Egorychev Method: A Hidden Treasure

Abstract

Egorychev method is a potent technique for reducing combinatorial sums. In spite of the effectiveness of the method, it is not well known or widely disseminated. Our purpose in writing this manuscript is to bring light to this method. At the heart of this method is the representation of functions as series. The chief idea in Egorychev method is to reduce a combinatorial sum by recognizing some factors in it as coefficients in a series (possibly in the form of contour integrals), then identifying the parts that can be summed in closed form. Once the summation is gone, the rest can be evaluated via one of several techniques, which are namely: (I) Direct extraction of coefficients, after an inspection telling us it is the generating function (formal power series) of a known sequence, (II) Applying residue operators, and (III) Appealing to Cauchy’s residue theorem, when the coefficients alluded to appear as contour integrals. We present some background from the theory of complex variables and illustrate each technique with some examples. In concluding remarks, we compare Egorychev method to alternative methods, such as Wilf–Zeilberger theory, Zeilberger algorithm, and Almkvist–Zeilberger algorithm and to the performance of computer algebra systems.

Appendix

We list here a few hypergeometric sums that are amenable to Egorychev method but for some reason or another the standard symbolic algebra systems could not reduce:⁸

A008279 (permutations of n things k at a time)

$$\begin{aligned}&\sum _{k=0}^n (-1)^k {n \atopwithdelims ()k} {n-m+k \atopwithdelims ()k} k^m = (-1)^n {n\atopwithdelims ()m} m!, \end{aligned}$$

A142150 (nonnegative integers interleaved with zeros)

$$\begin{aligned}&\sum _{k=0}^n (-1)^k {n-k\atopwithdelims ()k} {2k-n\atopwithdelims ()n} {2k\atopwithdelims ()n-1} = \frac{(-1)^{(n+1)/2}}{4} \, (n+1) \big (1+(-1)^{n+1}\big ), \end{aligned}$$

A011973 (irregular triangle of numbers read by rows)

$$\begin{aligned}&\sum _{k=0}^n (-1)^k {n-1+k\atopwithdelims ()n-m+k} {2m\atopwithdelims ()m+k} {m-1+k\atopwithdelims ()k} = {n-m-1\atopwithdelims ()m-1}, \end{aligned}$$

A144484 (a triangular array read by rows)

$$\begin{aligned}&\sum _{k=0}^{n-m} (-1)^k {n-m+k\atopwithdelims ()k} {2n-k\atopwithdelims ()n-m-k} {2n+1\atopwithdelims ()k}= {3n+1-m\atopwithdelims ()2n+1} (-1)^{n-m}, \end{aligned}$$

A001700 (number of monotone maps)

$$\begin{aligned}&\sum _{k=0}^{n-m} (-1)^k {2n-m\atopwithdelims ()n-m-k} {n-1+k\atopwithdelims ()k} {k-p+n\atopwithdelims ()p} = {2p-1\atopwithdelims ()p-1} (-1)^p, \end{aligned}$$

A141662 (a triangular array read by rows)

$$\begin{aligned}&\sum _{k=0}^n (-1)^k {n+m\atopwithdelims ()n-k-1} {m-1+k\atopwithdelims ()k} {m-1+k\atopwithdelims ()m} = (-1)^m (n-m^2), \end{aligned}$$

A155865 (Leibniz Harmonic triangle)

$$\begin{aligned}&\sum _{k=0}^n (-1)^k {n+k\atopwithdelims ()n-m-1} {n+1\atopwithdelims ()k+1} {2k\atopwithdelims ()m-1} = {n-2\atopwithdelims ()m-1} (n-1)(-1)^{m-1}, \end{aligned}$$

A243594 (coefficients in $(x+n)^n)$

$$\begin{aligned}&\sum _{k=0}^n (-1)^k {n \atopwithdelims ()k} {n-m+k\atopwithdelims ()k} {kn +m \atopwithdelims ()m} = (-1)^n {n\atopwithdelims ()m} n^m, \end{aligned}$$

A007318 (Pascal’s triangle by rows)

$$\begin{aligned}&\sum _{k=0}^{n-m} (-1)^k {p+k \atopwithdelims ()k} {2n-m\atopwithdelims ()n-m-k} {n-1 +k \atopwithdelims ()k} = (-1)^p{n-1\atopwithdelims ()p}. \end{aligned}$$

These identities were found by a computer search that pointed to entries in the Online Encyclopedia of Integer Sequences.

While a smooth reduction via Egorychev method is possible in each of these cases, computer algebra systems exhibit anomalous behavior. For instance, in several cases Maple© produces pages of expressions involving gamma functions or hypergeometric functions, or just goes into an infinite loop. In particular, for the identities in the entries A008279 and A243594, both Maple© and Mathematica© return the sums unevaluated, whereas a reduction via Egorychev method goes through without a hitch. In some cases Mathematica© gives something similar or even a wrong answer.

As a specific example, for the last identity (A007318) in the list above, Maple © delivers the hypergeometric answer

$$\begin{aligned} \left[ \begin{array}{c} {\frac{1}{2},1,1,1-n,n+1} \\ {2,2+m, 1 - \frac{m}{2}, \frac{3}{2} -\frac{m}{2}} \end{array} \, \Bigg |\, 1\right] . \end{aligned}$$

On the other hand, we can establish the following proof via Egorychev method.

The equality derived is valid for $n \ge m \ge 0$, and $p \le n-m$. We start with the usual preparation, representing binomial factors as coefficients in an expansion, using the variety of techniques in the main body of the paper, such as index shifting, switching the upper index of a binomial coefficient and extending the range of sums to infinity. We write

$$\begin{aligned} S_{n,m,p}&:= \sum _{k=0}^{n-m} (-1)^k {p+k \atopwithdelims ()k} {2n-m\atopwithdelims ()n-m-k} {n-1 +k \atopwithdelims ()k}\\&= \sum _{k=0}^{n-m} (-1)^k {p+k \atopwithdelims ()k} {2n-m\atopwithdelims ()n-m-k} {n-1 +k \atopwithdelims ()n-1}\\&= \sum _{k=0}^{n-m} (-1)^k {p+k \atopwithdelims ()k} [u^{n-m-k}] \, (1+u)^{2n-m}\, [v^{n-1}] \, (1+v)^{n-1+k}\\&= \sum _{k=0}^{n-m} (-1)^k {p+k \atopwithdelims ()k} [u^{n-m}] \, u^k (1+u)^{2n-m}\, [v^{n-1}]\, (1+v)^{n-1+k}\\&= [u^{n-m}] \, (1+u)^{2n-m}\, [v^{n-1}]\, (1+v)^{n-1} \sum _{k=0}^\infty {-p-1 \atopwithdelims ()k} \, u^k \, (1+v)^k\\&= [u^{n-m}] \, (1+u)^{2n-m} \, [v^{n-1}]\, (1+v)^{n-1} \frac{1}{\big (1+u(1+v)\big )^{p+1}}\\&= [u^{n-m}] \, (1+u)^{2n-m}\, [v^{n-1}]\, (1+v)^{n-1} \frac{u^{-p-1}}{\big (v+(1+u)/u\big )^{p+1}} \\&= [u^{n-m+p+1}] \, (1+u)^{2n-m}\, [v^{n-1}]\, (1+v)^{n-1} \frac{1}{\big (v+(1+u)/u\big )^{p+1}}. \end{aligned}$$

The contribution from v is

$$\begin{aligned} \underset{v\ }{\mathrm{res\ }} \frac{1}{v^n} (1+v)^{n-1} \frac{1}{\big (v +(1+u)/u)\big )^{p+1}}. \end{aligned}$$

The residue at infinity is 0, and we can evaluate the residue at $v=0$ as the negative of the residue at $v=-(1+u)/u$, which we can compute by Leibniz rule:

$$\begin{aligned}&-\underset{v={-(1+u)/u}\ }{\mathrm{res\ }} \frac{1}{v^n} (1+v)^{n-1} \frac{1}{\big (v +(1+u)/u) \big )^{p+1}} \\&\quad = - \frac{1}{p!} \times \frac{d^p}{d v^p} \Big (\frac{1}{v^n} (1+v)^{n-1}\Big ) \Big |_{v = - \frac{1+u}{u}}\\&\quad = - \frac{1}{p!} \sum _{q=0}^p {p \atopwithdelims ()q} (-1)^q \frac{n^{\overline{q}}}{v^{n+q}} \times (n-1)^{\underline{p-q}} \, (1+v)^{n-1-(p-q)}\Big |_{v = - \frac{1+u}{u}}\\&\quad = -\frac{1}{p!} \sum _{q=0}^p {p \atopwithdelims ()q} \, q!\, {n+q-1 \atopwithdelims ()q} \frac{(-1)^q}{v^{n+q}}\\&\qquad {} \times (p-q)!\, {n-1 \atopwithdelims ()p - q} (1+v)^{n-1-p+q} \Big |_{v = - \frac{1+u}{u}}\\&\quad = \sum _{q=0}^p {n+q-1 \atopwithdelims ()q} \frac{(-1)^{n+1} u^{n+q}}{(1+u)^{n+q}} {n-1 \atopwithdelims ()p - q} \frac{(-1)^{n-1-p+q}}{u^{n-1-p+q}}. \end{aligned}$$

Now we have

$$\begin{aligned} S_{n,m,p}&= [u^{n-m+p+1}] \, (1+u)^{2n-m} \sum _{q=0}^p {n+q-1 \atopwithdelims ()q}{n-1 \atopwithdelims ()p - q} \frac{(-1)^{p-q}}{(1+u)^{n+q}\, u^{-1-p}} \nonumber \\&= \sum _{q=0}^p [u^{n-m}] \, u^{-p-1}\, (1+u)^{n-m-q} {n+q-1 \atopwithdelims ()q} {n-1 \atopwithdelims ()p - q} \frac{(-1)^{p-q}}{\, u^{-1-p}}\nonumber \\&= (-1)^p\sum _{q=0}^p (-1)^q {n-m-q \atopwithdelims ()n-m} {n+q-1 \atopwithdelims ()q} {n-1 \atopwithdelims ()p-q}. \end{aligned}$$

(9)

With $n \ge m$, the leftmost binomial coefficient vanishes for all $q\ge 1$; only $q=0$ makes a contribution. We find

$$\begin{aligned} S_{n,m,p} = (-1)^p {n-m \atopwithdelims ()n-m} {n-1 \atopwithdelims ()0} {n-1 \atopwithdelims ()p} = (-1)^p{n-1 \atopwithdelims ()p}. \end{aligned}$$

The choice $p \le n- m$ is convenient to prevent the upper index of the leftmost binomial coefficient from going negative, producing a nonzero value. We can relax this condition to construct identities for values of p advancing a few steps beyond $n-m$. A few extra terms will appear upon writing out the expansion of (9).

For example, with $p=n-m+1$, all the binomial coefficients ${n-m-q \atopwithdelims ()n-m}$ have a positive upper index smaller than the lower index making that binomial coefficient 0, except the two terms $q=0$ and $q=n-m+1$. In the case $q=0$, we have ${n-m-q \atopwithdelims ()n-m} = {n-m \atopwithdelims ()n-m}$, and in the case in the case $q=n-m+1$,we have ${n-m-q \atopwithdelims ()n-m} = {-1 \atopwithdelims ()n-m} = (-1)^{n-m} {n-m \atopwithdelims ()n-m}$. We arrive at the reduction

$$\begin{aligned} S_{n, m,n-m+1}&= (-1)^{n-m+1} \Bigg ({n-1\atopwithdelims ()n-m+1}\\&\quad {} + (-1)^{n-m+1} (-1)^{n-m}{n-m \atopwithdelims ()n-m} {2n -m \atopwithdelims ()n-m+1} {n-1\atopwithdelims ()0}\Bigg )\\&= (-1)^{n-m+1} \Bigg ({n-1\atopwithdelims ()m-2} - {2n -m \atopwithdelims ()n-1} \Bigg ). \end{aligned}$$

This is identity can be found in the Online Encyclopedia of Integer Sequences as entry A007318 (Pascal’s triangle). It appears that this identity is a challenge for both Maple© Mathematica©, they both produce a hypergeometric function as an answer.

Note that more identities can be discovered this way. For instance, if we set $p=n-m+2$, three terms remain in (9), corresponding to $q=0$, $q=n-m+1$, and $q=n-m+2$. The latter two correspond to binomial coefficients with upper indices $-2$ and $-1$, respectively. So, $S_{n,m,n-m+2}$ is reduced to three binomial coefficients.