Question: ( 1 0 pts ) [ A simplification of Exercise 3 6 , Page 5 3 8 ] The Josephus survival number f ( n

(10 pts)[A simplification of Exercise 36, Page 538] The Josephus survival number f(n) for n in Z^(+) is defined recursively by f(1)=1 and
f(n)=2f([n/2])-(-1)^(n)
for n>=2. Prove by induction that
f(n)=2(n-2^(|logn|))+1
( 1 0 pts ) [ A simplification of Exercise 3 6 ,

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