WakeUp wrote: For a positive integer $k$, let $f_1(k)$ be the square of the sum of the digits of $k$. Define $f_{n+1}$ = $f_1 \circ f_n$ . Evaluate $f_{2007}(2^{2006} )$. Let $n(x)$ the number of digits of $x$ and $s(x)$ the sum of digits of $x$. $n(x)\le 1+\log_{10}(x)$ and so $s(x)\le 9n(x)=9(1+\log_{10}x)$ and $f(x)\le 81(1+\log_{10}x)^2$ So $f(2^{2006})\le 81(1+2006\log_{10} 2)^2$ $\le 81(1+\frac {2006}3)^2$ $<81\cdot 700^2 <10^8$ $f(f(2^{2006}))\le 81(1+8)^2 < 10^4$ $f(f(f(2^{2006}))) \le 81(1+4)^2=2025$ $f^{[4]}(2^{2006}) \le (1+9+9+9)^2= 784$ $f^{[5]}(2^{2006}) \le (6+9+9)^2= 576$ $f^{[6]}(2^{2006}) \le (4+9+9)^2= 484$ $f^{[7]}(2^{2006}) \le (3+9+9)^2= 441$ and this process enters then a loop on $441$ So we know that, from this point, $s(f^{[n]}(2^{2006}))\le 21$ But $2^{2006}\equiv 4\pmod 9$ and so it's clear that $s(f^{[n]}(2^{2006}))\equiv 4,7\pmod 9$ and so : $s(f^{[n]}(2^{2006}))\in\{4,7,13,16\}$ $\forall n\ge 7$ And since : $s(x)=4$ $\implies$ $f(x)=16$ whose sum of digits is $7$ $s(x)=7$ $\implies$ $f(x)=49$ whose sum of digits is $13$ $s(x)=13$ $\implies$ $f(x)=169$ whose sum of digits is $16$ $s(x)=16$ $\implies$ $f(x)=256$ whose sum of digits is $13$ we get that $f^{[n]}(2^{2006})\in\{169,256\}$ $\forall n\ge 10$ It remains to look at the values modulus $9$ to get that $f^{[2p]}(2^{2006})\equiv 4\pmod 9$ while $f^{[2p+1]}(2^{2006})\equiv 7\pmod 9$ And so $\boxed{f^{[2007]}(2^{2006})=169}$ since $2007$ is odd