Doepfer

Hi Dieter

> If I understand your problem it could be solved by the A-161 and the A-186-1
> as the A-161 has a separate gate output for each step available. To solve
> the 000/111 problem simply use output #2 of the A-161 as the first gate. The
> outputs of the A-161 can be combined by the A-186-1.
No, This only works for one stage:

Step 1 2 3 4 5 6 7 8
  1/1 .1.1.1.1.1.1.1.1.1.1.1.
  1/2 ...11..11..11..11..11..

Now you invert 1/2 and AND it with 1/1

Step 1 2 3 4 5 6 7 8
  1/1 .1.1.1.1.1.1.1.1.1.1.1.
  1/2 .11..11..11..11..11..11
      .1...1...1...1...1...1. = 1AND2
Looks perfect. But only as long as you do not proceed with 1/4:

Original
Step 1 2 3 4 5 6 7 8
  1/1 .1.1.1.1.1.1.1.1.1.1.1.
  1/2 ...11..11..11..11..11..
  1/4 .......1111....1111....

Now replace 1/2 by 1AND2 and invert 1/4 and do 2AND4

Step 1 2 3 4 5 6 7 8
  1/1 .1.1.1.1.1.1.1.1.1.1.1.
      .1...1...1...1...1...1. = 1AND2
  1/4 .111111....1111....1111
      .1...1.......1.......1. = 2AND4

As you can see, the 1/4 starts with two 1/2-Steps and then goes to 1/4. 
This is because the RS-Flopflop in the counter is clocked by Q instead of /Q

The dividers are like
clock---S  Q---S  Q---...
         R /Q   R /Q

But we would need something like:
clock---S  Q +-S  Q +-...
         R /Q-+ R /Q-+

Florian