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%I #11 Jun 21 2019 12:16:00
%S 12,48,192,144,576,3072,12288,9216,196608,5184,786432,36864,12582912,
%T 589824,82944,2359296,805306368,3221225472,331776,37748736,
%U 206158430208,746496,3298534883328,5308416,13194139533312,2415919104
%N Smallest positive integer for which the number of divisors is a product of 2 distinct primes: Min{x; d[x]=pq}.
%C Least solutions of d(x)=A000005(x)=pq equation, where 1<p<q, primes: x=(2^q)*(3^p) where the exponent of smaller base[=2] is the larger prime factor[=q].
%C Since 2^(pq-1)>2^(q-1)*3^(p-1) -> [(2^q)/3]^(p-1) holds for q>1, p>1, therefore these solutions are in fact minimal.
%H Amiram Eldar, <a href="/A061148/b061148.txt">Table of n, a(n) for n = 1..563</a>
%F a(n) = A005179(A006881(n)).
%e If d(x)=253=11*23, then 1<p=11<q=23,q-1=22,p-1=10, 2^22=4194304, 3^10=59049 so the smallest number x, which has 253 divisors is 4194304*59049=247669456896.
%Y Cf. A000005, A006881, A005117, A005179.
%K nonn
%O 1,1
%A _Labos Elemer_, May 30 2001
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