A note on a paper of G. H. Weiss and M. Dishon

Helmut Prodinger The John Knopfmacher Centre for Applicable Analysis and Number Theory School of Mathematics, University of the Witwatersrand, P. O. Wits 2050 Johannesburg, South Africa

                     helmut@gauss.cam.wits.ac.za

Yesterday^{1} I browsed through old volumes of The Fibonacci Quarterly (which I can recommend to everyone) and found [2]: In an attempt to improve on earlier result of Narayana and Kreweras, the authors prove that for r, s ≥ 1

[u^{r}v^{s}]\frac{1 − u − v − \sqrt{1 − 2(u + v) + (u − v)^{2}}}{2} = \frac{1}{r + s − 1}\binom{r + s − 1}{r}\binom{r + s − 1}{s}

(the notation [u^{r}v^{s}]f(u, v) refers to the coefficient of u^{r}v^{s} in the power series expansion of f(u, v)).

However, the really easy method in this context is the Lagrange inversion formula, as I will sketch now. Let S = (1 − u − v − \sqrt{1 − 2(u + v) + (u − v)^{2}} )/2, then S^{2} + (u + v − 1)S + uv = 0, or

u = \frac{S}{Φ(S)} with Φ(S) = \frac{v + S}{1 − v + S} .

Now the Lagrange inversion formula tells us (see, e. g., [1]) that

[u^{r}]S = \frac{1}{r}[S^{r−1}](Φ(S))^{r},

1October 16, 2001

or, with v = St,

[u^{r}v^{s}]S = \frac{1}{r}[S^{r−1}v^{s}](\frac{v + S}{1 − v + S} )^{r}

= \frac{1}{r}[S^{r+s−1}t^{s}](\frac{S(1 + t)}{1 − S(1 + t)})^{r}

= \frac{1}{r}[t^{s}](1 + t)^{r}[S^{s−1}](1 − S(1 + t))^{−r}

= \frac{1}{r}[t^{s}](1 + t)^{r+s−1}\binom{r + s − 2}{s − 1}

= \frac{1}{r}\binom{r + s − 1}{s}\binom{r + s − 2}{s − 1},

which is clearly equivalent to the statement to be proved.

References

1. R. Stanley, Enumerative combinatorics. Vol. 2, Cambridge University Press, Cambridge, 1999.

2. G. H. Weiss and M. Dishon, A method for the evaluation of certain sums involving binomial coefficients, Fibonacci Quarterly 14 (1976), 75–77.

A. M. S. Classification: 05A10