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I have trouble with a question i need to use the function of Ackermann, I need to show that Ack(2,3) = 9.

A(2,3)
A(1,A(2,2))
A(1,A(1,A(2,1)))
A(1,A(1,A(1,A(2,0))))
A(1,A(1,A(1,A(1,1)))))
A(1,A(1,A(1,A(0,A(1,0))))))
A(1,A(1,A(1,A(0,A(0,1))))))
A(1,A(1,A(1,A(0,2))))
A(1,A(1,A(1,3)))
A(1,A(1,A(0,A(1,2)))
A(1,A(1,A(0,A(0,A(1,1)))))
A(1,A(1,A(0,A(0,A(0,A(1,0))))))
A(1,A(1,A(0,A(0,A(0,A(0,1))))))
A(1,A(1,A(0,A(0,A(0,2)))))
A(1,A(1,A(0,A(0,3))))
A(1,A(1,A(0,4)))
A(1,A(1,4))
A(1,A(0,A(1,3)))
A(1,A(0,A(0,A(1,2))))
A(1,A(0,A(0,A(0,A(1,1)))))
A(1,A(0,A(0,A(0,A(0,A(1,0))))))
A(1,A(0,A(0,A(0,A(0,A(0,1))))))
A(1,A(0,A(0,A(0,A(0,2)))))
A(1,A(0,A(0,A(0,3))))
A(1,A(0,A(0,4)))
A(1,A(0,5))
A(1,6)
A(0,A(1,5))
A(0,A(0,A(1,4)))
A(0,A(0,A(0,A(1,3))))
A(0,A(0,A(0,A(0,A(1,2)))))
A(0,A(0,A(0,A(0,A(0,A(1,1))))))
A(0,A(0,A(0,A(0,A(0,A(0,A(1,0)))))))
A(0,A(0,A(0,A(0,A(0,A(0,A(0,1)))))))
A(0,A(0,A(0,A(0,A(0,A(0,2))))))
A(0,A(0,A(0,A(0,A(0,3)))))
A(0,A(0,A(0,A(0,4))))
A(0,A(0,A(0,5)))
A(0,A(0,6))
A(0,7)
A=8

Where did I go wrong? If anyone could find my mistake i'd be thankful for your help!

Thanks!

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2 Answers 2

up vote 3 down vote accepted

Where you simplify $A(1,A(1,A(0,4)))$ to $A(1,A(1,4))$. $A(0,4)$ is $5$. By the way, Wolfram Alpha can show you all the steps.

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Thanks alot for your help! Also thanks for the Wolfram Alpha tip. –  Jef Jan 12 '12 at 20:52
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The 17th line appears wrong $A(1,A(1,4))$ should be $A(1,A(1,5))$ since $A(0,4)$ is $5$.

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