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

## Contents

## 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 $F_m < n \le F_{m + 1}$

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $40$