The number of possible numeric outcomes of parenthesizations of 2^2^...^2 Consider an expression 2^2^...^2 with n operators ^. Operator ^ means exponentiation ("to the power of"). Assume that it has no default assosiativity, so the expression needs to be fully parenthesized to become unambiguous. The number of ways to parenthesize the expression are given by Catalan numbers C_n=(2n)!/(n+1)!/n!. 

Sometimes different parenthesizations give the same numeric outcome, for example (2^2)^(2^2)=((2^2)^2)^2, so the number of different possible numeric outcomes for a given n is less than C_n for all n&gt;1. The sequence starts as 1, 1, 2, 4, 8, ... as opposed to Catalan numbers 1, 2, 5, 14, 42, ...

The problem is to write the fastest program (or function) that accepts n as an input and returns the number of different possible numeric outcomes of the expression 2^2^...^2 with n operators ^. The performance should not significantly deteriorate as n grows, so direct calculation of high power towers is probably a bad idea.
Python 2.7

This approach takes advantage of the following considerations:

Any integer can be represented as a sum of powers of two.  The exponents in the powers of two can also be represented as powers of two.  For example:

8 = 2^3 = 2^(2^1 + 2^0) = 2^(2^(2^0) + 2^0)


These expressions that we end up with can be represented as sets of sets (in Python, I used the built-in frozenset):


0 becomes the empty set {}.
2^a becomes the set containing the set representing a.  E.g.: 1 = 2^0 -&gt; {{}} and 2 = 2^(2^0) -&gt; {{{}}}.
a+b becomes the concatenation of the sets representing a and b.  E.g., 3 = 2^(2^0) + 2^0 -&gt; {{{}},{}}


It turns out that expressions of the form 2^2^...^2 can easily be transformed into their unique set representation, even when the numerical value is much too large to be stored as an integer.



For n=20, this runs in 8.7s on CPython 2.7.5 on my machine (a bit slower in Python 3 and much slower in PyPy):

"""Analyze the expressions given by parenthesizations of 2^2^...^2.

Set representation:  s is a set of sets which represents an integer n.  n is
  given by the sum of all 2^m for the numbers m represented by the sets
  contained in s.  The empty set stands for the value 0.  Each number has
  exactly one set representation.

  In Python, frozensets are used for set representation.

  Definition in Python code:
      def numeric_value(s):
          n = sum(2**numeric_value(t) for t in s)
          return n"""

import itertools


def single_arg_memoize(func):
    """Fast memoization decorator for a function taking a single argument.

    The metadata of &lt;func&gt; is *not* preserved."""

    class Cache(dict):
        def __missing__(self, key):
            self[key] = result = func(key)
            return result
    return Cache().__getitem__


def count_results(num_exponentiations):
    """Return the number of results given by parenthesizations of 2^2^...^2."""
    return len(get_results(num_exponentiations))

@single_arg_memoize
def get_results(num_exponentiations):
    """Return a set of all results given by parenthesizations of 2^2^...^2.

    &lt;num_exponentiations&gt; is the number of exponentiation operators in the
    parenthesized expressions.

    The result of each parenthesized expression is given as a set.  The
    expression evaluates to 2^(2^n), where n is the number represented by the
    given set in set representation."""

    # The result of the expression "2" (0 exponentiations) is represented by
    # the empty set, since 2 = 2^(2^0).
    if num_exponentiations == 0:
        return {frozenset()}

    # Split the expression 2^2^...^2 at each of the first half of
    # exponentiation operators and parenthesize each side of the expession.
    split_points = xrange(num_exponentiations)
    splits = itertools.izip(split_points, reversed(split_points))
    splits_half = ((left_part, right_part) for left_part, right_part in splits
                                           if left_part &lt;= right_part)

    results = set()
    results_add = results.add
    for left_part, right_part in splits_half:
        for left in get_results(left_part):
            for right in get_results(right_part):
                results_add(exponentiate(left, right))
                results_add(exponentiate(right, left))
    return results


def exponentiate(base, exponent):
    """Return the result of the exponentiation of &lt;operands&gt;.

    &lt;operands&gt; is a tuple of &lt;base&gt; and &lt;exponent&gt;.  The operators are each
    given as the set representation of n, where 2^(2^n) is the value the
    operator stands for.

    The return value is the set representation of r, where 2^(2^r) is the
    result of the exponentiation."""

    # Where b is the number represented by &lt;base&gt;, e is the number represented
    # by &lt;exponent&gt; and r is the number represented by the return value:
    #   2^(2^r) = (2^(2^b)) ^ (2^(2^e))
    #   2^(2^r) = 2^(2^b * 2^(2^e))
    #   2^(2^r) = 2^(2^(b + 2^e))
    #   r = b + 2^e

    # If &lt;exponent&gt; is not in &lt;base&gt;, insert it to arrive at the set with the
    # value: b + 2^e.  If &lt;exponent&gt; is already in &lt;base&gt;, take it out,
    # increment e by 1 and repeat from the start to eventually arrive at:
    #   b - 2^e + 2^(e+1) =
    #   b + 2^e
    while exponent in base:
        base -= {exponent}
        exponent = successor(exponent)
    return base | {exponent}

@single_arg_memoize
def successor(value):
    """Return the successor of &lt;value&gt; in set representation."""
    # Call exponentiate() with &lt;value&gt; as base and the empty set as exponent to
    # get the set representing (n being the number represented by &lt;value&gt;):
    #   n + 2^0
    #   n + 1
    return exponentiate(value, frozenset())


def main():
    import timeit
    print timeit.timeit(lambda: count_results(20), number=1)
    for i in xrange(21):
        print '{:.&lt;2}..{:.&gt;9}'.format(i, count_results(i))

if __name__ == '__main__':
    main()


(The memoization decorator's concept is copied from http://code.activestate.com/recipes/578231-probably-the-fastest-memoization-decorator-in-the-/ .)

Output:

8.667753234
0...........1
1...........1
2...........1
3...........2
4...........4
5...........8
6..........17
[...]
19.....688366
20....1619087


Timings for different n:

 n    time
16    0.240
17    0.592
18    1.426
19    3.559
20    8.668
21   21.402


Any n above 21 results in a memory error on my machine.

I'd be interested if anyone can make this faster by translating it into a different language.

Edit: Optimized the get_results function.  Also, using Python 2.7.5 instead of 2.7.2 made it run a bit faster.