[seqfan] Re: 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 19:51:35 CET 2011
You're absolutely right, Franklin, I should've realized a prime divisor
would always come second with d in ascending order. Any integer of the form
3^x – 3^y with x > y + 1 will therefore be in A174658 but not come up in
this reinterpretation. Likewise numbers of the form 3^x – 3^y + z where z <
3^y – 3^(y – 1) — that last bound might be a little off, but you get the
idea.
Al
On Fri, Feb 25, 2011 at 9:29 PM, <franktaw at netscape.net> wrote:
> 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 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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