User blog comment:BlauesWasser/Hyper-Kefts Number/@comment-34700793-20180216030336/@comment-32783837-20180217060110

Still not distinguishable. But, since you seem to be dependent on such data, I see I have no choice but to contribute to this fruitless request.

Let's say (since it's close enough) that this number is in fact equal to the estimate of 3↑↑↑↑↑8.

That evaluates to:

3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3

3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑3↑↑↑3

3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑3↑↑3↑↑3

3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑↑3↑↑↑3↑↑3↑3↑3

3↑↑3 = 3↑3↑3 = 3↑27 = 7625597484987

3↑↑↑3 = 3↑↑7625597484987, also called tritri

Now, after about n=5, 3↑↑n can be estimated around (n-4) PT (6×10↑3638334640023). So tritri is roughly 7625597484983 PT (6×10↑3638334640023), where a PT (b) is defined as:

0 PT b = b

a PT b = 10↑(a-1 PT b) if a > 0

or a power tower of a 10's topped off with b, a notation used in Hypercalc. PT stands for power tower. And of course, 10↑b has b+1 digits. And that function merely represents tetrational growth, or in this case, a small pentational number.

And hexation is repeated pentation, so 3↑↑↑↑3 = 3↑↑↑3↑↑↑3, this is called grahal

Now, 3↑↑↑n would probably be around (3↑↑↑(n-1)) PT (6×10↑3638334640023), so the PT notation has been booked. We need a new notation, let's call this one TT (tetration tower).

0 TT b,c = b PT c

a TT b,c = (a-1 TT b,c) PT c

Then, 3↑↑↑↑3 is about (tritri-2) TT 7625597484983,(6×10↑3638334640023). Bam, hexation. But now we have to iterate THIS.

3↑↑↑↑n would be roughly (3↑↑↑↑(n-1)) TT 7625597484983,(6×10↑3638334640023), so we will continue with QT (pentation tower).

0 QT b,c,d = b TT c,d

a QT b,c,d = (a-1 QT b,c,d) TT c,d

Actually, never mind, this can be generalized:

tower(0,#) = tower(#)

tower(n) = n

tower(a,b,c,d,e,#) = tower(tower(a-1,b,c,d,e,#),c,d,e,#)

where # is any string of entries.

This is like basic Hyper-E, but the entries are backwards and the last entry can be non-integer.

3↑↑↑↑↑8, then, is very roughly: tower(8,1,1,7625597484983,(6×10↑3638334640023)).

That's the best I can do as far as helping you figure out how many digits there are.