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A255963 Number of empty sets in the n-th power set of the empty set. 0
1, 1, 2, 5, 41, 1343489 (list; graph; refs; listen; history; text; internal format)
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

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Last modified December 13 12:49 EST 2018. Contains 318086 sequences. (Running on oeis4.)