well after making a couple of squares of order 3, 5 ands 7, i see a pattern as to how one can choose the magic sum to one's liking for a specific odd order.
Let us say that the first number you start off with, as mentioned before, will be symbolised with b (base). let us state the order as x.
magic sum= b*x+x( x^2 -1 /2) = x(b+(x^2-1)/2)
so if you wanted to start the process that mark mentioned with the first number being 3, for an order of 5, the magic sum once it is done would be:
5(3+(5^2-1)/2) = 5(3+12)=75, whereas if the base number was 1 the magic sum would be 5(1+12)=65. kind of cute I suppose, reminds me of the equation of an arithmetic series although i don't recall it precisely so that may be inaccurate. if anyone has anything else to add to this fun fun investigation into the art form which are magic squares, they're great for parties, jump right in with a method for even orders
hehe