I think that rather depends on what 'max' one is looking for.

We were talking about the size of the largest range of integers that can accurately be represented by double-precision floats. That's given by floor(log2($upper-$lower+1)). And in this case, the upper bound of that range is 2**53. If the range was -1000..64535, then answer would be 16, not 9 or 15.

This is then rounded to even.

I didn't follow what you said in that paragraph, but there's no rounding involved in storing 2**53 in a float. It's stored precisely (Sign = 0, Exponent = 53, Mantissa = 0).

In reply to Re^8: 64-bit digest algorithms by ikegami
in thread 64-bit digest algorithms by BrowserUk

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