We are going to define two functions
$\quad \psi: \{3,4,5,6, \dots \} \to \{4,5,6,7, \dots \}$
and
$\quad \rho: \{3,4,5,6, \dots \} \to \{4,5,6,7, \dots \}$
where $\psi$ is defined using an algorithmic specification while $\rho$ is specified using recursion.
The specification for $\psi$:
We begin by introducing an auxiliary expression,
$\quad \lambda(n,k) = n^2 -kn + 1$
The function $\psi$ maps $k \in \{3,4,5,6, \dots \}$ as follows,
For any $k \in \{3,5,6, \dots \}$ form the graph $C_k = \{\big(n, \lambda(n,k)\big) \mid n \ge k + 1\}$.
Find the smallest $N$ in the domain of $C_k$ such that for all $t \ge N$ the statement
$\quad [\lambda(t,k) = ab] \land [1 \lt a \lt t] \land [1 \lt b \lt t]$
is false.
Set $\psi(k) = N$.
The specification for $\rho$:
$\quad \rho(3) = 4$
$\quad \rho(4) = 6$
$\quad \rho(5) = 9$
For $k \ge 6$
$\quad \text{If } \rho(k-1) - \rho(k-2) \ne \rho(k-2) - \rho(k-3) \text{ then } \rho(k) = 2 * \rho(k-1) - \rho(k-2) $
$\quad \text{else } \rho(k) = 2 * \rho(k-1) - \rho(k-2) + 1$
Are these two functions are equal?
If yes provide a proof.
If not find a $k$ where $\psi(k) \ne \rho(k)$.
My work
The definition of $\psi$ came about as result of my (amateurish) foray into number theory; see
$\quad$For $n \ge 4$ find a factorization $n^2 - 3n + 1 = ab$ where $a \lt n$ and $b \lt n$.
The definition of $\psi$ is kind of involute/esoteric and I can't relate it to any work I've done before in mathematics, but the patterns it exhibits (including the internal algorithmic patterns) seem worth investigating and this is a good start.
I wasn't sure what tags to use for this so I chose only one, recreational-mathematics. If a specialist in some area of math thinks other tags should be included then they should edit this question and add them in.