Representing the Stirling polynomials ( ) σ x n in dependence of n and an application to polynomial zero identities

: Denote by σ n the n - th Stirling polynomial in the sense of the well - known book Concrete Mathematics by Graham, Knuth and Patashnik. We show that there exist developments xσ x

Abstract: Denote by σ n the n-th Stirling polynomial in the sense of the well-known book Concrete Mathematics by Graham, Knuth and Patashnik.We show that there exist developments xσ x j q j x 2 with polynomials q j of degree j.We deduce from this the polynomial identities 1 Introduction In the context of research on a problem of random flights in dimension 5which we discuss briefly in Section 4the second author conjectured the identities in the abstract, for which the authors could not find many hints in the literature.The work on its proof led us to a (for us at least) surprising result about the behaviour of the coefficients of sequences of Stirling polynomials.Let σ x n ( ) be the nth Stirling polynomial in the sense of [6]; the precise definition is given in Section 2, but see Table 1 for the coefficients of the first few Stirling polynomials.The table tells us, for example, that xσ x x 0 0 3 ( ) = + − x x .
1 48 We agree to begin row and column indices both with 0 and then multiply the entries in column j of Table 1 with j 2 j !.We then get Table 2.The first main result proven in the current article can be expressed as saying that the jth diagonal, i.e., the sequence of numbers in positions j j j , 0 , + …of Table 2, gives the values of a poly- nomial of degree j on the nonnegative integers.
Accept this for the moment and denote the sequence by q n .j n 0 ( ( )) ≥ As is well known, see almost any text on numerical analysis, e.g.[2, p. 95ff], using j 1 + interpolation points of distinct abscissae, in our case n 0, 1, 2, = …, one can compute a unique polynomial of degree j ≤ whose graph passes through these points.From Table 2, one finds, for example, q n q n n q n n n q n n n n 1, 12 , 1 288 , 26 15 5 51840 .
The sequence of numbers in the jth diagonal of the original table is given by .
We can use its row n to determine the polynomial xσ x n ( ) for symbolic n.The leftmost coefficient is the beginning and hence at position 0 of diagonal n.So, it has value .
The coefficient of x 1 pertains to diagonal n 1.
− It is at position 1 of that diagonal so has value .
In general, the coefficient pertaining to x j is at position j of diagonal n j − and, therefore, has value .As it happens, the fact q 0 0 j ( ) = for j 1 ≥ implies n q n j | ( ) so that n j q n j .This means that by putting q n q n n ˜j j ( ) ( ) ≔ / and using the polynomials q q q , , 0 1 2 , and q 3 computed above, we find xσ x q j x j q n x n q n x n q n x n q n x n q j x j q n x n q n x n q n x n      1 which shows that we could also write xσ x − for all j n , 0 ∈ ≥ probably does not exist because it would, for example, via the identity B m m σ 0 m m ( ) = − !, imply a simple formula for the Bernoulli numbers.
In Section 2, we collect a number of results on Stirling numbers and Stirling polynomials.In Section 3, we assume the representation xσ x a x 1 is polynomial of degree k; a fact equivalent to the representation claimed for xσ x n ( ) given in the abstract.The latter should have some importance for refined asymptotic analyses of the Stirling numbers of the second kind.To obtain that result, we have to solve a first-order difference equation with polynomial coefficients.
In Section 4, we deduce the identities mentioned in the abstract.In Section 5, we provide reasons to conjecture that the second of the tables shown above is actually an example of a Riordan array and indicate other possibilities for perhaps further fruitful work.
More important than the particular polynomial identity, which we derive, might be the methods which we employ.They should be applicable in a number of similarly looking identities.But we admit that it would be desirable to first simplify our proof significantly.In this vein, we also note that by introducing the notation x x k k k [ ] ≔ / !, the identities assume a more convenient form.
The a and a + notations come from physics where they denote creation and annihilation operators in quantum mechanics, but, e.g. a x d d = / and a x = + .i.e., multiplication with x allow interpreta- tion as operators on polynomial space.See Kim and Kim [10,11] there hold for n 0 > the following recursions; for easy combinatorial explanations see [6]: Representing the Stirling polynomials  3 Jekuthiel Ginsburg discovered in 1928 that there is a way to meaningfully define, for n 0 as polynomials in x of degree n 2 (so that whenever x is an integer n > there occur the usual Stirling numbers).This is explained in [6], where it is also observed that when x n 0, 1, 2, , { } ∈ … , then these polynomials are zero, and hence, we find that with the exception of the case n 0, = the expressions The authors of [5] approach the topic of Stirling polynomials differently.They are interested in the , not so much for its own sake but for giving a combinatorial interpretation to the coefficients of the power series x f nx 1 . In In [5], it is the f x k ( ) that are called Stirling polynomials.
The "Stirling polynomials" of [6] and the "Stirling polynomials" of [5] are not the same but they are easily transformed to each other.By [6, p. 267], for all k n , ∈ , and then, by the usual polynomial argument on the formal level, that We should mention that we learned of at least two other notions of Stirling polynomials, which have a loose connection to the polynomials f x k ( ) or σ x m ( ).)We shall need the following recursion formula for the σ , n mentioned in [6, Exercise 6.18].
Substituting, for the left-and right-hand sides of this equation, respectively, the definitions of the σ n , we obtain Now, this is simply an instance of the recursion formula for Stirling polynomials of the first kind.□ In Section 4, we will also use the following known facts.
Proposition 2. Let a a x a x j j j 0 ( ) = =∑ ≥ be any polynomial and let p x xσ x 1 .
Then, (a) One has the following equivalent identities of finite sums For n and k nonnegative integers, there holds The sums are finite because the a j for j a deg > are 0 and because for a negative integer s, one has s 1 0. / != The left formula is then essentially mentioned for polynomials a x ( ) that are of the form x l as [6, formula (6.19)], namely, The formula given follows as any polynomial is a linear combination of monomials.The right formula follows by multiplying both sides with 1 m ( ) − and using that 1 1 .
Use the definition of p n to conclude the proof.□ Remark.In part (a) of the proposition, note that if a m deg < , then in the equalities of sums, the ones at the right-hand sides, and hence at the left-hand sides, are 0. In addition, if a x x l ( ) = , then the right-hand sides and l m { } , respectively.The identities in part (a) are usually proved by applying the forward difference operator.A multivariate generalisation based on completely different reasoning can be found in [17].
The following two lemmas will be necessary only in the first part of Section 4.
From the proof, it follows that here 0 , 0 occurring as a special case of a , l ν − has to be interpreted as 1, because only with this understanding is the use of the binomial theorem correct.
In accordance with the notation used in [6], in the following lemma we use for integer i 0 Hence, the left equality can be deduced as follows: The right equality follows by dividing by n!. □

The main result on the diagonals of the modified coefficient table of Stirling polynomials
We transform the recursion of Lemma 1 into a matrix equation for the coefficients.
With the understanding that a a 0 x n σ x a na x , and hence, since below the expression within the large parentheses is 0 if evaluated for l n = , In this proposition, σ n was fixed and a n k − is the coefficient of x n k − in σ n and hence the coefficient of Since we have to consider the dependence on n as well in the sequel, we define The matrix equation of the previous proposition says that, for k n 2, 3, , ,  = … Doing the proper replacements according to a a 1 − − , we obtain, after a rearrangement, the following equation that is valid for k n 2, , = … : The last two lines in the following formula being clear and writing now k for k 1 − , this can also be written as follows: The main line is valid at first for k n 1, , 1 = … − but as it happens it is also valid for k 0; = in which case, it reproduces that a n 1 2 .
! By systematically checking the nine cases nε k 1 and kε 0 2 for ε ε , , , ) ∈ satisfies exactly one of the three conditions given by the "for..."; and beginning with any n and k, the base of the recursion will satisfy the second or third condition.Thus, the recursion is well defined.
The recursion serves well if one would desire a rapid computation of the polynomials The following MATHEMATICA © code can be used to define a n k , (as a[n, k]).Then, e.g.Recall that one of our main goals is to show that the sequence Representing the Stirling polynomials  7 is the sequence of values at the integers larger or equal k of a polynomial of degree k.Motivated by this, one may feel that it is simpler to work with a recursion for f n k , instead of a .
n k ,

Multiplying the main line of the recursion above with
One more simplification now yields: Corollary 6.The numbers f n k , satisfy the following recursion: Our guiding principle for proving the theorem below was the following observation.
OBSERVATION.Assume p 11 and p 12 are two polynomials of degree 1 with the same leading coefficient and assume that q is a polynomial of degree k.Then, "in general" there will exist a particular solution , which is a polynomial of degree k.The "proof" for this goes as follows: assume say, p n a bn 11 ( ) = + and p n c bn, 12 ( ) = + then the equation can be written as a bn f f a c f q n .
Up to names of variables, this is a special case of the equation q x u q x u q x u q x Δ Δ in [14, p. 377].Here, q x i ( ) is supposed to be a polynomial of degree i ≤ , i n 0, 1, , = … ; q x ( ) is a polynomial of degree m; and Δ is the forward difference operator.In this case, the book says, we can "in general find a particular solution by assuming that and equating coefficients."This is true, but one needs to prove that the linear forms in the b i for the coefficients of the powers of x obtained on the left-hand side by substituting the mentioned ansatz are sufficiently generic to have a solvable linear system of equations.This is what we do next in our specific case.
Theorem 7. Let k be a nonnegative integer.Then, the sequence Proof.The main line of the recursion for the f n k , can be rewritten as follows: This is a necessary condition that the f , n k , uniquely and well defined by the recursion, must satisfy.We know that a n 2 , ! − and so by definitions, f 1 n,0 = for all n.(This can also be deduced from the recursion that reduces for the case k 0 = to f f n n ,0 1,0 = − and uses f 1. 0,0 = ) So, f n,0 is a polynomial of degree 0. We now fix k 0 > and assume that, for j k 0, 1, 2, , 1, = … − the sequences n f j nj , ∋ ↦ ≥ are polynomial of degree j.The right-hand side of the recursion shown is then a polynomial, and it must be of degree k since there exists only one polynomial sequence of degree k 1 − in the sum, namely n f , and all other sequences f n j , occurring have lower degree.We will denote the polynomial (sequence) defining the right-hand side by q n cn j k j j 0 ( ) = ∑ = and have c 0. k ≠ Now, we make the ansatz and (again) with the understanding that a a 0 and recalling 0, 0 Thus, the left-hand side of the recursion 1 * is − and so we have to solve, for a a a , , , , Solvability is obvious because the linear form l i actually depends only on a a , , So, in matrix form, the system would be upper triangular k k 1 1 + × + without zeros on the diagonal.
So far, we have shown that the equation 1 * at the beginning of the proof allows for a polynomial solution of degree k.As yet our reasoning did not take into account the hypothesis n k ≥ of any initial values.The general solution to the equation is obtained as the family of all sequences f f ˙, , is the polynomial sequence obtained above and f ˙n n ( ) ∈ is any solution to the homogeneous equation is well defined and nontrivial in the sense that for the i used, the subindices of a occurring are nonnegative and that the first one is larger than or equal to the second one and the factorial occurring is also nonnegative.By Theorem 7, the sequence with coefficients in the n by n ν i + − we obtain a polynomial expression p n ν i ( ) + − which we may view as a polynomial in i.Its leading coefficient as a polynomial in i will be a real number equal to ± its leading coefficient as a polynomial in n.In particular, its degree in i will be equal to its degree in n.Thus, in particular, the sequence and, for that matter, the sequence on n 0, , { } … above defined at the beginning of the proof will be polynomial of degree ν μ.
− The sum of the lemma is of the form , where q is a polynomials of degree m n ν μ ν μ m n n ( ) ( ) ≤ − − + + − = − < , and therefore, 0 as follows from the remark to Proposition 2.
Proof.It is sufficient to establish the claim by substituting the polynomial p k n i ν ( ) − + of degree n i ν ≤ − + in the sum by any term of this polynomial.By the definitions of a n l , and p n , we have p x a x .≤ In addition, note ν m n n 1. ≤ − ≤ − Thus, we have the computation where in the second and third equality we used Proposition 2(a) (with m n i = − and a k k n i μ ( ) = − + ) and (b) (replacing there n by μ, k by n i − ), and in the last equality we used Lemma 8.So we have what we wanted.□ A trivial, but at the end important, corollary is as follows: Proof.Take the sum over all ν m n 0, 1, , { } ∈ … − of the expression above and interchange the two outer sums obtained.□ 4 Origin, proof, and impact of the zero identities Our research originated in a new approach to the odd-dimensional uniform random flight problem: assume a particle at instant 0 at the origin of an odd dimensional Euclidean space jumps exactly one unit from its current position in a random direction at each tick of the clock.(Here, the directions are defined as position vectors to uniformly distributed points on the origin-centred unit sphere.)Question: What isas a function of r -the probability to encounter the particle after exactly n random jumps within the 0-centred ball of radius r?This question was solved by García-Pelayo [7] using advanced analytical tools like Fourier analysis and the Abel transform.The article [16] shows the main result that the mentioned probabilities are piecewise polynomial by more elementary means.Based on [7], Borwein and Sinnamon [8] gave explicit formulas but these are quite complicated.
In particular, in an attempt to find an alternative formula for the case of dimension 5, the second author was led to conjecture that, putting Let us cast the formulation of this proposition into a more manageable form.First, note that the sum is actually finite.Assume the expression for c t l , incorporated in the second sum.Then, we can speak of an outer and an inner sum.Recall that k 0 ∈ < , which implies k 1 0. / != Assume, we choose in the outer sum t n. > Then n t 0 − < and so at least one of l μ μ n l t , − + − − is negative.Hence, by the definition of multinomial coefficients, each of the multinomial coefficients associated with the inner sum is 0. Thus, since the roles of l and t are symmetrical we can limit the outer sum and assume it is written as . It follows that the inner sum can also be limited as μ t l n l t min , ( ) Finally, we may also replace x n + by x in the proposition above, and after dividing by n! we see that the conjecture can be rewritten as claiming the following theorem.
Theorem 11.Let n be an integer larger than 1.Then, there holds the identity The rest of this section is dedicated to a proof of this theorem.
Clearly, the expression above is a polynomial in x of degree at most n 3 1.− It is 0 as claimed if and only if all its coefficients are 0. The claim in the abstract is obtained simply by taking the s 1 ( ) − -st derivative of the identity shown.So, we focus on the identity shown here, which corresponds to the case s 1.
= Now, for any positive integer λ, we have that We use this for λ λ a c n a c , and as we have seen, this formula holds true for all t n 0, 1, 2, , 3 Hence, we write t for t new and have to prove S t S t 0, If t n < , then the inner sum is empty, so the whole sum is 0, hence S t S t 0. 1 Representing the Stirling polynomials  13 Note that the expression , then this factor is 0. In other words, summing over all values f κ ν i , , ( ), where ν κ , , and i are defined by the inequalities in the original representation of K is the same thing as summing over the values f n k ν k n i i , , ( ) − + + − , where ν k , , and i are defined by the inequalities in K′.Thus, using in the second step below that, by Proposition 2 Now, if ν 0 < , then the degree of p n i ν − + is smaller than n i − , which means that the rightmost sum is 0 by Proposition 2a and the remarks following it.Hence, With this result established, we hope to publish a note soon with the more natural formula for a uniform random flight in dimension 5 than the one given in [8].We also should mention that we have another conjecture similar (but even more complicated) than Theorem 1, which would yield probably a simplification of the formula in [8] for random flights in dimension 7 but for this conjecture to be established maybe it is worthwhile to first work towards a more transparent proof of Theorem 1.

Concluding remarks and open problems
After having finished the bulk of this article, we decided to comb once more through the combinatorial literature for results possibly related to ours.From this search arise some remarks and questions.Numerical experiments very soon, and somewhat surprisingly, revealed that Table 2 seems to be a Riordan array with the sequence A a , etc. (The explicit Riordan arrays published in the literature we have seen seem to be almost all integer tables and mostly serve purposes in enumerative combinatorics.)We then tried to prove that our array is indeed Riordan using the above characterisation.Although this might not be too hard, it seems not quite trivial.Perhaps one of the other characterisations given in [13] can help.Another way to approach the topic of Riordanicity of our array might be to have a characterisation of the Riordan arrays for which for each k the sequence d n k n n , 0 ( ) + ≥ is polynomial of degree k.Interestingly, as is not hard to see that if the A-sequence of a Riordan array with constant diagonal 0 ≠ satisfies a 0, 1 ≠ then its kth diagonal is necessarily a poly- nomial of degree k.But the converse is false.with 1s on the diagonal.By the cited fact from [5], its kth diagonal is a sequence that is polynomial of degree k 2 .Is there any close connection between this fact and the fact that the diagonals of our modified coefficient table are polynomials of degree k?An article by Carlitz [3] made us think so for a short while, but in the authors hands, it did not pan out.The question intends to spur an effort to find an "immediate" proof of Theorem 3.3 using some facts from the immense body of already known results on Stirling numbers or polynomials.
(c) The online encyclopaedia for integer sequences [15] tells us that the sequence of integers 1, 12, 288, 51,840, which the reader can see in the denominators of polynomials q q q , , 0 1 2 , and q 3 in Section 2 actually also occurs as the sequence of denominators of an asymptotic series of the Gamma function.Indeed, see, e.g.[14,Exercise 17,p. 269].So the full story concerning the q i has yet to be discovered.May these questions and remarks spur further research!

Representing the Stirling polynomials  5 Lemma 4 .
If n and k are nonnegative integers, then It is easy to see that these equations can be encoded in the above matrix equation: for example, for l n 2 = − we obtain a a is equivalent to the last encoded equation.(The case l n 1 = − need not be encoded since it expresses na b is a consequence of the known fact that the leading coefficients for σ n and σ n 1 − are, as mentioned earlier, the coefficient of x 2 in xσ x .5 ( )Alternatively, one may also use generating function approaches like the identity [6, (6.50)], which reads ze and the Series[...] command to obtain the polynomials σ . n

Lemma 8 .
n k = , this equation degenerates to kf ˙0 k = so that f ˙0.k = But then we see by that f ˙0 n = by putting successively n k k 1, 2, = + + ….Therefore, the only solution to the equa- tion 1 * possible for n k ≥ is the polynomial solution found.By putting n k = in that equation, we obtain k k It so happens that the recursion of Corollary 6 tells us f 0 k k , = for the k 1. ≥ Therefore, the sequence n f k nk , ∋ ↦ ≥ coincides indeed with the polynomial sequence of degree k found for . 1 * □ Representing the Stirling polynomials  9Recall that in Proposition 2, we introduced the polynomials p Let m n μ , , , and ν be integers for which n

−
the sum is finite, and under these conditions the four lower indices of the multinomial coefficients are nonnegative and define a composition of n, i.e., their sum is n.Now, assume that four nonnegative integers a b c , , , and d define a composition of n.Then, let μ a l a b t a c occurring in the double sum, in the context of what we wish to prove, it can evidently be replaced by 1 .
, which in the dynamic environment of the above sum is always n 2 1 0. ≥ − > It can happen, though, that a b 0 + = , and at the same time, λ a c t , 0 ( ) − ≤ .In this case, we obtain, within the sum, undefined terms of form 0 1 0 ⋅ / or 0 . 0The latter has to be interpreted as 1 for reasons mentioned after Lemma 3. To steer clear from any interpretation problems, we argue using the given case distinction.Having chosen any t n 0 the restrictions in the first sum S t 1 ( ) impose b − Otherwise, by nonnegativity of a b c , , , and d, the sum is empty and hence 0satisfied, via the binomial theorem applied to powers of 1 1 , = Theorem 11 is proved.
(a)  There is a probable connection with Riordan arrays.Proper Riordan arrays can be characterised as those infinite lower triangular tables of complex numbers d n k n k .g.He and Sprugnoli [9, Theorem 2.1] for a proof of this discovery attributed to D. G. Rogers in 1978.
these numbers by using the following scheme, which is a consequence of above equations and triangularity of the array:

of
Stirling numbers of the second kind is an infinite triangular array (d) The polynomials xσ x n ( ) occur in the series development of ze e 1 MATHEMATICA code in Section 3. In the study by Koparal et al. [12, Section 1 and Theorem 3], we learn that the numbers ρ n k , n k ≥ , equal to n kσ k n ( ) ! and that these same numbers and the (signed) Stirling numbers of the first kind s n k n are related to the Daehee numbers of order m, via the equality s n k ρ k m D easily convinces that Daehee numbers figure prominently in recent articles.

Table 1 :
Coefficients of the first few polynomials

Table 2 :
Result of multiplying column j of Table1with 2 j j!