SOLUTION TO PROBLEM 3

First we observe the following

Claim:        Perhaps after a non-repeating initial segment, the sequence $f_1(11),f_2(11),\ldots$ is periodic.

Proof of the claim.         Note that for k<1000 we have

$$f_1(k)\leq f_1(999)=(9+9+9)^2=729<1000.$$

Now we compute f_n(11) for the first few values of n, in the hope that the length of the period is short. This expectation is reasonable since the terms of the sequence are perfect squares, and since there are only 31 perfect squares less than 1000. We find that:

$$\begin{array}{l} f_1(11)=(1+1)^2=4,\\ f_2(11)=f_1(f_1(11))=f... ...^2=256,\\ f_6(11)=f_1(f_5(11))=f_1(256)=13^2=169. \end{array}$$

We stop at this point, since f_n(11) depends only on f_n-1(11), and hence the numbers 256 and 169 will continue to alternate. More precisely, for $n\geq 4$,

$$f_n(11)=\left\{\begin{array}{ll} 169 &\mbox{ if $n$\space is even,}\\ 256 &\mbox{ if $n$\space is odd.} \end{array}\right.$$

Since 2001 is odd, it follows that f₂₀₀₁(11)=256.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Wed Jan 24 19:30:18 CST 2001