Fwd: Ternary analogue of A094913?

[Maximilian Hasler]

On Dec 14, 2007 8:49 AM, David Wilson <davidwwilson at comcast.net> wrote:
> The sequence in question is not about binary number, it is a sequence about
> strings over a 2-character alphabet encoded as binary numbers, and does not
> concern the numerical values of the elements.  It is more akin to "necklaces
> having beads of 2 colors".

of course, cf an earlier reply of myself, and I don't have anything
against the binary (or 2-color) version, but if we go on to base b=3
(or b=3 colours) then we have to do it for
any number of colours b>2.
However, I think this is already done, if the formula I gave earlier
(Dec 11, 2007 8:48 AM) is correct:

A094913(n,base=2)=sum(k=1,n,min(base^k,n-k+1))

Splitting up the sum in 2 parts to get rid of the min() allows easily
to get an explicit analytic expression for the n-the term of the
sequence for any 'base' (= number of colours).

Maximilian