MathLabs

Problem 3

A function ff on positive integers is defined by f(1)=1f(1)=1, f(3)=3f(3)=3, f(2n)=f(n)f(2n)=f(n), f(4n+1)=2f(2n+1)−f(n)f(4n+1)=2f(2n+1)-f(n), and f(4n+3)=3f(2n+1)−2f(n)f(4n+3)=3f(2n+1)-2f(n). Determine the number of positive integers n≤1988n\le1988 for which f(n)=nf(n)=n.
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32−2=30,62+30=9232-2=30,\quad62+30=92
Detailed analysis

There are 32 fixed odd 11-bit numbers through 2047. The two exceeding 1988 are 2015 and 2047, leaving 30; hence 62+30=92.