On the Expansion of (XV)^{n} and Bessel Numbers

Abdelhay Benmoussa Secteur Scolaire Ansis, Unité Alanfoukt Tiqqi 80324, Agadir Ida Outanane Morocco

                 abdelhay.benmoussa@taalim.ma

Abstract We expand the operator (XV)^{n}, where X is multiplication by x and V is the Volterra operator. The resulting coefficients are shown to be the Bessel numbers (OEIS \underline{A001498}). We also present two applications and a conjectural generalization.

1 Introduction

Let

(Xf)(x) = xf(x), D = \frac{d}{dx}.

A well-known result in combinatorics states that [3]

(XD)^{n} = ∑_{k=0}^{n} S(n, k)X^{k}D^{k},

where S(n, k) denote the Stirling numbers of the second kind (OEIS \underline{A008277}).

In this note, we investigate an alternative setting obtained by replacing the derivative operator with the Volterra operator

V (f)(x) = ∫_{0}^{x} f(t) dt.

We show that the expansion of the operator (XV)^{n}, defined by

(XV)^{n}(f)(x) := \begin{cases}f(x), & if n = 0; \\ x ∫_{0}^{x}(XV)^{n−1}(f)(t) dt, & if n ≥ 1, \end{cases}

leads naturally to an explicit formula whose coefficients are the Bessel numbers (OEIS \underline{A001498}), given by

a(n, k) = [x^{k}] y_{n}(x) = \frac{(n + k)!}{2^{k} k! (n − k)!},

where y_{n}(x) is the n-th Bessel polynomial [2]. Then we present two applications, and conclude with a conjecture that generalizes the obtained formula.

2 Main result

Theorem 1. Let n ≥ 1. Then

(XV)^{n} = ∑_{k=0}^{n−1}(−1)^{k} a(n − 1, k) X^{n−k}V ^{n+k}.  (1)

Proof. Applying (XV)^{n} to a function f gives

(XV)^{n}(f)(x) = x ∫_{0}^{x} x_{n−1} ∫_{0}^{x_{n−1}} ⋯ x_{1} ∫_{0}^{x_{1}} f(t) dt dx_{1} ⋯ dx_{n−1}
= x ∫_{0}^{x} f(t) (∫∫∫_{t≤x_{1}≤···≤x_{n−1}≤x} x_{1}x_{2} ⋯ x_{n−1} dx_{1} ⋯ dx_{n−1}) dt.

Now, by induction we have

∫∫∫_{t≤x_{1}≤···≤x_{n−1}≤x} x_{1}x_{2} ⋯ x_{n−1} dx_{1} ⋯ dx_{n−1} = \frac{(x^{2} − t^{2})^{n−1}}{2^{n−1}(n − 1)!}.

Thus

(XV)^{n}(f)(x) = \frac{x}{2^{n−1}(n − 1)!} ∫_{0}^{x} f(t)(x^{2} − t^{2})^{n−1} dt

= \frac{x}{2^{n−1}(n − 1)!} ∫_{0}^{x} f(t)(x − t)^{n−1}(x + t)^{n−1} dt

= \frac{x}{2^{n−1}(n − 1)!} ∫_{0}^{x} f(t)(x − t)^{n−1} ∑_{k=0}^{n−1} \binom{n − 1}{k}(t − x)^{k}(2x)^{n−1−k} dt

= \frac{x}{2^{n−1}(n − 1)!} ∑_{k=0}^{n−1}(−1)^{k}\binom{n − 1}{k}(2x)^{n−1−k} ∫_{0}^{x} f(t)(x − t)^{n+k−1} dt

= ∑_{k=0}^{n−1}(−1)^{k} \frac{1}{2^{k}(n − 1 − k)!k!}x^{n−k} ∫_{0}^{x} f(t)(x − t)^{n+k−1} dt.

Applying Cauchy’s formula for repeated integration, we obtain

(XV)^{n}(f)(x) = ∑_{k=0}^{n−1}(−1)^{k} \frac{(n + k − 1)!}{2^{k}(n − 1 − k)!k!}x^{n−k} V^{n+k}(f)(x)

= ∑_{k=0}^{n−1}(−1)^{k}a(n − 1, k) x^{n−k}V^{n+k}(f)(x)

= (∑_{k=0}^{n−1}(−1)^{k}a(n − 1, k) X^{n−k}V^{n+k}) (f)(x).

3 Two applications

3.1 Power functions

Let α > 0. We first compute

(XV)^{n+1}(t^{α−1})(x) = x ∫_{0}^{x} \frac{(x^{2} − t^{2})^{n}}{2^{n}n!} t^{α−1} dt

= \frac{x}{2^{n}n!} ∫_{0}^{1} (x^{2}(1 − u))^{n}(xu^{1/2})^{α−1} \frac{x}{2}u^{−1/2} du (t = x\sqrt{u})

= \frac{x}{2^{n}n!} · \frac{x}{2} x^{2n}x^{α−1} ∫_{0}^{1}(1 − u)^{n}u^{(α/2)−1} du

= \frac{x^{α+2n+1}}{2^{n+1}n!} B(α/2, n + 1)

= \frac{x^{α+2n+1}Γ(α/2)}{2^{n+1} Γ(α/2 + n + 1)}

= \frac{x^{α+2n+1}}{α(α + 2) ⋯ (α + 2n)},

where B(x, y) is the Beta function [1].

On the other hand,

(XV)^{n+1}(t^{α−1})(x) = ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} V^{n+1+k}(t^{α−1})(x)

= ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} ∫_{0}^{x} \frac{(x − t)^{n+k}}{(n + k)!} t^{α−1} dt

= ∑_{k=0}^{n} \frac{(−1)^{k}a(n, k)}{(n + k)!} x^{n+1−k} ∫_{0}^{1} x^{n+k}(1 − u)^{n+k}x^{α−1}u^{α−1} x du (t = xu)

= ∑_{k=0}^{n} \frac{(−1)^{k}a(n, k)}{(n + k)!} x^{2n+α+1} ∫_{0}^{1}(1 − u)^{n+k}u^{α−1} du

= x^{2n+α+1} ∑_{k=0}^{n} \frac{(−1)^{k}a(n, k)}{(n + k)!}B(α, n + k + 1)

= x^{2n+α+1} ∑_{k=0}^{n} \frac{(−1)^{k}a(n, k)Γ(α)}{Γ(α + n + k + 1)}

= x^{α+2n+1} ∑_{k=0}^{n} \frac{(−1)^{k} a(n, k)}{α(α + 1) ⋯ (α + n + k)}.

By comparing the two expressions, we obtain the generating function

\frac{1}{α(α + 2) ⋯ (α + 2n)} = ∑_{k=0}^{n} \frac{(−1)^{k} a(n, k)}{α(α + 1) ⋯ (α + n + k)}.  (2)

3.2 Exponential function

Next, we consider the exponential function. Using Formula , we obtain

(XV)^{n+1}(e^{t})(x) = ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} V^{n+1+k}(e^{t})(x)

= ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} ∫_{0}^{x} \frac{(x − t)^{n+k}}{(n + k)!} e^{t} dt

= e^{x} ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} \frac{γ(n + k + 1, x)}{(n + k)!}

= e^{x} ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} (1 − e^{−x} ∑_{j=0}^{n+k} \frac{x^{j}}{j!} )

= e^{x} ∑_{k=0}^{n}(−1)^{k}a(n, k) x^{n+1−k} − ∑_{k=0}^{n} ∑_{j=0}^{n+k}(−1)^{k}a(n, k) \frac{x^{n+j+1−k}}{j!} ,

Here, γ(x, y) is the lower incomplete gamma function [1]. Rosengren proved [4] that

∑_{k=0}^{n} ∑_{j=0}^{n+k}(−1)^{k}a(n, k) \frac{x^{n+j−k}}{j!} = ∑_{k=0}^{n}(−1)^{n−k} \frac{(2(n − k) − 1)!!}{(2k)!!} x^{2k}.

Hence

(XV)^{n+1}(e^{t})(x) = x^{n+1}e^{x} y_{n}(−\frac{1}{x}) − ∑_{k=0}^{n}(−1)^{n−k} \frac{(2(n − k) − 1)!!}{(2k)!!} x^{2k+1}.  (3)

Evaluating at x = 1 yields the following Dobinski-like identity for the sequence a(n) = y_{n}(−1) (OEIS \underline{A000806}):

a(n) = \frac{1}{e} ((XV)^{n+1}(e^{t})(1) + ∑_{k=0}^{n}(−1)^{n−k} \frac{(2(n − k) − 1)!!}{(2k)!!}) .  (4)

4 Generalization

Using Cauchy’s formula for repeated integration, we obtain

XV − V X = V^{2}.

Motivated by this identity, consider the general commutation relation [3]

uv − vu = hv^{s},  h ∈ ℂ \ {0}, s ∈ ℝ.  (5)

We conclude with the following conjecture.

Conjecture 2. Let u and v be two variables satisfying (5). Then, for any integer n ≥ 1,

(uv)^{n} = ∑_{k=1}^{n} 𝔖_{s;h}(n, k) u^{k} v^{s(n−k)+k},  (6)

where 𝔖_{s;h}(n, k) are the generalized Stirling numbers introduced by Mansour and Schork [3].

5 Acknowledgments

The author thanks the referee and the editor for their valuable comments, which improved the manuscript.

References

[1] M. Abramowitz and I. A. Stegun, editors. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover.

[2] E. Grosswald. Bessel Polynomials, Springer, 1978.

[3] T. Mansour, M. Schork and M. Shattuck. The generalized Stirling and Bell numbers revisited. J. Integer Sequences 15 (2012), Article 12.8.3.

[4] H. Rosengren. Proving that a certain factorial double sum collapses to a double– factorial series. MathOverflow posting (2025). Available at https://mathoverflow.

   net/questions/499957.

2020 Mathematics Subject Classification: Primary 11B83; Secondary 11C08, 26A36. Keywords: Bessel number, Volterra operator.

(Concerned with sequences \underline{A000806}, \underline{A001497}, \underline{A001498}, \underline{A008277}, \underline{A122848}, and \underline{A122850}.)

Received July 26 2025; revised versions received July 30 2025; July 31 2025; December 6 2025; December 7 2025; May 19 2026. Published in Journal of Integer Sequences, May 20 2026.

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