This is an interesting problem.

We know that if the fifth son gets x5 then the fourth son gets x4 = 5*(x5 + 1)/4 – 1 = x5 + (x5+1)/4 and third son gets x3 = 5*(x4 + 1)/4 – 1 = x4 + (x4+1)/4 and so on.  Representing what each son gets in terms of what the fifth son gets gives:

x3 = x5 + (x5+1)/4 + ((x5+(x5+1)/4))/4
= x5 + (x5+1)/4 + 5(x5+1)/16
x2 = x5 + (x5+1)/4 + 5(x5+1)/16 + 25(x5+1)/64
x1 = x5 + (x5+1)/4 + 5(x5+1)/16 + 25(x5+1)/64 + 125(x5+1)/256

From this we know that the fifth son (x5) must get a number of coins that can be expressed in the form x5 = 256y – 1.

The other requirement is that x5 must be a multiple of 5, because each daughter gets 4/5 of what the 5th son gets.  That means that the final digit of y must be either 1 or 6 so that the final digit of 256y is 6.  The smallest such value for y is 1, making x5 = 255.

Plugging in 255 for x5 gives

x1 = 624
x2 = 499
x3 = 399
x4 = 319
x5 = 255
with each daughter getting 4/5 * 255 = 204 and the manager getting 5 gold coins

204 * 5 + 255 + 319 + 399 + 499 + 624 + 5 = 3121 = 624 * 5 + 1.  The minimum number of coins is 3121.

The second smallest number of coins possible is 18,746, which results from setting y = 6.