# Need help finding equation that's close to geometric series sum but not.

#### LexFox

Hello! I need help finding an equation that takes two variables.

Problem in word form.
Billy has n game tokens to play arcade machines that take 1 game token each. After playing m games, the arcade gives him 1 game token back. How many total games t can Billy play with an inital n game tokens?

Work done so far
It looks a lot like a geometric sum, but sometimes it's less. Here's the equation I've been using so far:
Simplify:
$$\displaystyle \frac{1-n}{1-m}$$
With m=2 this works.
 n m t-n (1-n)/(1-m) 0 2 0 -1 1 2 0 0 2 2 1 1 3 2 2 2 4 2 3 3 5 2 4 4 6 2 5 5 7 2 6 6 8 2 7 7 9 2 8 8 10 2 9 9
But with m>2 it doesn't always. m=3 (values of n where my equation works are in bold):
 n m t-n (1-n)/(1-m) Difference 0 3 0 -0.5 -0.5 1 3 0 0 0 2 3 0 0.5 0.5 3 3 1 1 0 4 3 1 1.5 0.5 5 3 1 2 1 6 3 2 2.5 0.5 7 3 3 3 0 8 3 3 3.5 0.5 9 3 4 4 0 10 3 4 4.5 0.5 11 3 4 5 1 12 3 5 5.5 0.5 13 3 5 6 1 14 3 5 6.5 1.5 15 3 6 7 1 16 3 7 7.5 0.5 17 3 7 8 1 18 3 8 8.5 0.5 19 3 9 9 0 20 3 9 9.5 0.5 27 3 13 13 0 81 3 40 40 0
And m=4...
 n m t-n (1-n)/(1-m) Difference 0 4 0 -0.33 -0.33 1 4 0 0.00 0.00 2 4 0 0.33 0.33 3 4 0 0.67 0.67 4 4 1 1.00 0.00 5 4 1 1.33 0.33 6 4 1 1.67 0.67 7 4 1 2.00 1.00 8 4 2 2.33 0.33 9 4 2 2.67 0.67 10 4 2 3.00 1.00 11 4 2 3.33 1.33 12 4 3 3.67 0.67 13 4 4 4.00 0.00 14 4 4 4.33 0.33 15 4 4 4.67 0.67 16 4 5 5.00 0.00 17 4 5 5.33 0.33 18 4 5 5.67 0.67 19 4 5 6.00 1.00 20 4 6 6.33 0.33 64 4 21 21 0 256 4 85 85 0

Last edited:

#### romsek

MHF Helper
Let

$n$ be the original number of tokens

$m$ be the number of games played to get an additional game token

$N$ be the total number of games Billy gets to play

$K=\log_m(n)$

$N=\displaystyle{\sum_{k=0}^K}\left \lfloor \dfrac n {m^k} \right \rfloor$

#### LexFox

Thanks for the amazingly fast reply! Is it possible to write that formula without the sigma notation though?

#### romsek

MHF Helper
Thanks for the amazingly fast reply! Is it possible to write that formula without the sigma notation though?
well...

it's approximately

$\dfrac {1-(\frac 1 m)^{K+1}}{1-\frac 1 m}$

but not exact due to the floor function.