🧵 Untitled Thread
Anonymous at Tue, 9 Apr 2024 22:04:46 UTC No. 16121511
Hello there
I have found a way to structure numbers on a grid
It looks like this:
1...3...5...7...9..11.13.15.17.19.2
2...6..10.14.18.22.26.30
4..12.20.28
8..24
16
A step to the right adds double the starting value
A step down doubles the previous value
The grid forms a parabolic layout adding numbers in counting order
It can also be represented on a graph, see pic related.
Whereas I highlighted the apparent different sequences following a predictive pattern
Only the first sequence contains uneven numbers
Subesequent sequences are formed by adding the starting value of the next sequence, which is double the starting value of the previous sequence.
I.e. first sequence starts with 1, add 2 each step. Second sequence starts with 2, add 4 each step. Third sequence starts with 4, add 8 each step. Fourth sequence starts with 8, afd 16 each step. Etc.
I'm not very smart but I have a feeling this can be plotted along a spiral or fractal.
Does this grid have any algebraic use or am I just being retarded
Anonymous at Tue, 9 Apr 2024 22:54:58 UTC No. 16121547
Cool, you found a way to subdivide all positive natural numbers in a series of infinite sequences.
For n from 0 to inf you define
{2^0 + n*2^1}
{2^1 + n*2^2}
{2^2 + n*2^3}
{2^3 + n*2^4}
...
{2^m + n*2^(m+1)} with m arbitrarily big
Intuition says they combine to cover N, prove left as exercise
Anonymous at Tue, 9 Apr 2024 23:32:02 UTC No. 16121592
>>16121547
The odd numbers are covered by 1 + 2n.
So let x = 2n. let 2 divide n q times, then x = 2 * 2^q * m for some m, and m is odd, so m = 2z + 1. thus x = 2*(2^q)*(2z + 1) = (2^(q + 2))z + 2^(q + 1) = 2^y + z*2^(y + 1)
Anonymous at Tue, 9 Apr 2024 23:39:03 UTC No. 16121600
>>16121511
Isn't this a grid to determine subnets in a computer network?