Subdivide-Sum Sequence
Python, 82 78
def f(b):G=b^b&amp;b-1;return sum(b**(b/G-i-1)*(G*i+(G-1)*b/2)for i in range(b/G))

Huh?

The number of digit groups that the subdivison yields, G, is simply the greatest power of two that divides the number of digits (i.e. the base), b.
It's given by G = b ^ (b &amp; (b - 1)), where ^ is bitwise-XOR.
If you're familiar with the fact that n is a power of two iff n &amp; (n - 1) = 0 then it should be pretty easy to see why.
Otherwise, work out a few cases (in binary) and it'll become clear.

The number of digits per group, g, is simply b / G.

The first digit group, 012...(g-1), as a number in base b, is .

The next group, g(g+1)...(2g-1), as a number in base b, is the sum
.

More generally, the n-th group (zero-based), as a number in base b, an, is .

Recall that there are G groups, hence the sum of all groups is



which is what the program calculates.