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%I #19 Apr 11 2020 06:10:50
%S 3,6,8,9,12,15,18,21,24,27,30,33,36,39,40,42,45,48,51,54,56,57,60,63,
%T 64,66,69,72,75,78,81,84,87,88,90,93,96,99,102,104,105,108,111,114,
%U 117,120,123,125,126,129,132,135,136,138,141,144,147,150,152,153
%N List of positive integers whose prime tower factorization, as defined in comments, contains the prime 3.
%C This set is the complement of A182337.
%C The prime tower factorization of a number can be recursively defined as follows:
%C (0) The prime tower factorization of 1 is itself
%C (1) To find the prime tower factorization of an integer n>1, let n = p1^e1 * p2^e2 * ... * pk^ek be the usual prime factorization of n. Then the prime tower factorization is given by p1^(f1) * p2^(f2) * ... * pk^(fk), where fi is the prime tower factorization of ei.
%H Amiram Eldar, <a href="/A182338/b182338.txt">Table of n, a(n) for n = 1..10000</a>
%H Patrick Devlin and Edinah Gnang, <a href="http://arxiv.org/abs/1204.5251">Primes Appearing in Prime Tower Factorization</a>, arXiv:1204.5251v1 [math.NT], 2012-2014.
%p # The integer n is in this sequence if and only if
%p # containsPrimeInTower(3, n) returns true
%p containsPrimeInTower:=proc(q, n) local i, L, currentExponent; option remember;
%p if n <= 1 then return false: end if;
%p if type(n/q, integer) then return true: end if;
%p L := ifactors(n)[2];
%p for i to nops(L) do currentExponent := L[i][2];
%p if containsPrimeInTower(q, currentExponent) then return true: end if
%p end do;
%p return false:
%p end proc:
%p select(x-> containsPrimeInTower(3,x), [$1..160])[];
%t indic[1] = 1; indic[n_] := indic[n] = Switch[f = FactorInteger[n], {{3, _}}, 0, {{_, _}}, indic[f[[1, 2]]], _, Times @@ (indic /@ (Power @@@ f))];
%t Select[Range[200], indic[#] != 1&] (* _Jean-François Alcover_, Jul 11 2018 *)
%Y Complement of A182337. Cf. A182318.
%K nonn
%O 1,1
%A _Patrick Devlin_, Apr 25 2012
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