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A116549 a(0) = 1. a(m + 2^n) = a(n) + a(m), for 0 <= m <= 2^n - 1. 4

%I #34 Jul 23 2019 04:25:00

%S 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,

%T 14,15,6,7,8,9,9,10,11,12,10,11,12,13,13,14,15,16,10,11,12,13,13,14,

%U 15,16,14,15,16,17,17,18,19,20

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

%C 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

%H Reinhard Zumkeller, <a href="/A116549/b116549.txt">Table of n, a(n) for n = 0..10000</a>

%F For n > 0: a(n) = a(A000523(n)) + a(A053645(n)). - _Reinhard Zumkeller_, Aug 27 2014

%e From _Gus Wiseman_, Jul 22 2019: (Start)

%e 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).

%e 0: o

%e 1: (o)

%e 2: ((o))

%e 3: (o(o))

%e 4: (((o)))

%e 5: (o((o)))

%e 6: ((o)((o)))

%e 7: (o(o)((o)))

%e 8: ((o(o)))

%e 9: (o(o(o)))

%e 10: ((o)(o(o)))

%e 11: (o(o)(o(o)))

%e 12: (((o))(o(o)))

%e 13: (o((o))(o(o)))

%e 14: ((o)((o))(o(o)))

%e 15: (o(o)((o))(o(o)))

%e 16: ((((o))))

%e 17: (o(((o))))

%e 18: ((o)(((o))))

%e 10: (o(o)(((o))))

%e (End)

%t 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 *)

%t bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];

%t dab[n_]:=1+Total[dab/@(bpe[n]-1)];

%t Array[dab,30,0] (* _Gus Wiseman_, Jul 22 2019 *)

%o (Haskell)

%o import Data.Function (on); import Data.List (genericIndex)

%o a116549 = genericIndex a116549_list

%o a116549_list = 1 : zipWith ((+) `on` a116549) a000523_list a053645_list

%o -- _Reinhard Zumkeller_, Aug 27 2014

%Y Cf. A000523, A053645.

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

%K easy,nonn

%O 0,2

%A _Leroy Quet_, Mar 16 2006

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Last modified April 17 22:02 EDT 2024. Contains 371767 sequences. (Running on oeis4.)