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March 7th, 2017, 02:35 AM  #1 
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23  Monkey Root of $X^X$ numbers
To know if a number $P\in\mathbb{N^+}$ is of the form $X^X$ you can make the recoursive difference from P of: $P1^1 = R_1$ $R_12^2=R_2$ ... till you've the first negative value $R^_n$. If the negative value is equal to: $\displaystyle R^_n = \sum_{X=1}^{X1} X^X$ than $P=X^X$ It require only n step example: $P= 27$ $R_1=271^1=26$ $R_2=262^2=22$ $R_3=223^3=5$ $1^1+2^2 = 5 = R_3 $ than $P=27=3^3$ From the trivial identity: $\displaystyle \sum_{X=1}^{X1}X^X = X^X\sum_{X=1}^{X}X^X $ Last edited by complicatemodulus; March 7th, 2017 at 02:37 AM. 
March 7th, 2017, 03:02 AM  #2  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,557 Thanks: 2148 Math Focus: Mainly analysis and algebra  Quote:
Your example shows you subtracting $1^1$, $2^2$ and $3^3$ from 27. If I already know $3^3$, why do I need your process? It is quicker to calculate $k^k$ for $k=1,2,\ldots$ until $k^k \ge p$ with equality if $k^k = p$. Also, calculating $k^k$ is order $k$, so the algorithm is order $k^k$. Last edited by v8archie; March 7th, 2017 at 03:59 AM.  
March 7th, 2017, 03:08 AM  #3 
Senior Member Joined: Feb 2016 From: Australia Posts: 989 Thanks: 350 Math Focus: Yet to find out.  How to test this if I don't know what X is?
Last edited by skipjack; March 7th, 2017 at 10:09 AM. 
March 7th, 2017, 05:19 AM  #4  
Senior Member Joined: Dec 2012 Posts: 925 Thanks: 23  Quote:
 
March 7th, 2017, 06:54 AM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,555 Thanks: 597 Math Focus: Wibbly wobbly timeywimey stuff.  

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