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