Anonymous ID: 145639 Nov. 13, 2020, 5:34 a.m. No.11625107   🗄️.is 🔗kun   >>5136

>>11625079

True this!

Why geometry was so precise, resonant parameters.

And materials so important, shipped from distances.

Always wondered, what was the Ark filled with? Water with Monotomic Gold? Just air?

Did initiates submerge themselves?

Anonymous ID: 145639 Nov. 13, 2020, 6 a.m. No.11625356   🗄️.is 🔗kun

>>11625261

11625261 = 3 * 3875087

 

There is a formula for the probability of a number in the integer sequence being prime.

It decreases as a log function as the value of the integer increases.

 

Solution for Low Numbers

The solution to this problem is straightforward for low numbers x. All that we need to do is simply count the numbers of primes that are less than or equal to x. We divide the number of primes less than or equal to x by the number x.

 

We're in the range of 10M posts (11.6M)

There are 664,579 primes less than 10M

664,679 / 10,000,000 = 6.6%

And dropping..

Anonymous ID: 145639 Nov. 13, 2020, 6:04 a.m. No.11625392   🗄️.is 🔗kun   >>5411 >>5511

>>11624864

Regarding tetrahedral numbers, andThe GRID, an anon did some interesting calcs (not this anon's work, pic related).

 

"I've been looking more at the tetrahedral numbers (4,10,20), and I may have found something. Look at these grids. Basically the white columns are the seeds, and each value is equal to the sum of the one above and one to the left of it. By default, the first row is a 1. Then the top left box is just regular numbers seeded by a 1. The highlighted column in that box are the tetrahedral numbers. The highlighted column in the next is square numbers. Then the next is pentagonal numbers, then hexagonal, heptagonal, etc. If you look at the highlighted row, then you can see that there is a simple pattern to get from grid to grid. (ie, from 1 to 2 you subtract 1, from 2 to 3 you subtract 2, from 3 to 4 you subtract 3) so you can easily navigate through these grids. Also for each box n, the first column is all the numbers where mod n is the seed. So maybe we could start with a zero seed (cuz then its divisible) or something like that. Or we could use the identity n*n = T(n) + T(n-1) where T(n) is the nth triangular number. Get the number in the square grid and translate it to a different grid and do stuff. Also these grids can be extended into the negative (I'll get on that).

 

Basically the idea is to get our number. E would be the seed and D would be the start column. Then we can use some of these identities to navigate around to the grid where we are in the A column with a 0 seed."