Here is a lovely generalized continued fraction!

While working with Stephen Glasby on some problems (see Some nice combinatorics for your viewing pleasure), I found a different way to write a partial sum of binomial coefficients. (Here’s my chance to use LaTeX in a post.) The problem involved the following function of r and m:

s_m(r) := \sum_{i=0}^r{m \choose i}.

Stephen and I were optimizing certain functions involving this term. After checking some examples, I came up with some terms for a_i and b_i and conjectured the following relation:

s_m(r+1)/s_m(r) = \frac{m-r+1}{r+1} + \frac{1}{r+1}\mathcal{K}_{i=1}^r b_i/a_i.

The notation with the curly K stands for a continued fraction. I think it looks better than b_1/[a_1 + b_2/[a_2 + b_3/[a_3 + … + b_r/a_r] …] . I found some nice things to put in for a_i (m -2r+3i) and b_i (2i(r+1-i)).

It took about three weeks and over thirty sheets of paper (both sides) before I found a proof. This may be in the literature, but it was new to me and Stephen. A similar statement,

\sum_{i=0}^r {m \choose i} = {m \choose r}(m - r)/(m - 2r + \mathcal{K}_{j=1}^r\frac{2j(r+1-j)}{m-2r+3j}).

almost follows from the proof and gives a representative form which works for many values of r. I hope to use this in other problems involving partial sums of binomial coefficients.

Update 2022.03.09: Work on the earlier problem got accepted for publication in the Electronic Journal of Combinatorics. (Yay!) I’ve learned a little more about generalized continued fractions. However, the easy route (Euler’s continued fraction theorem) seems not to lead to the representation here. Perhaps it really is new?

One thought on “Here is a lovely generalized continued fraction!

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