Calculate the longest series of 1's in an integer's binary value
JavaScript (ES6),  33  32 bytes
f=x=&gt;x&amp;(k=x&gt;&gt;1)?f(x&amp;k):k&amp;&amp;1+f(k)

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JavaScript (ES6),  41 40 36  34 bytes
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f=x=&gt;(k=x&amp;x/2)?f(k):Math.log2(x)|0

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How?
General case x &gt; 0
We recursively AND the input x with x / 2 which progressively reduces the patterns of consecutive set bits until only the rightmost bit of the sequence remains. We keep a copy of the last non-zero value and determine the position of its most significant bit by floor-rounding its base-2 logarithm.
Below are the steps we go through for x = 750 (1011101110 in binary).
    1011101110 --.
,----------------'
'-&gt; 1011101110
AND 0101110111
    ----------
=   0001100110 --.
,----------------'
'-&gt; 0001100110
AND 0000110011
    ----------
=   0000100010 --.  --&gt; output = position of MSB = 5  (0000100010)
,----------------'                                         ^
'-&gt; 0000100010
AND 0000010001
    ----------
=   0000000000

Edge case x = 0
The special case x = 0 immediately leads to Math.log2(0) | 0. The logarithm of 0 evaluates to -Infinity and the binary bitwise OR forces a coercion to a 32-bit integer. In compliance with the specification of the abstract operation ToInt32, this gives the expected 0:

If number is NaN, +0, −0, +∞, or −∞, return
+0