Further investigation strongly supports n_0 = k^2 - 5/3 - (16/45)/k^2 + O(1/k^4) WFL On 6/27/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
What happens when row n of the Pascal triangle is shifted by k then added to itself? Or in continuous terms, two binomial distributions, identical apart from the shift along the x-axis, are summed?
Consider k fixed and n increasing: initially
n_C_i + n_C_j where i+j = n+k
is bimodal; but at some point n_0 it becomes unimodal and remains so for n > n_0.
Numerical experiments suggest that
n_0 ~ k^2 - 5/3 ;
indeed this bound estimate is remarkably good, being only <0.1 too large at k = 2 , and <0.001 too large at k = 20 .
No doubt probability wonks know all about this stuff already: so can anyone point me at a reference to a proof of the bound; or alternatively explain why it's perfectly obvious?
Fred Lunnon