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

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=22713,
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
limnnumber of such integers 2nxn=f(x)
or find c such that
limnnumber of such integers 2nn=c.

My guess is that,
in the latter case,
c=0 or 1,
but I have no idea which is true.
Maybe its
11e.

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

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