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

[Leroy Quet]

I went ahead and submitted the first 10 terms of
this sequence, as calculated by Maximilian
Hasler..
I assume that the question as to whether EVERY
positive integer n leads to a solution is an open
problem. (Or did I miss something in the
discussion?)
I forgot to add the comment that there is no
solution for any ODD positive integer n to
d(m)=d(m+n) = n.

Maybe someone else should submit the
d(m)=d(m+n)=2n sequence, since I myself did not
come up with that variation.

When my sequence below finally appears on the
EIS, I encourage anyone who wants to post
extensions/comments to do so.

Thanks always,
Leroy Quet

PS: It has been an unusually long time since I
posted this sequence for me to not yet have
received an automated reply.
Are things at the on-line EIS working okay? (It
IS Sunday, one of the days things tend to break
down, since those who would fix any problems have
taken the weekend off...)

%I A139416
%S A139416 3, 6, 12, 70, 600281, 60, 1458, 264,
450, 266875
%N A139416 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.
%C A139416 Does this sequence have a term for
every positive integer n, or are there no
solutions for some n?

First 10 terms calculated by Maximilian Hasler.
%e A139416 For a(4) we want the smallest integer
m such that d(m) = d(m+8) = 8. The positive
integers that have 8 divisors each form the
sequence: 24, 30, 40, 42, 54, 56, 66, 70, 78, 88,
102, 104, 105, 110,...(A030626)
The first (not necessarily adjacent) pair of
integers with 8 divisors each that is separated
by exactly 8 is (70,78). So a(4) is the least
element of this pair, which is 70.
%O A139416 1
%K A139416 ,more,nonn,




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