10<=a(12)<=15. If there were 16 such consecutive integers, two would be consecutive multiples of 8. One would have the form 32p and the other the form 8q^2 with odd primes p and q; this implies that 8q^2 is congruent to 24 or 40 (mod 64), which is impossible. It is likely that a(12)=15; this would follow from Dickson's conjecture.
a(14)=3. If there were 4, two would be consecutive even numbers. One would have the form 64p and the other the form 2q^6 with odd primes p and q. Since 2q^6 == 2 (mod 16), this implies that 2q^6 = 64p+2, so p = (q^31)(q^3+1)/32 is prime, which is impossible.
a(16)=7. If there were 8, one would be congruent to 4 (mod 8), which is impossible.
Schinzel's conjecture H would imply that:
a(2p) = 3 for all prime p > 3;
a(2pq) = 3 for all primes p, q such that gcd(p1,q1) > 4;
a(6p) = 5 for all odd prime p;
a(n) = 7 for all n > 4 such that n is divisible by 4 and nondivisible by 3.  Vladimir Letsko, Jul 18 2016
99949636937406199604777509122843 starts a run of 13 consecutive integers each having 12 divisors. Therefore 13 <= a(12) <= 15.  Vladimir Letsko, Jul 07 2017
8100239725694207838698666538353341829610974940 starts a run of 18 consecutive integers each having 48 divisors. Therefore 18 <= a(48) <= 31.  Vladimir Letsko, Dec 24 2020
