with
v = coin value
c = multiplier (from 1 to 3.5)
h = current number of heads (1 to 9)

the formula is v * (c^(h-1)), always rounded up

This even applies to factor c=2.5, 5th heads h = 5, where v * (2.5^(5-1)) = v * 39.0625 ~ 40

This leads to the following table:



I tested factor 1 and 1.5 for 2 and 3 heads, factor 2 to 3 for 2 to five heads, and 3.5 to the end

Since your run ends at Head #10, there is no further amount for 10 heads.