## Integer Factorizations: Notes for the week of Jan 8-14, 2023

Consider odd positive composite integer m, with adjusted ceiling square c2 and remainder r. There will be a least positive integer s such that m=s2+t2 for some positive t<s. This value of s will be greater than or equal to the ceiling root c corresponding to the ceiling square. Define the ascent of m to be s-c. (I chose the term because it is how far the ceiling must “ascend” to produce the s2-t2 expression needed for “Fermat factorization.”

It is somewhat easy to see that when m is a discrete semiprime with a relatively small prime factor, it will have a larger ascent than if the smallest prime factor is closer to the square root of m. Discrete semiprimes m=pq with p and q close to each other often have ascents of 0, and the zone, so to speak, in which this happens widens as m gets larger. The ascent being small when p and q are large and relatively close together helps with ceiling-square-based factorization runtimes for larger discrete semiprimes, I have found.

The ascent will always be even, since the ceiling square and any value of s ascended from that value will match the evenness/oddness of (m+1)/2.

Here is a snapshot of my current work table and its 3D graph (produced via data export to Graphing Calculator 3D by Runiter Company – Dr. Saeid Nourian, www.runiter.com – and used with thanks)

The CSV file data I calculated and imported from the (ceiling square, remainder, ascent) triplets for positive odd composite integers <= 169=132 in this graph follows:

``````1,0,0
3,0,0
4,1,0
5,4,0
5,0,0
6,9,0
7,16,0
6,1,0
8,25,0
7,4,0
7,0,0
8,13,2
8,9,0
9,24,2
8,1,0
9,16,0
9,12,4
10,25,0
9,4,0
9,0,0
11,36,0
10,13,6
10,9,0
11,28,6
10,5,2
10,1,0
11,16,0
12,33,8
12,29,2
11,4,0
12,25,0
11,0,0
12,21,10
13,44,2
13,40,10
13,36,0
12,9,0
13,28,12
12,1,0
13,24,4
14,49,0
13,16,0
14,41,4
14,37,14
13,8,2
13,4,0
13,0,0``````