Recursively Defined Sequence/Examples/Minimum over k of Maximum of 1 plus Function of k and 2 plus Function of n-k

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Theorem

Consider the integer sequence $\left\langle{f \left({n}\right)}\right\rangle$ defined recusrively as:

$f \left({n}\right) = \begin{cases} 0 & : n = 1 \\ \displaystyle \min_{0 \mathop < k \mathop < n} \max \left({1 + f \left({k}\right), 2 + f \left({n - k}\right)}\right) & : n > 1 \end{cases}$


$f \left({n}\right)$ has a closed-form expression:

$f \left({n}\right) = m$ for $F_m < n \le F_{m + 1}$

where $F_m$ denotes the $m$th Fibonacci number.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:

$f \left({n}\right) = m$ for $F_m < n \le F_{m + 1}$


Basis for the Induction

$P \left({1}\right)$ is the case:

\(\displaystyle 0\) \(=\) \(\displaystyle \)
\(\displaystyle F_0\) \(<\) \(\, \displaystyle 1 \, \) \(\, \displaystyle \le\, \) \(\displaystyle F_{0 + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\) Definition of Fibonacci Number: $F_1 = 1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle m\) \(=\) \(\displaystyle 0\)

Thus $P \left({1}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.


So this is the induction hypothesis:

$f \left({r}\right) = m$ for $F_m < r \le F_{m + 1}$


from which it is to be shown that:

$f \left({r + 1}\right) = m$ for $F_m < {r + 1} \le F_{m + 1}$


Induction Step

This is the induction step:



\(\displaystyle \) \(=\) \(\displaystyle \)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \)

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 1}: f \left({n}\right) = m$ for $F_m < n \le F_{m + 1}$

$\blacksquare$


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