lpf(n1) < lpf(n3),[seq(2*k,k=3..1000)]); # _Robert Israel_, Jul 15 2014
%t lpf[n_] := FactorInteger[n][[1, 1]];
%t Reap[For[n = 6, n <= 300, n += 2, If[lpf[n1] < lpf[n3], Sow[n]]]][[2, 1]] (* _JeanFrançois Alcover_, Feb 25 2019 *)
%Y Cf. A020639, A242719, A242720, A243937.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, Jul 10 2014
%E More terms from _Peter J. C. Moses_, Jul 10 2014
