Non arithmetic, non geometric series starting with 4 and 8. My son (7th grade) had this homework problem last night: 
Come up with a function for a non arithmetic, non geometric series starting with 4 and 8.
In his math book they define a series as a function $F(n)$ with $n=1,2,3,4,...$
He'd been out sick, and I wasn't sure what process he was supposed to go through to solve it, but I went with a system of equations, where we would have one more unknown than equations.
I'll include my solution below.
Is there a more straightforward way for an algebra I student?
 A: We start with $F(1) = 4$ and $F(2) = 8$.  Quadratic solutions are neither arithmetic nor geometric, so we also start with $ax^2+bx+c$.
$F(1) = a(1^2) + b(1) + c = 4$
$F(2) = a(2^2) + b(2) + c = 8$
Equals:
$1a + 1b + c = 4$
$4a + 2b + c = 8$
Subtract twice the first from the second:
$2a + 0b - c = 0 \to a = c/2$
Substituting back,
$b = 4-3c/2$
I then explained there were an infinite number of c's to choose from, but asked him to think of any c that would give integer answers when divided by two.  So we picked 2.
$a = 1, b = 1 \to F(x) = x^2 + x + 2$
Is there a better more intuitive way of achieving this answer -- particularly for a beginning algebra student?
A: The squares $1,4,9,16,25\cdots$ are forming a non-arithmetic, non-geometric sequence.
If we can map $1$ to $4$ and $4$ to $8$, we are done. The linear relation
$$y=(8-4)\frac{x-1}{4-1}+4$$ achieves that.
Hence,
$$f_n=\frac{4n^2+8}3.$$

Alternatively, the factorials form a non-arithmetic, non-geometric sequence.
Now, we need to map $1!=1$ to $4$ and $2!=2$ to $8$, and get
$$f_n=4n!$$

Alternatively, let us combine the arithmetic series $1,2,3,4\cdots$ with an appropriate geometric series: the first two terms need to be $4-1=3$ and $8-2=6$, hence
$$f_n=n+3\cdot2^{n-1}.$$
And with the double arithmetic series,
$$f_n=2n+2^n,$$ which might be what the teacher expected most.

Alternatively, start from the first two terms and create the corresponding arithmetic or geometric series, $4n$ or $2^{n+1}$, and perturb it by a function that vanishes for $n=1$ and $n=2$:
$$f_n=4n+(n-1)(n-2)=n^2+n+2,$$ (like you found) or $$f_n=2^{n+1}+(n-1)(n-2).$$

And finally, for the lazy ones,
$$f_1=4,\\f_2=8,\\f_n=0.$$
