OFFSET
1,1
COMMENTS
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.
Conjecture: a composite number is in this sequence if and only if it is the product of 2 consecutive terms of A050376.
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
J. Lowell, Dec 04 2021
EXTENSIONS
More terms from Jon E. Schoenfield, Dec 04 2021
STATUS
approved