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%I #24 Dec 05 2021 10:46:14
%S 3,5,6,7,11,12,13,17,19,20,23,29,31,35,37,41,43,47,53,59,61,63,67,71,
%T 73,79,83,89,97,99,101,103,107,109,113,127,131,137,139,143,149,151,
%U 157,163,167,173,179,181,191,193,197,199,208,211,223,227,229,233,239
%N Start with b(0)=4. a(n) is the smallest integer such that b(n)=a(n)*b(n-1) has twice as many divisors as b(n-1) has.
%C If b(0)=4 is replaced by b(0)=1 we get numbers of the form p^(2^k) where p is prime; that is, A050376. The number of divisors of 4*the n-th partial product of sequence terms is 3*2^n.
%C Conjecture: a composite number is in this sequence if and only if it is the product of 2 consecutive terms of A050376.
%e 4 has 3 divisors. a(1) must be a number by which you can multiply 4 to get a number with 3*2=6 divisors. It can't be 2 because 4*2=8 has 4 divisors. It can be 3 because 4*3=12 has 6 divisors.
%o (PARI) lista(nn) = {my(b = 4); for (n=1, nn, my(a=1,d=numdiv(b)); while(numdiv(a*b) != 2*d, a++); print1(a, ", "); b = b*a;);} \\ _Michel Marcus_, Dec 05 2021
%Y Cf. A050376.
%K nonn
%O 1,1
%A _J. Lowell_, Dec 04 2021
%E More terms from _Jon E. Schoenfield_, Dec 04 2021
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