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⋅2⋅7⋅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

limn→∞number of such integers ≥2≤nxn=f(x)

or find c such that

limn→∞number of such integers ≥2≤nn=c.My guess is that,

in the latter case,

c=0 or 1,

but I have no idea which is true.

Maybe its

1−1e.Note: I have modified this

to not allow 1 as a divisor.

**Answer**

Every even number has consecutive factors: 1 and 2.

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

The probability is 1/2.

**Attribution***Source : Link , Question Author : marty cohen , Answer Author : ajotatxe*