A072507

[David Wilson]

a(10) = 3.

Proof:

The smallest exemplar showing a(10) >= 3 is 7939375.

To show a(10) <= 3, suppose there were 4 numbers n in a row with d(n) = 10.

Two of these numbers, a and b = a+2, would be even, with d(a) = d(b) = 10.

Since d(a) = d(b) = 10, a and b must each be of the form p^9 or pq^4 with p 
and q prime.

If either a or b is of the form p^9, then, being even, it must be equal to 
2^9 = 512. But then the other does not have 10 divisors.

So a and b are both of the form pq^4.

a and b are even, so they must be of the form 2p^4 or 16p, with p and q odd 
primes.

This implies that a and b must each be congruent to 2 or 16 modulo 32.

This is inconsistent with a+2 = b.

Hence, no two consecutive even integers n have d(n) = 10.

Therefore, at most three consecutive integers n have d(n) = 10.

This gives a(10) <= 3, so a(10) = 3.

Maybe this argument could be generalized for a(2p)?