Home ยป Termination of the sixth Goodstein Sequence

The actual value is much higher

On September 17th, 2010 Deedlit says:

You've underestimated the true value by quite bit.

To get the value of the Goodstein function at n, you take n, write it in hereditary base 2, then replace every appearance to with the infinite ordinal $ \omega $. Call the result R(n). The value of G(n) is then $$H_{R(n)}(3) - 2$$ where $ H_a(x) $ is the Hardy hierarchy, defined by

$ H_0(x) = x $

$ H_{a+1} (x) = H_a (x+1) $

$ H_a (x) = H_{a[x]} (x) $ for limit ordinals a

So to find G(6), we write $ 6 = 2^2 + 2 $, so $ R(6) = \omega^\omega + \omega $. Hence, $$ G(6) = H_{\omega^\omega + \omega} (3) - 2 = H_{\omega^\omega}( H_{\omega} (3)) - 2 = F_{\omega} (F_1 (3)) - 2 = F_{\omega} (6) - 2 = F_6 (6) - 2 $$

where $ F_a(x) $ is the fast-growing hierarchy, defined by

$ F_0(x) = x+1 $

$ F_{a+1} (x) = F_a^x (x) $

$ F_a (x) = F_{a[x]} (x) $ for limit ordinals a

(or you could just leave the answer in terms of the Hardy hierarchy, I just changed to the fast-growing hierarchy because the answer is a little simpler.)

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