The error diminishes faster than my english marks

We can look at all the possible choices of positive $a, b<50$ that can result in the 2nd largest number 3325 and see what is the next number. The $9$ th to $14$ th Fibonacci numbers are $34, 55, 89, 144, 233, 377$

$34a+55b=34\cdot25+55\cdot45=3325$
$55\cdot25+89\cdot45=5580$

$55a+89b=55\cdot20+89\cdot25=3325$
$89\cdot20+144\cdot25=5580$

$89a+144b=89\cdot5+144\cdot20=3325$
$144\cdot5+233\cdot20=5580$

$144a+233b=144\cdot15+233\cdot5=3325$
$233\cdot15+377\cdot5=5580$

Thus, Houdini had all the bases covered. $5580$ is the only possible largest number, given the conditions.

Let us list out the terms in Khalifa's list in increasing order: a1, a2, a3, a4, etc.

The terms of the sequence are defined by this relation: $a_{n}=a_{n-1}+a_{n-2}$

Finding the characteristic polynomial, we find that the general term is of the form: $a_{n}=a(\varphi )^n+b(1-\varphi )^n$ where phi represents the golden ratio.

Consider the ratio between $a_{n}$ and $a_{n-1}$ as n grows large: $\frac{a(\varphi )^n+b(1-\varphi )^n}{a(\varphi )^{n-1}+b(1-\varphi )^{n-1}}$

Because $\left |1-\varphi \right |$ is less than 1, $b(1-\varphi )^n$ and $b(1-\varphi )^{n-1}$ quickly converge to zero, so the ratio between $a_{n}$ and $a_{n-1}$ converges to $\varphi$ .

Houdini simply multiplied 3325 by $\varphi$ and rounded to the nearest integer (5380). Houdini is an A+ chad.