Turning a linear sequence in a numpy array into a geometric progression sequence

You can use np.add.reduceat and int.bit_length.

Example:

# make example sequence
a = np.arange(100.0)

# form sums
np.add.reduceat(a,(1<<np.arange(a.size.bit_length()))-1)
# array([   0.,    3.,   18.,   84.,  360., 1488., 2997.])

This adds items 0; 1+2; 3+4+5+6; etc. which your example suggests is what you really want.

from math import log, floor
a = [12.0, 3.4, 9.2, 7.7, 4.9, 3.8, 6.9]
print([sum(a[2**i-1:2**(i+1)-1]) for i in range(floor(log(len(a), 2)) + 1)])

This outputs:

[12.0, 12.6, 23.3]

The indices used for the sums are:

[(0, 1), (1, 3), (3, 7), (7, 15), (15, 31), ...]

The last sum can be shortened when the length of the list isn't exact one less than a power of 2.

seq = [12.0, 3.4, 9.2, 7.7, 4.9, 3.8, 6.9]

r = 0       # this is the binary power (2^r)
g_seq = []  # list to contain your final answer
i = 0       # index of start position for summing at each iteration  

while i < len(seq):   # once i reaches the last element, stop looping
    n = 2**r          # number of elements to sum in each iteration
    j = i + n         # index of end position for summing
    r+= 1             # increment power for next iteration
    subseq = seq[i:j] # slice of your input sequence to be summed
    g_seq.append(sum(subseq))   # sum the slice and append to answer list
    i = j             # j becomes the i for the next iteration

print(g_seq)
--------------------
[12.0, 12.6, 23.300000000000004]

Edit: added comments