What proportion of positive integers have two factors that differ by 1?

Every even number has consecutive factors: $1$ and $2$.

No odd number has, because all its factors are odd.

The probability is $1/2$.

What kind of numbers have this property?

• All multiples of 6 (because 6 = 2 × 3). So that's 1/6 of the integers.
• All multiples of 12 (12 = 3 × 4), but these have already been counted as multiples of 6.
• All multiples of 20 (20 = 4 × 5), so add 1/20 of the integers. But we've double-counted multiples of 60 (LCD of 6 and 20), so subtract 1/60. This gives us 1/6 + 1/20 - 1/60 = 1/5.
• All multiples of 30 (5 × 6) or 42 (6 × 7), but again, these have already been counted as multiples of 6.
• All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105.
• All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6.
• All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310.

Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what.