
COMMENTS

Conjecture: this sequence is finite.
Comments from Owen Whitby, Jul 10 2008 (Start): If 2,3,...,q are all of the primes <= q (q=19 is sufficient for n<=22) and if a=2^a2*3^a3*...*q^aq, where ai>=0 for all i, then f(a)=(a2+1)(a3+1)...(aq+1) is the number of divisors of a and similarly f(k)=(k2+1)(k3+1)...(kq+1).
Hence f(k*a)=r(a;k)*f(a) where r(a;k)=[(a2+k2+1)(a3+k3+1)...(aq+kq+1)]/[(a2+1)(a3+1)...(aq+1)]. For a(n) find a2,a3,...,aq to minimize a (the product of the prime powers) while satisfying each of the inequalities r(a;k)>r(a;k1) for k=2,3,...,n.
After simplification, each inequality involves only a small number of ai and examining the inequalities sequentially is fairly tractable up to at least n=20.
Number of digits in a(1) to a(20) is 1,1,1,2,2,3,3,12,12,12,12,14,14,39,51,51,51,66,66,120. a(21), a(22), a(23) exist and are <= 5.3 10^128. (End)


EXAMPLE

720 is valid for a(6) because the number of divisors for the first 6 multiples of 720 are 30, 36, 40, 42, 45, 48
