Predicting increasing income within a given time I have this game where player earns "gold" and "reputation". With higher reputation, the player nets more income (gold).


*

*Each reputation points yields 1% bonus in income (gold)

*For every 20 gold received, player receives 1 reputation points

*Player starts with receiving 10 gold every second. This increases as reputation point increases.


How much gold and reputation will the player receive after 1000 seconds?
So I'm asking for an equation/formula and how you came up with it. I tried simplifying the example so I can explain it easily. I think I need a quadratic equation of some sort, but i'm just speculating.
I'm not really that good with providing my own formula :(
EDIT:
Here's an excel simulation of what I described above: https://www.dropbox.com/s/vz5wizqg4x43599/offline%20progression%20simulation.xlsx?dl=0
The first row holds the initial value (Time = 0)
 A: Let $G(n)$ be the gold after $n$ seconds and $R(n)$ the reputation after $n$ seconds.  Our system is $$G(n)=G(n-1)+(1+.01R(n-1))10\\R(n)=R(n-1)+G(n-1)/20\\R(0)=G(0)=0$$  
Please check that these match your description.  In particular, the gain in rep comes one turn after receiving the gold and the gain in rate of gold comes one turn after the gain in rep as I have written it.  Also I didn't consider the gains indivisible-maybe you want floor functions around some things.  
For small $n$, say up to a few thousand, it is easy to make a spreadsheet and calculate the values.  If you do it in a column, copy down is your friend.  I get $R(100)\approx 46264.3,\  G(100) \approx 65568.9,\  R(1000) \approx 6.549E31,\  G(1000) \approx 9.262E31$.  The long term growth rate for both is proportional to $1.070711^n$, which you can find by computing the eigenvalues of the matrix, or just dividing the values at time $n$ by those at time $n-1$  
Added:  In your sheet, the rep is computed from current gold, not added, so the equations become $$G(n)=G(n-1)+(1+.01R(n-1))10\\R(n)=G(n-1)/20\\R(0)=G(0)=0$$  You actually round (it appears, though the function is called floor) $R$ to an integer.  That turns out not to make much difference.  Taking out the $R(n-1)$ from my earlier equations, we agree very closely.  Now we can write the second as $R(n-1)=G(n-2)/20$ and substitute into the first, getting $$G(n)=G(n-1)+10+G(n-2)/200$$  In the long run, $G(n)$ will increase as $1.005^n$, but we aren't there yet by 100 seconds.  Alpha gives values for the recurrence that verify my spreadsheet.  The $1.005$ comes from the larger root of the characteristic equation $x^2=x+100$
A: Let $G(n)$ and $R(n)$ denote the amounts of gold and reputation after $n$ seconds, respectively. Then $G(0)=R(0)=0$ and $R(n)=\lfloor\tfrac{G(n)}{20}\rfloor$ and 
$$G(n+1)=G(n)+10(1+0.1R(n))=G(n)+10(1+0.01\lfloor\tfrac{G(n)}{20}\rfloor).$$
If we ignore the floor function, this becomes $G(n+1)=1.005G(n)+10$, which can be written as
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}
\begin{pmatrix}
G(n)\\1
\end{pmatrix}.$$
This gives us an explicit expression for $G(n+1)$ in terms of $n$;
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}^{n+1}
\begin{pmatrix}
0\\1
\end{pmatrix}$$
Luckily, powers of this matrix are easy to compute because $\tbinom{0\ 10}{0\ \hphantom{1}0}^2=0$, so by the binomial theorem
\begin{eqnarray*}
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}^{n+1}
&=&\left(
\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}
+\begin{pmatrix}
0&10\\
0&0
\end{pmatrix}\right)^{n+1}\\
&=&\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}^{n+1}
+(n+1)\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}^n
\begin{pmatrix}
0&10\\
0&0
\end{pmatrix}\\
&=&
\begin{pmatrix}
1.005^{n+1}&10(n+1)1.005^n\\
0&1
\end{pmatrix}
\end{eqnarray*}
This shows that
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005^{n+1}&10(n+1)1.005^n\\
0&1
\end{pmatrix}
\begin{pmatrix}
0\\1
\end{pmatrix},$$
and hence that
$$G(n+1)=10(n+1)1.005^n.$$
In particular, after $n=1000$ seconds this gives us
$$G(1000)\approx1458464.$$
Keep in mind that we have ignored the floor function at the start; in this model you do not gain $1$ reputation for every $20$ gold in increments, but you simply gain $1/20$-th of your gold in reputation.
