Donut in three packs

I do not know any formal solution but I did experimentally with a gf:

$\frac{1}{\left(1-x^5\right) \left(1-x^9\right) \left(1-x^{13}\right)}$

Finding the taylor series of which gives you:

$\cdots+4 x^{50}+4 x^{49}+3 x^{48}+2 x^{47}+3 x^{46}+4 x^{45}+3 x^{44}+2 x^{43}+2 x^{42}+3 x^{41}+3 x^{40}+2 x^{39}+2 x^{38}+2 x^{37}+3 x^{36}+2 x^{35}+x^{34}+2 x^{33}+2 x^{32}+2 x^{31}+x^{30}+x^{29}+2 x^{28}+2 x^{27}+x^{26}+x^{25}+x^{24}+2 x^{23}+x^{22}+x^{20}+x^{19}+2 x^{18}+x^{15}+x^{14}+x^{13}+x^{10}+x^9+x^5+1$

Looks like everything higher than 21 is possible.

Read on for a proof that all integer quantities > 21 can be expressed as a multiple sum of these quantites:

22 = 13 + 9

23 = 13 + 5 + 5

24 = 9 + 5 + 5 + 5

25 = 5 + 5 + 5 + 5 + 5

26 = 13 + 13

To get a number higher than 26, you just need to add 5's to the extreme right of the sequences...

Analytical Solution:

When the value of the coins are in arithmetic progression, this formula works:

$Frobenius(a,a+d,a+2d+...a+sd)=\left(\left\lfloor \frac{a-2}{s}\right\rfloor +1\right) a+(d-1) \ (a-1)-1$

Plugin a=5,d=4,s=2. The answer is then 21

Thanks to @bobbym none for the analytical solution.