product k^mu(n+1-k)

[Max Alekseyev]

On 5/21/07, Max Alekseyev <maxale at gmail.com> wrote:

> btw, for a similar sequence
> b(n) = product{k=1 to n} k^mu(k) = A130086(n) / A130087(n)
> I can prove that there is no such n (i.e., all b(n) are distinct).

Oops, this statement is obviously wrong since b(n)=b(n+1) as soon as
n+1 is not square-free.
I may have a proof that this is the only possible case, meaning that
for all n<m, either b(n)=b(n+1)=b(n+2)=...=b(m); or b(n) <> b(m).
But I did not carefully check all the details.

Max