Home ยป Termination of the sixth Goodstein Sequence

Solution

On June 4th, 2010 kagidab says:

$ k(a,n)= $ amount of steps to reduce $ (a-1)*a^n+(a-1)a^{n-1}+...(a-1)a^0 $ to -1

$ (a-1)a^1+a-1\rightarrow(a-1)(a+a)^1-1\rightarrow(a-2)(2a)^1-(2a-1) $

If you do this a - 1 times: $ (a - 1) a ^ 1 + a - 1 -> a * 2^{a - 1} - 1 $

Which means it is reduced to a 0th power and will take $ a*2^{a-1} $ steps to finish.

So total steps $ =a(2^{0}+2^{1}+2^{2}+...+2^{a-2})+a*2^{a-1}=a*(2^{a}-1)=k(a, 1)+1 $

For $ (a-1)*a^n+(a-1)a^(n-1)+... +(a-1) a^0 $:

$ k(a,n) = k(k...k(a,n-1)...)) - 2 $ [k a times]

$ f(n)=g(h(n)-2,h(n), h(n))-2 $

Where $ g(o) = g(0, n, o) = o * 2^o $, $ g(m, n, o)=g(m - 1, n, g(g(...g(o)))...) $ [n copies of g] and $ h(n) $ is the first number in the sequence in the form $ (a - 1) * a^(a-1)+(a - 1) a ^ (a - 2) + ... (a - 1) $ $ h([3,4,5,6,7])=[2, 3, 4, 6, 8] $

E.g.

$ f(4) = g(1, 3, 3) - 2 = g(0, 3, g(g(g(3)))) - 2 = 3 * 2^{402653211} - 2 $

Using $ g(n)\sim2^{n} $:

$ f(6) = g(4,6,6)-2\sim g(3, 6, 2\uparrow\uparrow 7})\sim2\uparrow\uparrow25 $

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