Gottfried Wrote:Dear Gianfranco (and all) - the proposals are nice.

However .... for iterated exponentiation and decremented iterated exponentiation, with which I'm involved, I always need three parameters ... . So I ... announce:

y = x{4,b}h for iterated exponentiation beginning at x: b^...^b^b^x

(and from earlier discussions)

y = x{3,b}h for iterated multiplication beginning at x: x*b^h

y = x{2,b}h for iterated addition beginning at x: x+b*h

Hi, Gottfried !

As a matter of fact, I promised you to react to some interesting proposals of yours on this matter, that you kindly sent me sometime ago. I am sorry for my delay in answering to you about these important issues. The reason is that I was (... am !) trying to imagine a notation, compatible with the tetrational notation, which we (GFR & KAR) proposed for storing very large numbers, similar to the scientific "floating point" number notation used in physics and in engineering. We need a small change in it, for several reasons, but particularly because that notation is actually based on a ternary operation (base, height, and initial/final exponent) and this may create problems during "manipulations" (if we are not careful).

Keeping aside for a moment this particular problem, I think that your proposals sound good and reasonable. I should only like to examine them in the light of the

Slashenquasigroup hyperops notation that pointed out in my recent discussions with Henryk.

In fact, noting that in a general hyperop of rank s, we may write:

y = b[s]n, with s: rank, b: base, n: hyperexponent,

the n-iterated [s] hyperop (on b) could be represented as follows:

y = b[s]<n>b = b[s+1](n+1)

In case of n-iteration of b[s] on a number x >< b, we should have:

y = b[s]<n>x.

The implementation of that, for h iterations of hyperops of ranks 3, 2, 1 , compared with your proposal and using your notations, could be (priority to the right ... whenever necessary) :

y =

x{4,b}h = b[3]<h>x = b^...^b^b^x = x@b[4]h = x@b#h, (@ is the "tower extension"),

y =

x{3,b}h = b[2]<h>x = b*...b*b*b*x = x*b[3]h = x*b^h,

y =

x{2,b}h = b[1]<h>x = b+...b+b+b+x = x+b[2]h = x+b*h,

... (and I stop here, for ... avoiding problems ...)

Obviously:

b[3]<h>b = b[4](h+1)

b[2]<h>b = b[3](h+1)

b[1]<h>b = b[2](h+1).

But this is another story.

Now, your proposal is more compact and my proposal seems more graphically "extended". Nevertheless, my proposal could also be compatible with some other "exponential-type" notations of the iterations (with <h> or °h as exponents of "b[s]").

GFR