So f(4)=4.

(2) Take an even number x∈pn and divide x by 2. If the quotient is still even, divide it by 2 ... After k times, the quotient must be odd. At this time, the quotient is m,

So x=m? 2k, where m is an odd number and k∈N*

According to the conditions, if m∈A, then x∈A,? K is an even number.

If m? A, then x∈A? K is an odd number

So whether X belongs to A depends on whether M belongs to A, and let Qn be the set of all odd numbers in Pn.

So f(n) is equal to the number of subsets of Qn, and when n is even (or odd), the number of odd numbers in Pn is (or).

∴

(2) Take an even number x∈pn and divide x by 2. If the quotient is still even, divide it by 2...k times, and the quotient will be odd. At this point, the quotient is m, so x=m? 2 k, where m is an odd number, and k∈N* is known from the conditions. If m∈A, then x∈A,? K is an even number. If m? A, then x∈A? K is an odd number, so whether X belongs to A depends on whether M belongs to A. Let Qn be the set of all odd numbers in Pn, so f(n) is equal to the number of subsets of Qn. When n is even (or odd), the number of odd numbers in Pn is n/2 (or n+ 1/2).