User blog comment:Wythagoras/All my stuff/@comment-10429372-20130718073726/@comment-7484840-20130718081309

Yes, that is the conclusion we came to below, that ¥(n) for any sufficiently large n is bigger than Hollom's number, because it can also be defined as 'the biggest non infinite integer', as the function just escalates as close to infinity as it can. For example:

One possibility for ¥(about 93) is: "the output of z(1000), where z(n) is the largest number definable in 2n+3 symbols of English." Then, if the z function does not refer to any kind of definability function, this is a perfectly legitimate and finite output.

However, if the z function then referred to z(z(z(z(.....)))) as many times as it could, and none of these used any kind of definability function, this would also be a completely legitimate and finite output.

Then, if the outermost of those z functions could refer to z(z(...)), and none of these referred to the z function.... And so on. Can you see why the ¥ function doesn't really work?