A116549 a(0) = 1. a(m + 2^n) = a(n) + a(m), for 0 <= m <= 2^n - 1.

1, 2, 3, 4, 4, 5, 6, 7, 5, 6, 7, 8, 8, 9, 10, 11, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 10, 11, 12, 13, 13, 14, 15, 16, 14, 15, 16, 17, 17, 18, 19, 20

Offset: 0

Author

Leroy Quet, Mar 16 2006

Keywords

• easy
• nonn

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

Examples

			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)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000523, A053645.

Cf. A000081, A000120, A004111, A029931, A048793, A061775, A070939, A072639, A276625, A279861, A326031.

Programs

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

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 *)

Formula

For n > 0: a(n) = a(A000523(n)) + a(A053645(n)). - Reinhard Zumkeller, Aug 27 2014