A255963 Number of empty sets in the n-th power set of the empty set.

1, 1, 2, 5, 41, 1343489

Offset

0,3

Comments

Starting with the empty set {}, we can repeatedly take the power set, say, n times, to obtain the n-th power set. If we write out the n-th power set, a(n) is the number of occurrences of {} in this written form. For n>0, this corresponds with the total number of empty sets contained in the n-th power set in any level of the set hierarchy.

a(6) = a(5)*2^65535 + 1 is too large to display in full. - N. J. A. Sloane, Mar 31 2015

Links

Table of n, a(n) for n=0..5.

Formula

Let t(n) = 2^2^2^...^2 be an exponentiation tower, n-1 2s high. The n-th element of the sequence a(n) is then given by the recurrence a(n) = 1 if n=0 or n=1, a(n) = (a(n-1)*t(n))/2 + 1 if n>1.

Example

For n=0, we take the power set of {} 0 times, which yields {}. In the written form, there is one occurrence of {}, so a(0) = 1.
For n=2, the 2nd power set of {} is { {}, {{}} }. In the written form there are 2 occurrences of {}, so a(2) = 2. Also in all levels of the set hierarchy together, this set contains 2 empty sets. Indeed the recurrent formula yields a(2) = (a(1)*t(2))/2 + 1 = (1*2)/2 + 1 = 2.

Prog

(Python)
def empty_sets(n):
.    if n==0:
..        return 1
.    if n==1:
..        return 1
.    else:
..        t = 2
..        for i in range(n-2):
...            t = 2**t
..        return ((empty_sets(n-1)*t)/2 + 1)
# Kesava van Gelder, Mar 12 2015

Crossrefs

Sequence in context: A054859 A076725 A059917 * A093625 A042447 A176266

Adjacent sequences: A255960 A255961 A255962 * A255964 A255965 A255966

Keyword

nonn,easy

Author

Kesava van Gelder, Mar 12 2015

Status

approved