[seqfan] Do all numbers of A174658 occur as balanced ternary reinterpretations of mu(d) sequences?
Alonso Del Arte
alonso.delarte at gmail.com
Sat Feb 26 02:39:51 CET 2011
We all know that Sum_{d | n} mu(d) = 0 with the exception of n = 1.
What happens if we take the sequence of mu(d) but instead of adding it up,
we reinterpret it as a balanced ternary representation? From the foregoing,
we know that with the exception of a(1), all a(n) will be members of
A174658. Also, it is easy to see that a(p) = 2 since the sequence of mu(d)
is {1, -1}. But, can all terms of A174658 occur in this way, or is it
possible some terms of A174658 will never occur?
This is taking the list mu(d) with d sorted in ascending order. I have also
considered d in descending order. Thus, a(p) = -2, but not all other terms
are terms of the other sequence multiplied by -1. For example, a(4) = 6 in
the former but -2 in the latter. However, a good reason to prefer ascending
order is that non-squarefree numbers don't give sequences with leading
zeroes.
The Mathematica program I'm using is this:
fromBalTernDigits[digitList_/;Complement[digitList, {-1, 0, 1}] == {}] :=
Plus@@Table[Reverse[digitList][[n]] * 3^(n - 1), {n, Length[digitList]}];
Table[fromBalTernDigits[MoebiusMu[Divisors[n]]], {n, 50}]
(and of course you just stick in a Reverse command to consider the sequence
mu(d) with d in descending order).
Al
P. S. In regards to my question a few weeks ago of whether 1 is squarefree,
I think the reason I like the best is that it enables us to say "If a number
is squarefree, then so are all its divisors."
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