A333447 a(n) is the integer corresponding to a bit-string representation of the von Neumann ordinal representation of n, with largest sets listed first, and with '{' represented by the bit 1, '}' represented by the bit zero, and ignoring commas.

2, 12, 228, 62052, 4180832868, 18201642257939067492, 338021701687178649306251838479209230948, 115407456979036362321626052309736660160730393295399201179594209600531491615332

Offset: 0

Author

Kit Scriven, Mar 21 2020

Keywords

• nonn
• base

Comments

Since the von Neumann ordinals begin with 0={}, it seems appropriate to have an offset of 0.

The sequence grows super-exponentially.

Similar to A092124, except that A092124 reverses the order of the elements in the ordinal.

The binary expansion of a(n-1) has length 2^n and consists of n 1's followed by the leading terms of A308187. - Andrey Zabolotskiy, Mar 21 2020

Examples

			A table demonstrating the von Neumann ordinals of the first three integers, their corresponding bit strings, and their sequence values is as follows:
  n   set notation       bit string         a(n)
  0   {}                 10                 2
  1   {{}}               1100               12
  2   {{{}}{}}           11100100           228
  3   {{{{}}{}}{{}}{}}   1111001001100100   62052
		

Links

• Kit Scriven, Table of n, a(n) for n = 0..10
• Johann von Neumann, Zur Einführung der transfiniten Zahlen, Acta Litterarum AC Scientiarum Ragiae Universitatis Hungaricae Francisco-Josephinae, 1 (1923), 199-208.

Crossrefs

Cf. A079559, A092124, A308187.

Programs

Mathematica

With[{nmax=8},Map[FromDigits[#,2]&,NestList["1"<>#<>StringTake[#,{2,-2}]<>"0"&,"10",nmax]]] (* Paolo Xausa, Nov 21 2023 *)

Python

def fBinDigit(n):
    return 2**(2**(n+1) - 1)
def a333447(n):
    if n==0:
        return 2
    else:
        prevAsEle = a333447(n-1) * fBinDigit(n-1)
        restOfEle = a333447(n-1) - fBinDigit(n-1)
        return fBinDigit(n)+prevAsEle+restOfEle

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A333447(n): return (1<<((m:=1<A333447(n-1) if n else 2 # Chai Wah Wu, Nov 23 2023

Formula

a(0) = 2, a(n) = 2^(2^(n+1)-1) - 2^(2^(n)-1) + a(n-1)*(2^(2^(n)-1) + 1).