## Numbers representable by successive exponentiation

For the discussion of math. Duh.

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sevenperforce
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### Numbers representable by successive exponentiation

Over in this thread, I was speculating about ways that a super-intelligent alien might represent large numbers with a reasonable degree of precision (for example, expressing a time period on the order of several million years in units of Planck time, a number on the order of 1057). I thought perhaps the use of successive exponentiation in the form n1^n2^n3^n4 might be a possibility with very low information density.

I used Excel to autogenerate all possible combinations of n1^n2^n3^n4, using n = 2..9, and found that I could form the following values which were within an order of magnitude of 1057:
• 4.1359e56
• 7.8464e56
• 1.7970e57
• 6.2771e57
• 2.2980e58
Each of these could be formed in quite a few ways...for example, 6.2771e57 could be formed as 2^8^8^3, as 8^8^8, as 8^8^4^2, or several other combinations.

I wanted more precision, though, so I added another exponent slot and repeated the process using the form n1^n2^n3^n4^n5.

To my chagrin, although I found many more combinations that yielded the above values, I didn't find any new values in between them. It would seem, then, that for any maximum value nmax, there is only a very limited number of values which can be formed as the result of n1^n2^..., where n is a whole number, and this holds regardless of how many n-values you use.

Is this something that's already well-known and has an obvious name? Like, "Oh, right, the Pulling-Dolbrach numbers are those numbers which can be formed as a result of successively exponentiating whole numbers."

PeteP
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### Re: Numbers representable by successive exponentiation

I don't know about names.
Anyway since n1^n2^n3^n4= n1^(k) where k is a whole number you can only display a small subset of multiples of n1. More precisely since (a^m)^n= a^(m*n) you are calculating n1^(n2*n3*n4). And when you add more n you do add more possible results but at the same time you make other results impossible (for instance with n1^n2 you can display the number 4 but with (n1^n2)^n3)

Also writing the numbers as n1^n2^n3 is confusing since afaik you would usually solve that as n1^(n2^n3) I had to look at your result examples to know that you are using (n1^n2)^n3

Edit: "However, some computer systems may resolve the ambiguous expression differently. For example, Microsoft Office Excel evaluates a^b^c as (a^b)^c which is opposite of normally accepted convention of top-down order of execution for exponentiation. " so is it doing what you want it to do?

Yakk
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### Re: Numbers representable by successive exponentiation

((n1^n2)^n3)^n4 is just n1^(n2*n3*n4). So adding n5 doesn't really help. Heck, adding n3 and n4 didn't help much.

n1^(n2^(n3^n4)) is ... usually much bigger. It will help you reach medium sized numbers using very small constants. Adding n5 will make the size of number you can reach bit a touch bigger, but probably won't help you find more "small" numbers that well.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

sevenperforce
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### Re: Numbers representable by successive exponentiation

PeteP wrote:Since n1^n2^n3^n4= n1^(k) where k is a whole number you can only display a small subset of multiples of n1. More precisely since (a^m)^n= a^(m*n) you are calculating n1^(n2*n3*n4).

Ahah! I knew I was missing something. Yeah, that makes perfect sense; it's trivial to recognize that the set of possible results is limited to values obtainable as n1^k, and adding further values only further limits those options (though obviously without removing those already obtained) and so you converge to a small result set below a given limit.

Let's see...
Spoiler:
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^2^2 = 2^4 = 16
2^2^3 = 2^8 = 256
2^2^4 = 2^16 = 65,536
2^2^5 = 2^32 = 4,294,967,296
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 543
3^6 = 729
3^7 = 2,187
3^8 = 6,561
3^9 = 19,683
3^3^2 = 3^9 = 19,683

Yeah, it looks like adding additional exponentiation slots will certainly provide different ways to generate numbers, it's never going to generate new numbers lower than the maximum generated by a smaller number of exponentiation slots.

So if a^b with a,b = 1..9 generates set E2 with a maximum value of 9^9 (387,420,489), then the set E3 generated by a^(b^c) with a,b,c = 1..9 will not contain any numbers smaller than 9^9 which are not already contained in set E2. Or, to say the same thing in a different way, all the values in E3 \ E2 will be greater than the maximum value contained in E2.

PeteP
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### Re: Numbers representable by successive exponentiation

Not quite. If you set nmax=24 then 24^24>5^(5*5)
(You can't reach 5^25 with any n1^n2 no other number under 25 has only fives as prime factors)

sevenperforce
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### Re: Numbers representable by successive exponentiation

PeteP wrote:Not quite. If you set nmax=24 then 24^24>5^(5*5)

But is 5^(5^2) = 5^25 = 2.9802322e17 not a value contained within the set E[sub]2[sub] with a,b = 0...24?

(uses Excel to test)

Huh, apparently not.

PeteP
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### Re: Numbers representable by successive exponentiation

If you calculate n1^k the result contains the same prime factors with multiplicities (wiki says that is the term) that are k times greater. The only prime factor of 5^25 is 5 to reach that you need an n1=5^x otherwise other factors get introduced. Also 25/x=k.

>-)
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### Re: Numbers representable by successive exponentiation

On a tangential note, would it be impossible to do much better than just a decimal system from an information theory standpoint?

You might as well do (10*n1+n2)*10^(10*n3+n4) and now you can have, within a magnitude of 1E57

10*10^55
11*10^55
...
99*10^56
10*10^57

which is certainly a lot more accuracy

sevenperforce
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### Re: Numbers representable by successive exponentiation

>-) wrote:On a tangential note, would it be impossible to do much better than just a decimal system from an information theory standpoint?

You might as well do (10*n1+n2)*10^(10*n3+n4) and now you can have, within a magnitude of 1E57

10*10^55
11*10^55
...
99*10^56
10*10^57

which is certainly a lot more accuracy

But at a rather significant information theory cost.

Using only four digits with implied exponentiation is (if I'm doing this right)....104 or 10,000 or 213.29, which is barely more than 13 bits. What you have is eight digits with explicit multiplication and exponentiation, so that's 128 or 429,981,696 or 228.6, nearly 29 bits.

It's an interesting problem in information theory, I suppose. Let's say you need to be able to represent any value between 1e1 and 1e61 (roughly the number of Planck times since the Big Bang) to within 5% of its actual value. How many bits of information is that?

From 101 to 102 requires 46 values, which will be the same for (102..3), (103...104), and so forth. 46*60 is 2,760, so you apparently only need 2,760 values to represent all numbers from 10 to 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 with 95% accuracy. That's just over 211.4, or around a dozen bits. Theoretically, then, we should be able to come up with an very information-dense notation system that could represent all numbers in that range with 90% accuracy using an even smaller number of bits.

PeteP
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### Re: Numbers representable by successive exponentiation

(1/0.9)*1.1=1.22222222222222222…
If you use 1.222^x you can approximate everything with an error <10%. Not a very exiting way perhaps, but it should work fine.
Or take (1.1*1.1)^x depending if the lower value works as approximation as long as it isn't more than 10% smaller or if the approximated value can't be more than 10% bigger than the lower values.

Yakk
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### Re: Numbers representable by successive exponentiation

Along that same line:

11 bits as 1.1^n to represent everything from 1 to Plank Age of Universe within 5%.

Because 1.1^2047 is 5e84, and no value below 5e84 is further away than 5% from one of the values.

We could simplify.

We divide the values from 1 to 2 into 8 even subdivisions (2^(1/8) = 1.09).

So our value is a.b in octal, and we do 2^(a.b). b has 1 octal digit, and a has 11-3=8/3 = 3 octal digits.

That is 12 bits.

We can simplify further. 10^3 = 2^10. As we need 1/8th of a 2, this is the same as 3/10/8 = 3/80th of a 10. 2 decimal places is enough. So 10^a.b where a and b are each 2 decimal places will specify every value from 1 to 1e61 within 5%.

4 decimal digits is 4000 or 14 bits. If we require the last digit of the decimal to be even, 13 bits.

So for the cost of 2-3 bits, we get a more conventional notation. Probably worth it.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

sevenperforce
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### Re: Numbers representable by successive exponentiation

Yakk wrote:We can simplify further. 10^3 = 2^10. As we need 1/8th of a 2, this is the same as 3/10/8 = 3/80th of a 10. 2 decimal places is enough. So 10^a.b where a and b are each 2 decimal places will specify every value from 1 to 1e61 within 5%.

4 decimal digits is 4000 or 14 bits. If we require the last digit of the decimal to be even, 13 bits.

So for the cost of 2-3 bits, we get a more conventional notation. Probably worth it.

4 decimal digits is actually 10,000 but yeah, that's not too many bits.

Actually, I think the 10a.b notation maps fairly closely to regular scientific notation, doesn't it?

10a.b = (10a)b/100

Since xb/100 with b = 0..99 can be no more than 9.7x and no less than 1x, the "a" value serves to specify the order of magnitude and the "b" value serves to specify the quantity (though I think it's technically specifying the logarithm of the quantity, or something). There's probably some exact mapping here, like x*10y = 10y.log(x), but logarithms were never my strong suit.

Would a simple "ab" notation be better if a ran from 0.1 to 9.9 and b ran from 0 to 99? That's a range from 1e-99 to 3.7e98, still using only four digits.

Yakk
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### Re: Numbers representable by successive exponentiation

Yes, the joke was that regular scientific notation, with 2 digits before and after the decimal place, was about as good as the abstruse ones.

Throw in a sign bit, and you go down all the way.

ab is bad for approximating, because the a bits don't matter enough.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

sevenperforce
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### Re: Numbers representable by successive exponentiation

Yakk wrote:ab is bad for approximating, because the a bits don't matter enough.

But will the "a bits" allow you to hit your desired "within 10%" margin?

9.999 is 3.697e98. 9.899 is 1.353e98. Guess not.

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