How many copies of Ø occur in the expanded notation for Ƥ⁵(Ø) ?

(Hint: see Binary nature of power set)

Answer:

Let C(Ƥⁿ(Ø)) be the number of elements in Ƥⁿ(Ø)
Let E(Ƥⁿ(Ø)) be the number of Ø occurrence in Ƥⁿ(Ø)

Based on the observation in the Binary nature of power set,

1. when n > 0, C(Ƥⁿ(Ø)) = 2^C(Ƥn-1(Ø))
   
2. when n > 1, every element in Ƥ^n-1(Ø) is included in exactly half of all the element sets in Ƥⁿ(Ø)
   
3. there always exists an empty set in a non-empty power set
   

This translates to:

  E(Ƥn(Ø)) = E(Ƥn-1(Ø)) · C(Ƥn(Ø))/2 + 1 where n > 1

  Ƥ0(Ø) ≡ Ø
  Ƥ1(Ø) ≡ {Ø}
  Ƥ2(Ø) ≡ {Ø,{Ø}}
  Ƥ3(Ø) ≡ {Ø,{Ø},{{Ø}},{Ø,{Ø}}}
  ...

  C(Ƥ0(Ø)) = 0
  C(Ƥ1(Ø)) = 2C(Ƥ0(Ø)) = 1
  C(Ƥ2(Ø)) = 2C(Ƥ1(Ø)) = 2
  C(Ƥ3(Ø)) = 2C(Ƥ2(Ø)) = 4
  C(Ƥ4(Ø)) = 2C(Ƥ3(Ø)) = 16
  C(Ƥ5(Ø)) = 2C(Ƥ4(Ø)) = 65536

  E(Ƥ1(Ø)) = 1
  E(Ƥ2(Ø)) = E(Ƥ1(Ø)) · C(Ƥ2(Ø))/2 + 1 = 1 · 2 /2 + 1 = 2
  E(Ƥ3(Ø)) = E(Ƥ2(Ø)) · C(Ƥ3(Ø))/2 + 1 = 2 · 4 /2 + 1 = 5
  E(Ƥ4(Ø)) = E(Ƥ3(Ø)) · C(Ƥ4(Ø))/2 + 1 = 5 · 16 /2 + 1 = 41
  E(Ƥ5(Ø)) = E(Ƥ4(Ø)) · C(Ƥ5(Ø))/2 + 1 = 41 · 65536 /2 + 1 = 1343489

So there are 1,343,489 copies of Ø occur in the expanded notation for Ƥ⁵(Ø)