Sum Of (m/k), Over Some Divisors k

[Leroy Quet]

I wrote, in-part:

>....

>f(1) = 1;
>
>f(m) = sum{k|m, k not among {f(1),f(2),...,f(m-1)}}  (m/k)
>
>1, 1, 1, 3, 1,  4, 1, 5, 1,  6, 1, 7,...

>....


I just realized that this follows the simple pattern of
..., 1, k, 1, k+1, 1, k+2,  1,...

Sorry about that.

(I still think the c-sequence is interesting,  however.)
:)

thanks,
Leroy Quet