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  1. PO
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    <p>It is not clear what you are asking, but I will try to guess.</p> <p>We first solve the equation <code>z^z = a</code> for a real number <code>z</code>. Let <code>u</code> and <code>v</code> be <code>z</code> rounded down and up, respectively. Among the three candidates <code>(u,u)</code>, <code>(v,u)</code>, <code>(u,v)</code> we choose the largest one that does not exceed <code>a</code>.</p> <p>Example: Consder the case <code>a = 2000</code>. We solve <code>z^z = 2000</code> by numerical methods (see below) to get an approximate solution <code>z = 4.8278228255818725</code>. We round down an up to obtain <code>u = 4</code> and <code>v = 5</code>. We now have three candidates, <code>4^4 = 256</code>, <code>4^5 = 1023</code> and <code>5^4 = 625</code>. They are all smaller than <code>2000</code>, so we take the one that gives the largest answer, which is <code>x = 4</code>, <code>y = 5</code>.</p> <p>Here is Python code. The function <code>solve_approx</code> does what you want. It works well for <code>a &gt;= 3</code>. I am sure you can cope with the cases <code>a = 1</code> and <code>a = 2</code> by yourself.</p> <pre><code>import math def solve(a): """"Solve the equation x^x = a using Newton's method""" x = math.log(a) / math.log(math.log(a)) # Initial estimate while abs (x ** x - a) &gt; 0.1: x = x - (x ** x - a) / (x ** x * (1 + math.log(x))) return x def solve_approx(a): """"Find two integer numbers x and y such that x^y is smaller than a but as close to it as possible, and try to make x and y as equal as possible.""" # First we solve exactly to find z such that z^z = a z = solve(a) # We round z up and down u = math.floor(z) v = math.ceil(z) # We now have three possible candidates to choose from: # u ** zdwon, v ** u, u ** v candidates = [(u, u), (v, u), (u, v)] # We filter out those that are too big: candidates = [(x,y) for (x,y) in candidates if x ** y &lt;= a] # And we select the one that gives the largest result candidates.sort(key=(lambda key: key[0] ** key[1])) return candidates[-1] </code></pre> <p>Here is a little demo:</p> <pre><code> &gt;&gt;&gt; solve_approx(5) solve_approx(5) (2, 2) &gt;&gt;&gt; solve_approx(100) solve_approx(100) (3, 4) &gt;&gt;&gt; solve_approx(200) solve_approx(200) (3, 4) &gt;&gt;&gt; solve_approx(1000) solve_approx(1000) (5, 4) &gt;&gt;&gt; solve_approx(1000000) solve_approx(1000000) (7, 7) </code></pre>
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    1. POFind maximal x^y smaller than number
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    1. USAndrej Bauer
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    1. COBut that doesn't fulfil `A = 33 => x^y = 2^5`. How the possible conflict between maximising `x^y` and minimising `|x - y|` is to be resolved isn't specified, however.
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      1. USDaniel Fischer
    2. COIndeed, it is not specified, and since I cannot read minds, I did it so that the difference between 'x' and 'y' is always at most 1. I am sure there are other readings of the problem.
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      1. USAndrej Bauer
 

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