>>This leads to the total of
>
>>2·2(n - 1) + 1 = 4n - 3,
>
>>which is less than your estimate.
>
>I don't agree with you.
>In your case you need one
>more extra call.

What for?

>And the
>more groups you divide you'll
>need more extra calls.

To see how it works, try small numbers like 4, 8, etc.

For 4 you have two groups: 2 and 2. The first fellow in each group makes a call to the other fellow. In the notations of the original poster:

1, 2, 3, 4
12, 12, 34, 34

Then the first fellows talk to each other:

1234, 12, 1234, 34

Then it takes 2 more intragroup calls:

1234, 1234, 1234, 1234

In all, 5 calls for n = 4. Your estimate would give 2(4 - 1) = 6 calls, right?

>2n-2 is less than 4n-3 or
>8n-5.
>
>Am I right??

Yes, of course, but this is irrelevant here. n is different from expression to expression. In words your formula says "Twice the number minus 1." If the number is n, then the result is 2(n - 1). If the number is 2n, the result is 2(2n - 1), and so on.

In the previous post, I considered two cases of the number being 2n and 4n.