I remember learning how to create a formula of form $a^x + b^y + c^z$ where the variables will create any other set of chosen numbers as the first terms of a sequence. But I don't remember how to do that anymore. How do I?
To generate a sequence like $1, 5, 6,$ or any three chosen numbers, how would I create a formula which outputs those three numbers as the first three terms of the sequence?
I will walk you most of the way there, but given that this is a pretty open-ended question, I will ask of you do to the rest.
Basically, for a function to output $n$ desired outputs, you need a formula of $n$ unknown variables over an independent variable (like $x$), then do some algebra to determine what they (the unknowns) need to be.
Suppose, like in your case of $n=3$, we want some function $f(x)$ such that $f(0)=a$, $f(1)=b$, and $f(2)=c$ (where $a,b$ and $c$ are what you want the first three outputs to be). They don't have to be inputs at $0$, $1$, and $2$, so I'm assuming that's what you want; but know that you can make them whatever you want.
Now, we need some arbitrary formula that has three unknowns (let's call them $c_0,c_1$ and $c_2$) over $x$. Not all choices for $f$ will be able to be solved for nicely, so that just takes experience of knowing what generally works and what doesn't. The point, however, is that you do have a choice here. A common choice would be a polynomial $f(x)=c_0+c_1x+c_2x^2$. This choice is nice because it generalizes to any $n$.
Thus, we essentially now have three equations
$$c_0+c_1(0)+c_2(0)^2=a,$$ $$c_0+c_1(1)+c_2(1)^2=b,$$ $$c_0+c_1(2)+c_2(2)^2=c.$$
From here, we have all the information to deduce the function. If you notice, this practically gives us $c_0$, as the first equation simplifies to $c_0=a$. We can use that to solve the next two equations.
Can you go from here?
Remember, $a$, $b$ and $c$ are what we have already chosen $f$ to output, so you will be solving for $c_0$, $c_1$ and $c_2$.