Why does Java's hashCode() in String use 31 as a multiplier?

Question

Per the Java documentation, the hash code for a String object is computed as:

s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
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Answer 1

By multiplying, bits are shifted to the left. This uses more of the available space of hash codes, reducing collisions.

By not using a power of two, the lower-order, rightmost bits are populated as well, to be mixed with the next piece of data going into the hash.

The expression n * 31 is equivalent to (n << 5) - n.