Weird Equality I Wasn't Expecting
I am trying to program Maple and I stumbled onto something interesting.
Given a triangular arrangement of m dots, where m is the nth triangular number and the dots are arranged on the integer points of a cartesian graph’s first quadrant, I tried determining how many planar trees I could form from a series of n adjacent dots in the triangle, as long as the planar trees had at least one point on the x-axis and one on the y-axis (in other words, as pushed into the origin as possible. The optimal case would include one of the points on the origin). This came down to a simple matter of (m choose n). (m choose n) is equal to m!/[(m-n)!n!]. Since m is given by (n^2+n)/2, this became ((n^2+n)/2)!/(((n^2-n)/2)!*n!). I graphed it and found that it blew up quickly. I was curious and wanted to know what the ratio (m choose n)/(m* choose (n-1)) is (where m* is the (n-1)th triangle number). I noticed that it approximated a line at large n values. I divided the just previously mentioned ratio by n and got a horizontal asymptote of (e^2)/2. Converting the new ratio into a series of gamma functions, I determined that the limit as n goes to infinity of G((n^2+n+2)/2)*G((n^2-3n+4)/2)/(n^2*G((n^2-n+2)/2)^2) = (e^2)/2, where G(z) is the gamma function of z.
I wonder if there is some equation which I could use to show me that…
Given a triangular arrangement of m dots, where m is the nth triangular number and the dots are arranged on the integer points of a cartesian graph’s first quadrant, I tried determining how many planar trees I could form from a series of n adjacent dots in the triangle, as long as the planar trees had at least one point on the x-axis and one on the y-axis (in other words, as pushed into the origin as possible. The optimal case would include one of the points on the origin). This came down to a simple matter of (m choose n). (m choose n) is equal to m!/[(m-n)!n!]. Since m is given by (n^2+n)/2, this became ((n^2+n)/2)!/(((n^2-n)/2)!*n!). I graphed it and found that it blew up quickly. I was curious and wanted to know what the ratio (m choose n)/(m* choose (n-1)) is (where m* is the (n-1)th triangle number). I noticed that it approximated a line at large n values. I divided the just previously mentioned ratio by n and got a horizontal asymptote of (e^2)/2. Converting the new ratio into a series of gamma functions, I determined that the limit as n goes to infinity of G((n^2+n+2)/2)*G((n^2-3n+4)/2)/(n^2*G((n^2-n+2)/2)^2) = (e^2)/2, where G(z) is the gamma function of z.
I wonder if there is some equation which I could use to show me that…