Cantor Expressions: Solving A-F (2,7,19,87,1000,1M)

raross
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What is the cantor expansion of:

A. 2
B. 7
C. 19
D. 87
E. 1000
F. 1,000,000

The algorithm to solve these small problems is the most difficult for me.
The algorithm that I came up with states:
Asub(n) N! + Asub(n-1) (n-1)! +...+ Asub(2)2! + Asub(1)1!, where
Asub1 is an integer with 0 <= Asubi <= i for i = 1,2,...n,

I have tried to find other information on google, and have failed. It seems there is not much help with cantor expressions. So any help would be grateful!
 
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Have you seen http://www.rose-hulman.edu/mathjournal/2004/vol5-n1/paper4/v5n1-4do.doc ?

2.2 Example
23 = 3*3!+2*2!+1*1!
24 = 23 + 1 = 3*3!+2*2!+1*1! + 1 = 4!

This has an algorithm to convert a decimal number to Cantor.
 
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So do these answers make sense or am I doing it wrong?

2 = 1*1! + 1
7 = 2*2! + 1*1! + 2
19 = 2*2! + 1*1! + 14
87 = 3*3! + 2*2! + 1*1! + 64


Thanks for your help!
 
Im not sure how many digits you can put at the end, but it only makes sense I guess. Anyone help?
 
2 = 1*1! + 1
7 = 2*2! + 1*1! + 2
19 = 3*3! + 1
87 = 4*4! - 9

Hrmm, how about these answers?
Anyone have a clue?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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