How Much Should a 19th-Century French Bastard Inherit

Doron ZEILBERGER ^{1}

Dedicated to Gerry Ladas on his 60th Birthday.

Abstract: Catalan’s formula, for the portion of the inheritance that a legitimate child of a 19thcentury deceased French gentleman should receive, is given a new proof (using Difference Operators), and generalized. Another, more computationally efficient, formula is also derived.

Catalan’s Formula

Chrystal’s classical text[Ch] (II, p. 416) contains the following charming exercise, due to Catalan[Ca].

By French law an illegitimate child receives one-third of the portion of the inheritance that he would have received had he been legitimate. If there are l legitimate and n illegitimate children, show that the portion of inheritance 1 due to a legitimate child is

\frac{1}{l} − \frac{n}{3l(l + 1)} + \frac{n(n − 1)}{3^{2}l(l + 1)(l + 2)} − … ± \frac{n(n − 1) … 2.1}{3^{n}l(l + 1) … (l + n)} .

Let’s call the portion due to a legitimate child a(l, n). If the condition would have been that each illegitimate child receives one-third of what a legitimate child is to receive, this would have yielded the simple algebraic equation la(l, n) + n(1/3)a(l, n) = 1 and hence a(l, n) = 3/(3l + n). But since the bastard is to receive one-third of what he would have received had he been legitimate, we have a difference (or recurrence) equation:

la(l, n) + (n/3)a(l + 1, n − 1) = 1 .  (∗)

Introducing the shift operators Lf(l, n) := f(l + 1, n) and Nf(l, n) = f(l, n + 1), this can be rewritten:

(1 + \frac{n}{3l} LN^{−1})a(l, n) = \frac{1}{l} ,

yielding:

a(l, n) = (1 + \frac{n}{3l} LN^{−1})^{−1} \frac{1}{l} = ∑_{r=0}^{∞}(−\frac{n}{3l} LN^{−1})^{r} \frac{1}{l}

= ∑_{r=0}^{∞}(−1)^{r}(\frac{n}{3l} \frac{n − 1}{3(l + 1)} … \frac{n − r + 1}{3(l + r − 1)}L^{r}N^{−r})\frac{1}{l} = ∑_{r=0}^{n} \frac{(−1)^{r}n(n − 1) … (n − r + 1)}{3^{r}l(l + 1) … (l + r)} .

1 Department of Mathematics, Temple University, Philadelphia, PA 19122, USA. zeilberg@math.temple.edu ;

http://www.math.temple.edu/~zeilberg . Supported in part by the NSF. Feb. 6, 1997.

Another Formula

While Catalan’s formula is elegant, it is not any better than using the recurrence (∗) directly, by starting with a(l + n, 0) = 1/(l + n) and successively computing a(l + n − 1, 1), a(l + n − 2, 2), …, a(l, n), requiring O(n) operations and O(1) storage. The recurrence (∗) enables us to quickly find a(n + 1, l − 1) or a(n − 1, l + 1), once we know a(n, l), in case the status of one of the children changes. However, neither the recurrence (∗), nor Catalan’s formula, are efficient if we have already computed a(l, n), and all of a sudden another illegitimate child, that was previously unknown to the family, shows up at the lawyer’s office. By using zeillim in the package EKHAD that accompanies [PWZ], we find the recurrence

a(l, n + 1) = \frac{1}{l + n + 1} + \frac{2(n + 1)}{3(l + n + 1)}a(l, n) ,

that implies the alternative formula:

a(l, n) = \frac{n!(2/3)^{n}}{(n + l)!} ∑_{b=0}^{n} \frac{(l + b − 1)!}{b!}(3/2)^{b} .

The Multi-Mistress Case

A typical well-to-do 19th-century French gentleman must have had plusieurs amantes. It is very unlikely that he liked them equally well. Let’s say that he had m of them, and he wished to leave to every one of the n_{i} children of Mistress i, x_{i} of what he would have left them had they been legitimate (i = 1, …, m). Suppose that he also had l legitimate children. What portion of the inheritance should each of these l legitimate children receive? Let’s call this quantity a(l; n_{1}, …, n_{m}). Then,

(1 + ∑_{i=1}^{m} \frac{n_{i}x_{i}}{l} LN_{i}^{−1}) a(l; n_{1}, …, n_{m}) = \frac{1}{l} , and hence

a(l; n_{1}, …, n_{m}) =

(1 + ∑_{i=1}^{m} \frac{n_{i}x_{i}}{l} LN_{i}^{−1})^{−1} \frac{1}{l} = ∑_{r_{1},...,r_{m}≥0}^{∞}(−1)^{r_{1}+...+r_{m}} \frac{(r_{1} + … + r_{m})!}{r_{1}! ⋯ r_{m}!} (∏_{j=1}^{m}(\frac{x_{j}n_{j}}{l} LN_{j}^{−1})^{r_{j}} ) \frac{1}{l}

= ∑_{r_{1},...,r_{m}≥0}^{∞}(−1)^{r_{1}+...+r_{m}} \frac{(r_{1} + … + r_{m})!}{r_{1}! ⋯ r_{m}!} (∏_{j=1}^{m} x_{j}^{r_{j}} ) (∏_{j=1}^{m}(n_{j}N_{j}^{−1})^{r_{j}} ) ((1/l)L)^{r_{1}+...+r_{m}} \frac{1}{l}

= ∑_{r_{1},...,r_{m}≥0}^{∞}(−1)^{r_{1}+...+r_{m}} \frac{(r_{1} + … + r_{m})!}{r_{1}! ⋯ r_{m}!} \frac{1}{(l)(l + 1) ⋯ (l + r_{1} + … + r_{j})} ∏_{j=1}^{m} x_{j}^{r_{j}} ∏_{j=1}^{m} \frac{n_{j}!}{(n_{j} − r_{j})!} .

Hence

a(l; n_{1}, …, n_{m}) = ∑_{r_{1}=0}^{n_{1}} … ∑_{r_{m}=0}^{n_{m}} \frac{∏_{j=1}^{m}(−x_{j})^{r_{j}} \binom{n_{j}}{r_{j}}}{(1 + r_{1} + … + r_{m})\binom{l+r_{1}+...+r_{m}}{l−1}} .

REFERENCES

[Ca] E.C. Catalan, Nouv. Ann., ser. II, t.2.

[Ch] G. Chrystal, “ALGEBRA”, reprinted by Chelsea, New York, NY, 1964.

[PWZ] M. Petkovsek, H.S. Wilf, and D. Zeilberger, “A=B”, A.K.Peters, Wellesley, MA, 1996. The accompanying Maple package EKHAD may be downloaded from the author’s website.