Solve for $lambda(x): mathbb N to mathbb N$ given two cases.
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Solve for $lambda:Bbb NtoBbb N$ in $lambda(x)=begin{cases} x+1 & xin2Bbb N \ 2lambda(lfloor x/2rfloor) & xin2Bbb N+1 end{cases}$ I know that an answer is $lambda(x)=x+1$ , but is it the only one? We know it is true for $xin2Bbb N$ , but what about $xin 2Bbb N+1$ ? Let $x=2k+1$ for some $kin Bbb N$ . Then $lambda(2k+1)=2lambda(lfloor (2k+1)/2rfloor)=2lambda(lfloor k+1/2rfloor)=2lambda(k)$ But now we are in the same scenario, having to check the parity of $k$ to be able to progress. Is there a way to prove this, maybe inductively? Thanks.
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edited Dec 16 '18 at 22:58