Dividing a cake between $n-1$, $n$, or $n+1$ guests 
A housewife is waiting for guests and has prepared a cake. She doesn't know how many guests will come, but it will be $n-1$, $n$, or $n+1$. 
  What is the minimal number $f(n)$ of pieces the cake should be cut to make it possible to divide between guests equally?

For $n=2$, $f(n)=f(2)=4$:

The problem was posed 16.10.2018 by Oleksandr Maksymets on page 76 of Volume 2 of the Lviv Scottish Book.
The prize: Cooked duck or lunch + beer!

$f(7)=15$. 
$f(7)\ge15$ follows from a comment of Fedor Petrov on the original question, so it suffices to find a way to cut the cake into $15$ pieces so as to serve $6$, $7$, or $8$ guests. 
Let the size of the cake be $168$ (so that all the following computations involve only whole numbers). Let the $15$ pieces be of sizes $1,2,4,5,7,8,10,11,13,14,16,17,19,20,21$ (that is, every size not a multiple of $3$ up to $20$, and $21$). Then 
$$1+20=2+19=4+17=5+16=7+14=8+13=10+11=21,$$
$$4+20=5+19=7+17=8+16=10+14=11+13=1+2+21(=24),$$
$$7+21=8+20=4+5+19=11+17=2+10+16=1+13+14(=28).$$
Note that this disproves my conjecture $f(n)=[5n/2]-1$ which evaluates to $16$ when $n=7$. 
