A number theory problem by Ilham Saiful Fauzi

First observe that $f(2^k)=2^{k-1}+2^k=3.2^{k-1}$

$f^2(2^k)=3.2^{k-2}+2^{k-1}$

Continuing like this it can be simplified as : $f^m(2^k)=2^{k-m}.(2 m+1)$

Now observe the following pattern :

$f(1)=2 \\ f^2(1)=3 \\ f^3(1)=4 \\ f^4(1)=6 \\ f^5(1)=5 \\ f^6(1)=8 \\ .... \\ f^{10}(1)=16 \\ .... \\ f^{15}(1)=32$

The powers of $2$ occurs at the values of terms of an series whose common difference is an A.P.The series is $(1,3,6,10,15,....)$ .Its general term is $\frac{n(n+1)}{2}$

Now $2016=32 \times 63=2^5 \times (2.31+1)$

From the above equation $2 m+1=63 \\ m=31 \\ k-m=5 \\ k=31+5=36$

So, we need $f^m(2^{36})$ .The value of $m=36.37/2=666$

Now $f^{666}(1)=2^{36}$

Now consider $f^{666+l}(1)=2^{36-l}.(2 l+1)$ .From the above factorization we get $l=31$

Now the equation becomes $f^{666+31}(1)=2^{36-31}.(2.31+1) \\ f^{697}(1)=2^5.63=2016$

So, the value of $k= \boxed{697}$

Nice question