Currently, the code uses Py_ABS(Py_SIZE(x))*PyLong_SHIFT to estimate the upper-bound of the number of bits of the number x. It's a raw estimation, the difference can be up to 29 bits. We may try to compute the exact number of bits, x.bit_length().

Python 3.5 estimate the number of "decimalbase" (10**9) digits using:

def decimalbase_digits1(x):
    bits = size(x) * PyLong_SHIFT
    return 1 + bits // (3 * _PyLong_DECIMAL_SHIFT)

I wrote a test to compute how many digits are overallocated (and unused): 552961 for this function. I'm not sure that "1+" is needed, since 3.0 is already lower than log2(10) (3.32...). If we compute the exact number of bits using the Python 3.5 function, it's a little bit better:

def decimalbase_digits2(x):
    bits = x.bit_length()
    return 1 + bits // (3 * _PyLong_DECIMAL_SHIFT)

=> 546250 digits (1% less). You propose a better estimation:

def decimalbase_digits3(x):
    digits = size(x)
    d = (33 * _PyLong_DECIMAL_SHIFT) // (10 * PyLong_SHIFT - 33 * _PyLong_DECIMAL_SHIFT)
    return 1 + digits + digits // d

With your estimation, only 504243 are overallocated (9% less than Python 3.5 function). But why only using 2 digits for log2(10) estimation? We can more digits:

def decimalbase_digits4(x):
    bits = size(x) * PyLong_SHIFT
    return bits * 10000 // (33219 * _PyLong_DECIMAL_SHIFT)

=> 491908 digits (11% less)

According to my tests, the best function uses the number of bits and the better estimation of log2(10):

def new_decimalbase_digits5(x):
    bits = x.bit_length()
    return bits * 10000 // (33219 * _PyLong_DECIMAL_SHIFT)

=> 483424 digits (13% less)


See attached for my tests.