[seqfan] Re: Do all numbers of A174658 occur as balanced ternary reinterpretations of mu(d) sequences?

franktaw at netscape.net franktaw at netscape.net
Sat Feb 26 03:29:19 CET 2011

Certainly not all members of A174658 can occur. The divisors of n > 1 
always start (1, p), and so the mu values start (1,-1). Thus, for 
example, 8 (1,0,-1) and 32 (1,1,-1,-1) cannot occur.

There are other sequences that can't occur; for example, we can't start 
with 1,-1,1. And any sequence that includes a 0 must end with 0.

Yet further; if the number of prime divisors omega(n) = A001221(n) = k 
> 0, the number of 1's and the number of -1's will each be 2^(k-1).

It would be interesting to see a complete description of what sequences 
can be obtained in this way.

Franklin T. Adams-Watters

-----Original Message-----
From: Alonso Del Arte <alonso.delarte at gmail.com>

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 
we reinterpret it as a balanced ternary representation? From the 
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 
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 
considered d in descending order. Thus, a(p) = -2, but not all other 
are terms of the other sequence multiplied by -1. For example, a(4) = 6 
the former but -2 in the latter. However, a good reason to prefer 
order is that non-squarefree numbers don't give sequences with leading

The Mathematica program I'm using is this:

 fromBalTernDigits[digitList_/;Complement[digitList, {-1, 0, 1}] == {}] 
Plus@@Table[Reverse[digitList][[n]] * 3^(n - 1), {n, 
Table[fromBalTernDigits[MoebiusMu[Divisors[n]]], {n, 50}]

(and of course you just stick in a Reverse command to consider the 
mu(d) with d in descending order).


P. S. In regards to my question a few weeks ago of whether 1 is 
I think the reason I like the best is that it enables us to say "If a 
is squarefree, then so are all its divisors."


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