Hyperoperators [n] basics for large n
We could inductively define hyperoperators as follows:
a, b, n being positive integers:
a[1]b := a + b
n > 1 -> a[n]1 := a
a[n+1](b + 1) := a[n](a[n+1]b)
From this some lemmas can be proven:
1. a[2]b = a * b
2. a[3]b = a ^ b
3. 2[n]2 = 4
4. n > 2 -> 1[n]b = 1
5. a > 1 -> a[n](b + 1) > a[n]b
6. (a + 1)[n]b > a[n]b
7. ((a > 2 or b > 2) and a > 1 and b > 1) -> a[n+1]b > a[n]b
8.
1 < a < b
c = \( ^a\log(b) \) rounded up to integer
m > 0, k >= 0
Then: a [4] m >= c * (b + k) -> a [4] (m + k + 1) >= b [4] (k + 2)
Is this the common definition here?
Have proofs been given somewhere for the lemmas?
I wrote them down long time ago, and I was about to do it again before I discovered this forum.
[edit]Some minor corrections made in this post[/edit]
We could inductively define hyperoperators as follows:
a, b, n being positive integers:
a[1]b := a + b
n > 1 -> a[n]1 := a
a[n+1](b + 1) := a[n](a[n+1]b)
From this some lemmas can be proven:
1. a[2]b = a * b
2. a[3]b = a ^ b
3. 2[n]2 = 4
4. n > 2 -> 1[n]b = 1
5. a > 1 -> a[n](b + 1) > a[n]b
6. (a + 1)[n]b > a[n]b
7. ((a > 2 or b > 2) and a > 1 and b > 1) -> a[n+1]b > a[n]b
8.
1 < a < b
c = \( ^a\log(b) \) rounded up to integer
m > 0, k >= 0
Then: a [4] m >= c * (b + k) -> a [4] (m + k + 1) >= b [4] (k + 2)
Is this the common definition here?
Have proofs been given somewhere for the lemmas?
I wrote them down long time ago, and I was about to do it again before I discovered this forum.
[edit]Some minor corrections made in this post[/edit]