Is the following number bigger than Graham's number or Tree (3)??

Question

Take every nanosecond (1 billionth of a second) that the universe has existed (nearly 14 billion years), and from the moment of the big bang, begin counting to to a certain number every revolution (beginning at one nanosecond - one billionth of a second). For instance, in the first revolution of the universe's existence (again, the length of a nanosecond), count to one trillion, in the second revolution, count to one trillion times one trillion (septillion), in the third revolution, count to one septillion times one septillion, and on and on, multiplying the number by itself each revolution.

Additionally, every revolution, the length of the revolutions are also shortened. For instance, after the first revolution (count to one trillion in the span of a nanosecond), the second resolution would last one billionth of a billionth, the third resolution, one billionth of a billionth of a billionth, and on and on it would go.

You would continue this until you reach the present day.

Which number is bigger - this number, or Graham's Number? This number, or Tree (3)???

Answer 1

The first problem with your method is you would never get to present day. In fact, you'd never get to even 2 nanoseconds after the big bang.



Your first time period is 1 billionth of a second. The next time period is 1 billionth billionth of a second and so on.



That's equivalent to 1/10^9 + 1/10^18 + 1/10^27 + ...



The total time would end up being 1/10^9 / (1 - 1/10^9) = 1/999,999,999. That's barely longer than 1 nanosecond which is 1/1,000,000,000 but would have an infinite number of cycles.



So even if you were just counting to 1 each time, you've already got a method that would have an infinite number of cycles within it and you'd exceed any finite number you can imagine.



If you don't see that, imagine a more mundane case where in the first second of the universe, you count 1. Then in the next 1/2 second you count 2, then in the next 1/4 second you count 3, etc. You'll never even make it to 2 seconds, but you'll have an infinite number of terms.



Answer:

You've created an number that cannot be counted in the time specified and would be infinite. That's bigger than any finite number you can construct (Graham's number, Tree(3), etc.)

Answer 2

Probably larger than Graham's Number and TREE(3), Smaller than The little bird of Svithjod on steroids.

Easily.

Because you're working unrestricted for an arbitrarily large number.



I'm not expert on Graham's Number, but TREE(3) is special not because it's an impossibly large number, but because it is an impossibly large number of combinations following a pattern where TREE(1) and TREE(2) are actually quite mundane.



I can make arbitrarily larger numbers than both too! Graham's Number^2 and TREE(4)



And if we're just looking purely for big numbers, like a child realizing they can put as many zeroes after the one and keep making the number bigger as long as they have paper, there's The little bird of Svithjod.



"High up in the north, in the land called Svithjod, there stands a rock. It is a hundred miles high and a hundred miles wide. Once every thousand years a little bird comes to this rock to sharpen its beak. When the rock has thus been worn away, then a single day of eternity will have gone by."



So a thought experiment not unlike your own, but a lot less unworldly and potential larger if it was using terms like nanosecond and trillion^2

Answer 3

Not sure I'm following your Tree(n), but it being only exponentials of exponentials, etc, it still seems as Graham's number is still wayyyyy bigger.



Update: Ok, Im quite sure I understood what you mean. You are considering Tree = T, T(1), T(2), T(3), T(4), ans so on. That goes to infinity. Like, construct the following: 1,2,3,4,5,6,.... the integers. If you go far enough, its "infinite", you will surpass G's number.



But for your T(3) is definitely way smaller than G's number...wayyyyy smaller. Even T(100000000000) is way smaller that G's number.

The interesting questions would be, for which N does T(N) > G. If I have time, I will look into it.



As Puzzling pointed out, you cant use decreasing revolutions the way you are (call them periods from now on) because it doesnt add up to the age of our universe. Nonetheless, in the same vane of your question:



Let N be the inverse of your time interval ( a "revolution"). For instance, for a nanosecond = 10^-9, N = 10^9.

There are M = ≈14Billion * 365 * 24 * 3600 * N such intervals since the BB < N * 5E17 intervals ( = 5E26 if you use nanoseconds).

Call B your "counting number", the Quantity, your "trillion" say. So the first quantity is B, then B², then B^4 then B^8 etc... You do this M times. That is, you have a Total T = B + B² + B^4 + ... + B^(2^M) < M*B^(2^M).



So, your total counts is always < M*B^(2^M).

Any number *you* can choose & write down for M & B (or N & B) in the expression M*B^(2^M) will always be smaller than G's number.



Like if N (or M) = 10^(10^google) and B = googleplex^googleplex, then this M*B^(2^M) is still way smaller than G's number.