OFFSET
0,2
COMMENTS
Consider the following bijection between the natural numbers and hereditarily finite sets. For each n, write out n in binary. Assign to each set already given a natural number m the (m+1)-th digit of the binary number (reading from right to left). Let the set assigned to n contain all and only those sets which have a 1 for their digit. Then a(n) gives the number of pairs of braces appearing in the n-th set written out in full, e.g., for 3, we have {{{}}{}}, with 4 pairs of braces. - Thomas Anton, Mar 16 2019
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
FORMULA
EXAMPLE
From Gus Wiseman, Jul 22 2019: (Start)
A finitary (or hereditarily finite) set is equivalent to a rooted identity tree. The following list shows the first few rooted identity trees together with their corresponding index in the sequence (o = leaf).
0: o
1: (o)
2: ((o))
3: (o(o))
4: (((o)))
5: (o((o)))
6: ((o)((o)))
7: (o(o)((o)))
8: ((o(o)))
9: (o(o(o)))
10: ((o)(o(o)))
11: (o(o)(o(o)))
12: (((o))(o(o)))
13: (o((o))(o(o)))
14: ((o)((o))(o(o)))
15: (o(o)((o))(o(o)))
16: ((((o))))
17: (o(((o))))
18: ((o)(((o))))
10: (o(o)(((o))))
(End)
MATHEMATICA
Nest[Append[#1, #1[[#3 + 1]] + #1[[#2 - 2^#3 + 1]] & @@ {#1, #2, Floor@ Log2@ #2}] & @@ {#, Length@ #} &, {1}, 63] (* Michael De Vlieger, Apr 21 2019 *)
bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]], 1];
dab[n_]:=1+Total[dab/@(bpe[n]-1)];
Array[dab, 30, 0] (* Gus Wiseman, Jul 22 2019 *)
PROG
(Haskell)
import Data.Function (on); import Data.List (genericIndex)
a116549 = genericIndex a116549_list
a116549_list = 1 : zipWith ((+) `on` a116549) a000523_list a053645_list
-- Reinhard Zumkeller, Aug 27 2014
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, Mar 16 2006
STATUS
approved