4 ms·
Given that Graham's Number is an (over-)extension of the Ackermann-function that means that it's still computable, correct? Given the series leading up to Grah
by infogulch 11y ago
Given that Graham's Number is an (over-)extension of the Ackermann-function that means that it's still computable, correct?
Given the series leading up to Graham's Number, G = g_64 (g_1, g_2, ...), BB(n) will outgrow g_n, right? If so what's the smallest n such that BB(n) > g_n?
- infogulch 11y agoAfter a bit of research I found that there is a proof [0][1] that BB(12) > g_1. But that's still a ways from g_2, let alone g_64. [0]: "A Lower Bound on Rado's Sigma Function for Binary Turing Machines" by Milton Green (1964) [1]: https://en.wikipedia.org/wiki/Busy_beaver#Known_values_for_.CE.A3_and_S https://en.wikipedia.org/wiki/Busy_beaver#Known_values_for_....