When d(m) = d(m+n) = n

[Leroy Quet]

I have yet to submit this sequence:

a(n) = smallest positive integer such that
d(a(n)) = d(a(n)+2n) = 2n,
where d(m) is the number of positive divisors of
m.

The sequence begins: 3,6,12,70,...

--
Aside: It is easily proved that there are no
positive integers m where
d(m) = d(m+n) = n, for n = any odd positive
integer.
Proof:
If d(m) is odd, then m is a square.
If two squares, m and m+n in this case, are
separated by n, then m must be <= ((n-1)/2)^2.
(For larger m, the gaps between successive
squares are > n.)
Now, the number of positive divisors of any m is
< 2*sqrt(m).
(Worst case scenario, every positive integer <=
sqrt(m) divides m, such as when m = 12.)
So the largest number of positive divisors of m
that is possible when m is <=  ((n-1)/2)^2 is
n-1. So it is not poissible for m to have n
divisors when n is odd.
QED.
--

Back to the original sequence (when the
difference and number of divisors is even).
Does this sequence contain a term for every
positive even difference/number-of-divisors?

If not, does the sequence have an infinite number
of terms, or are there only a finite number of n
where d(a(n)) = d(a(n)+2n) = 2n?


Thanks,
Leroy Quet



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