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

Question

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

This question occurred to me while trying to figure out why there are 7 days in a week.

I looked at 364, the number of days closest to a year (there are about 364.2422 days in a year, iirc). Since $364 = 2\cdot 2 \cdot 7 \cdot 13$, the number of possible number that evenly divide a year are 2, 4, 7, 13, 14, 26, 28, and larger.

Given this, 7 looks reasonable - 2 and 4 are too short and 13 is too long.

Anyway, I noticed that 13 and 14 are there, and wondered how often this happens.

I wasn't able to figure out a nice way to specify the probability (as in a Hardy-Littlewood product), and wasn't able to do it from the inverse direction (i.e., sort of a sieve with n(n+1) going into the array of integers).

Ideally, I would like an asymptotic function f(x) such that $\lim_{n \to \infty} \dfrac{\text{number of such integers } \ge 2 \le nx}{n} =f(x) $ or find $c$ such that $\lim_{n \to \infty} \dfrac{\text{number of such integers } \ge 2 \le n}{n} =c $.

My guess is that, in the latter case, $c = 0$ or 1, but I have no idea which is true. Maybe its $1-\frac1{e}$.

Note: I have modified this to not allow 1 as a divisor.

Answer 1

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

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

The probability is $1/2$.

Answer 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.