A364725 Starting with a plane on which two parallel lines and two additional lines have been drawn such that the four lines form two non-congruent triangles, a(n) is the total number of lines on the plane after the n-th step, where each step consists of drawing lines that connect every intersection of two lines.
4, 6, 7, 11, 69, 176404
Offset: 1
Links
- Colin Linzer, Doodle Sequence (a desmos graph).
Formula
A recursive formula for an upper bound:
a(n+1) <= (a(n)^4 - a(n)^2)/8 + (a(n) - a(n)^3)/4 which is equivalent to
a(n+1) <= binomial(binomial(a(n),2),2) (proven).
The proof of the above formula comes from the fact that if there are m lines on a graph, then there are at most (m^2-m)/2 intersections. If there are o points on a graph then at most (o^2-o)/2 lines that can be drawn between them, substituting and simplifying leads to the former upper limit.
A non-recursive formula for a lower bound:
a(n) >= 4 + 61*floor(n/5) + 2*[n>=1 (mod 6)] + [n>=2 (mod 6)] + 4*[n>=3 (mod 5)] + 50*[n==4 (mod 5)] (proven).
The proof for the latter formula comes from the fact that in term 6, the new lines form an object congruent to the initial position. That "replica" will "grow" at a rate that must be greater than or equal to that of the original due to the extra lines not part of it, and that it will also produce another objects congruent to the original that will produce another... This formula comes directly from the difference between the terms, so you start with four, then add 2, then 1 ... then when the replicas are produced every 5 terms, you repeat the process. The growth is actually quadratic or higher in order as there are multiple replicas created each time, but that is harder to calculate and has not been done yet.
Extensions
a(6) from Jinyuan Wang, Aug 09 2023
Definition corrected by Colin Linzer, Sep 12 2023
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