User blog comment:DrCeasium/new hyperfactorial array notation/@comment-5150073-20130415141127/@comment-7484840-20130416175323

okay, i'll check the thing with the X's out. Until then, I have one question and one theory:

Question: would any array at all, for example an  X^^4 & 3 array, have a number of entries equal to the numberical value of X^^4, ie if X was 3, as in a 3 quadrimensional array, would the number of entries in the array be the value of 3^^4?

Theory: steering clear of arrays bigger than dimensional for the moment, I have come up with a smaller version of my previous theory on easier to picture arrays. The theory is this: due to BEAF only changing its power if the pilot is the first entry in a row, with the only other factors being the actual numbers and the amount of them, it appears to not matter how you arrange the rows (as long as they stay in order). Therefore, the theory is that an a^b dimensional array = an a x (the value of)a^(b-1) planar array. This simplifies down to:

If a dimensional dividor in BEAF describes dimensional space, and so is in the form (n), the array {#(n)#} = {#(1)#}. For example, a 2^4 array of m's, or

{m,m(1) m,m (2) m,m(1) m,m (3) m,m(1) m,m (2) m,m(1) m,m }

This would be exactly equal in value to a 2 by 8 array of m's, or

{m,m(1) m,m (1) m,m(1) m,m (1) m,m(1) m,m (1) m,m(1) m,m },

because both have the same number of entries and rows, the value should be the same, because there are no rules in BEAF about rows in dimensional space being any more powerful than rows in planar space, other than it is easier to define more of them.