A beautiful journey into the heart of R-sequence

I’ve been thinking a bit of the rationale for the R-sequence and noticed (this has not been confirmed) that every L-sided number (ie. number on the “left side” of the R-sequence in \Xi matrix) multiplied by a prime is yet another L-sided number.

Then, I thought about claiming that multiplication of two L-sided numbers would give an L-sided number.

Were this true, the left side of the \Xi matrix would create a specific specimen of links between all the L-sided numbers, pushing thoughts toward the similarity between this idea and Pascal’s triangle.

From what we know, 2\Xi[x-1,y-1]=X[x,y] for all n-blockers (but 3^n) and 2\Xi[x,y]=\Xi[x+1,y] for all n-blockers. We also notice that all n-blockers (but 3^n) are even and 3^n blocker is always the smallest of n-blockers (for a fixed y).

We notice that last (column before 2x columns), for n-blockers, we have {{\Xi[x+1,y]}\over{\Xi[x,y]}}:1,42;1,92;1,86;1,42;1,92;1,90;1,95;1,80;1,92;1,94;1,69;1,98;1,65;1,92;1,90;1,95;1,98;1,94;1,86;1,97, never 2, which has been conjectured earlier, as a conjecture equivalent to Riemann Zeta conjecture.

Based on all these, a couple of other things, vision of goemetry of numbers from me and Ramanujan, and vision of new topology from Poincare, I’m trying to work on a more beautiful Pascal triangle, ie. the triangle that covers combinatorics not directly by the use of C(k,n), but direct connections between numbers. This is also going to explain the Pascal triangle.

About misha

Imagine a story that one can't believe. Hi. Life changes here. Small things only.
This entry was posted in Mathematics. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s