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