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A242059 lpf_3(A242057(n)-1), where lpf_3(n) = lpf(n/3^t) (cf. A020639) such that 3^t (t>=0) is the maximal power of 3 which divides n.

%I A242059 #18 Aug 20 2014 02:32:31
%S A242059 1,5,7,5,1,11,5,13,5,7,5,7,5,23,5,1,5,7,5,11,5,37,5,7,5,43,7,5,47,11,
%T A242059 5,17,5,53,7,5,13,5,61,5,7,5,67,7,5,11,71,5,13,5,7,5,1,11,5,7,5,7,5,
%U A242059 31,5,5,7,5,103,5,11,17,5,7,37,5,113,11,7,5,13,5
%N A242059 lpf_3(A242057(n)-1), where lpf_3(n) = lpf(n/3^t) (cf. A020639) such that 3^t (t>=0) is the maximal power of 3 which divides n.
%C A242059 An analog of A242033.
%H A242059 Jens Kruse Andersen, <a href="/A242059/b242059.txt">Table of n, a(n) for n = 1..10000</a>
%t A242059 lpf[n_]:=lpf[n]=First[First[FactorInteger[n]]];
%t A242059 lpf3[n_]:=lpf3[n]=If[#==1,1,lpf[#]]&[n/3^IntegerExponent[n,3]]
%t A242059 Map[lpf3[#-1]&,Select[Range[4,300,2],lpf3[#-1]<lpf3[#-3]&]](* _Peter J. C. Moses_, Aug 13 2014 *)
%o A242059 (PARI) lpf3(n)=m=n/3^valuation(n, 3); if(m>1, factor(m)[1,1], 1)
%o A242059 apply(n->lpf3(n-1), select(n->lpf3(n-1)<lpf3(n-3), vector(200, x, 2*x))) \\ _Jens Kruse Andersen_, Aug 19 2014
%Y A242059 Cf. A245024, A243937, A242057, A242058, A242033, A242034, A020639.
%K A242059 nonn
%O A242059 1,2
%A A242059 _Vladimir Shevelev_, Aug 13 2014
%E A242059 More terms from _Peter J. C. Moses_, Aug 13 2014