Hmmm. (1,1) is one of the most important cells. So I started there looking for patterns. Interesting find for a *b values moving diagonally in the Grid from (1,1). Found the following combos:
(4,4) a * an
(9,9) BigN * c
(25,25) an * c
(49,49) a * c
Hmmm. (1,1) is one of the most important cells. So I started there looking for patterns. Interesting find for a *b values moving diagonally in the Grid from (1,1). Found the following combos:
(4,4) a * an
(9,9) BigN * c
(25,25) an * c
(49,49) a * c
I just noticed a cool pattern.
Starting at the (na transform) element,
a[t] (e) - a[t] (-f) = an increasing pattern.
128-127=1
100-97=3
76-71=5
56-49=7 which is (n-1)
So the difference between a[t] (e,1) compared to a[t] (-f,1) moves upward in an ascending pattern until the first element in a given e column.
for odd e, (e+1)/2 = a[1]
So this info greatly limits our search area.
Thinking out loud over here.
We limit the search using the (na transform) element and the a[1] element.
To be clear, Iโm not suggesting iterating by x.
Iโm suggesting using every available piece of info to limit the search area.
BigN is also a limiting piece of info
Along with 2(sqrt((f-1)/8)-1
Set boundaries, then factor
More patterns.
This one has (-f,1) a[t]=c and its equivalent element in (e,1) is distance of (a) apart, 287 - 270= 7 = correct a value
Also, the a[t] = (bn) value = 328
in (e,1) 328 - 287 = 41 = b
so c, a, and b are available/calculable in a triangle pattern right next to each other in adjacent elements.
Here's a diagram.
Once you c it you canโt un-c it?? Why does a[t]=c appear in many of our cases?
for c287
2(sqrt(d))-1 = 7 = (n-1)
Must be a fluke, but still analyzing for patterns over here.
Lol, for c6107 2(sqrt(d))-1= 15 and correct (n-1)=35
Shared factor of 5. Interesting.
and 2(sqrt((f-1)/8)-1 = 7 which is another factor for 35
Hello PMA, Jan, and 5DAnon can I please request your assistance?
Could it be a characteristic of ONLY semiprime c values that we always have one a[t]=c element?
Makes sense that c always appears at a[t] since the prime factors have to show up again.
And if that element exists for a given semiprime c, we can solve the problem with the adjacent elements. (maybe!)
You guys know how I work, calculator in hand with a pencil at the ready. If the idea sucks, that's fine. Working from small examples over here.