Generalized Stirling Numbers and Hyper-Sums of Powers of Binomial Coefficients

We work with a generalization of Stirling numbers of the second kind related to the boson normal ordering problem (P. Blasiak et al.). We show that these numbers appear as part of the coefficients of expressions in which certain sequences of products of binomials, together with their partial sums, are written as linear combinations of some other binomials. We show that the number arrays formed by these coefficients can be seen as natural generalizations of Pascal and Lucas triangles, since many of the known properties on rows, columns, falling diagonals and rising diagonals in Pascal and Lucas triangles, are also valid (some natural generalizations of them) in the arrays considered in this work. We also show that certain closed formulas for hyper-sums of powers of binomial coefficients appear in a natural way in these arrays.


Introduction
Along the years, Stirling numbers have demonstrated to be a fundamental tool ('of the greatest utility' [17]) for dealing with combinatorial problems.We can find now many research works pursuing generalizations (in some sense) of Stirling numbers of both kinds (some of them accompanied with the corresponding combinatorial interpretations).We mention some: L. Carlitz [6] studied a type of λ-weighted Stirling numbers/polynomials which have demonstrated to be useful to understand different mathematical problems (see for example [12] and [18]); W. Lang [21] considered a (one-parameter) generalization of Stirling numbers, by using the language of 'infinitesimal transformations' (with some flavor from Lie groups theory).Lang himself [20] studied the corresponding combinatorial interpretation of his generalized Stirling numbers.One more generalized version of Stirling numbers appeared in the work of T. Mansour et al. [22], and one yet more in the work of M. Dziemiańczuk [11].We mention also the work of Tian-Xiao He [16], which presents a generalization that includes some other known generalizations of Stirling numbers.(See also Remark 1 at the end of this section.)Among all these works, there is one concerning Stirling numbers of the second kind, which has a natural connection with certain products of binomial coefficients.Having this connection as starting point we construct some number arrays which are natural generalizations of Pascal and Lucas triangles (and also of the corresponding results of the previous work [24]).To study these number arrays is the main theme of this article.The mentioned generalization of Stirling numbers of the second kind appears in the works of P. Blasiak et al. [2,4,23] (see also [1]).(Two works relating this generalization with other combinatorial-physics objects are [26,27]; two works about the combinatorial interpretations of these generalized Stirling numbers are [3,10].)We devote the rest of this section to present the basic facts of Blasiak's generalized Stirling numbers of the second kind.
Throughout the work, p denotes a given natural number.To keep the original author's notation, we use r and s to denote positive integers such that r s.The (r, s)-Stirling numbers of the second kind S r,s (p, k) (or simply generalized Stirling numbers of the second kind ), are introduced in the P. Blasiak's main work [2] as certain coefficients appearing in the so-called boson normal ordering problem.Roughly speaking, the numbers S r,s (p, k) are determined by the following equation: For example, if we set r = 4, s = 3 and p = 2, formula (1) looks as Letting act both sides of (2) on a C 6 (R) function y = y (x), the left-hand side becomes x 4 d 3 dx 3 x 4 d 3 y dx 3 = x 2 24x 3 d 3 y dx 3 + 36x 4 d 4 y dx 4 + 12x 5 d 5 y dx 5 + x 6 d 6 y dx 6 , so we have that S 4,3 (2, 3) = 24, S 4,3 (2, 4) = 36, S 4,3 (2, 5) = 12 and S 4,3 (2, 6) = 1.By convention one takes S r,s (p, 0) = δ p,0 .It can be shown that S r,s (p, j) = 0 if and only if s j sp, and that S r,s (p, ps) = 1 for p ∈ N. The standard Stirling numbers of the second kind S (p, j) correspond to the case r = s = 1.An explicit formula for S r,s (p, j) is S r,s (p, j) = (−1) j j! j i=s (−1) i j i where m s = s−1 l=0 (m − l) is the falling factorial.From (3) one can see at once that S s,s (p, s) = (s!) p−1 .(Note that when r = s = 1, formula (3) is the known explicit formula for Stirling numbers of the second kind S (p, k) = (−1) k k! k i=1 (−1) i k i i p .)It is possible to see from (1) that p j=1 (x + (j − 1) (r − s)) s = ps k=s S r,s (p, k) x k . (4) Observe that the case s = r of ( 4) is and that the case r = 1 of ( 5) is the well-known expression in which the Stirling numbers of the second kind S (p, k) appear as the connecting coefficients when x p is written as a linear combination of the falling factorials x k , k = 1, 2, . . ., p. Formula (4.56) -corrected-from Blasiak [2], gives us the recurrence relation (When r = s = 1 we have the known recurrence S (p + 1, k) = S (p, k − 1) + kS (p, k) for the usual Stirling numbers of the second kind.) An interesting particular case is when r = 2s.We claim that Clearly this formula is true for p = 1.Assuming the formula is true for a given p ∈ N, we have according to (7) that By using the induction hypothesis together with some algebraic manipulations, we can write (9) as But identity (6.47) from Gould's book [13] gives us that so we conclude finally from (10) and (11) that proving our claim (8).(The case s = 1 of ( 8) In [2] there are tables for S r,s (p, k) with r = 1, 2, 3, 1 s r and some small values of p and the corresponding s k ps.Some S 4,s (p, k) for 1 s 4 are the following:    Recall that the convolution of the sequences a n and b n is defined as the sequence a n * b n = n t=0 a t b n−t ; in particular a n * 1 is the sequence of partial sums of a n , and for a non-negative integer l, the sequence a n * l 1 is defined recursively as a n * 0 1 = a n and Beginning with the sequence n p (where p ∈ N is given), written as (see formula (6.10) in [15]) and the partial sums of it, namely n p * l 1, we studied in [24] the number arrays formed by the coefficients resulting when we write these sequences as linear combinations of binomial coefficients.We called them p-arrays.It turns out that these arrays are natural generalizations of Pascal's triangle (case p = 1) and Lucas triangle (case p = 2).We showed that many of the known properties of Pascal and Lucas triangles (the expected generalizations of those properties) are valid for the p-arrays, including the property on rising diagonals (recall that the sums of rising diagonals of Pascal's triangle are Fibonacci numbers, and in the Lucas triangle are Lucas numbers).We showed that the sums of the rising diagonals in a p-array is equal to some constant (the Stirling-Bernoulli transform of Fibonacci numbers, that depends only on p; see [25, A050946]) times a Fibonacci number, when p is odd, and times a Lucas number, when p is even.In this work we generalize the above results.Beginning with the sequence p j=1 n+(j−1)(r−s) s , where r, s ∈ N are given, and 1 s r, we show that a similar formula to ( 13) is valid, namely Observe that in the case r = s = 1, formula ( 14) is (13), and in the case r = s formula ( 14) looks as which gives us the p-th power of the binomial coefficient n s written as a linear combination of the binomials n j , s j sp.Moreover, we consider the partial sums of the sequence p j=1 n+(j−1)(r−s) s and show that they can be written as linear combination of binomial coefficients according to where k pr − s + 1.With the coefficients a (r,s) we form the (r, s, p)-arrays (with k for lines and j for columns), which generalize the p-arrays of [24], and then also generalize the Pascal's triangle (case r = s = p = 1) and Lucas triangle (case r = s = 1, p = 2).Observe that the case r = s of (16) give us the 'hyper-sum' (partial sums) of powers of binomial coefficients expressed as a linear combination of binomial coefficients, namely Expressions like (18) will appear several times in this work, with variants in the righthand side.What we have in the left-hand side of ( 18) is a (partial) sum of (partial) sums of the sequence of p-th powers of the binomial coefficients n s , in which s is fixed and n is the running index.To find closed formulas for sums of powers of binomial coefficients of the type n k=0 n k p are difficult problems (it is known that in some cases these formulas do not exist; see for example [5]).Our sums of powers of binomial coefficients do not belong to the difficult ones, but we believe that they are new results.
In section 2 we prove the main results of the work (formulas ( 14) and ( 16), in propositions 2 and 3, respectively).We also exhibit in this section the first few rows and columns of some (r, s, p)-arrays.In section 3 we mention some of the 'easy' properties of the (r, s, p)-arrays, related to their rows, columns, and falling diagonals.The proofs of most of these properties are left as easy exercises to the reader.In section 4 we consider the rising diagonals of the (r, s, p)-arrays.As in the previous work [24], this is the most important section of the article, and justifies that our (r, s, p)-arrays can be seen as generalizations of Pascal and Lucas triangles.In this section we consider only (r, s, p)-arrays in which r = s (see Remark 9).We prove that in the sums of rising diagonals in a (r, r, p)array appear Fibonacci numbers when r is even or r and p are odd (this includes the case r = 1 and p odd of [24]), or Lucas numbers when r is odd and p is even (this includes the case r = 1 and p even of [24]), multiplied by a constant that depends only on r and p.The corresponding sequences of these constants (with p ∈ N and r is fixed) generalize the sequence of Stirling-Bernoulli transform of Fibonacci numbers mentioned before.Finally, in section 5 we use a result of section 4 to obtain a different version of (15).It turns out that this new version can be simplified to a different expression for n s p similar to (15) but simpler than it.
Remark 1.We would like to do some comments on the direction presented in [8] for a generalization of Stirling numbers (see also the previous works related to this generalization [9,19], and the later work [7] containing q-versions of it).Certainly what is presented in [8] is different to the direction contained in Blasiak's works, but we think it is worth to mention it since we can see some similar mathematical ingredients in both (besides the non-mathematical fact -but a coincidence, after all-that the generalization presented in [8] is also a two parameters one, and these parameters are also denoted by r and s -with nothing to do with Blasiak's parameters r and s-).
The classical definition of Stirling numbers of the second kind S (n, k) in ( 6) can be generalized if the 'central' x (= x+0) of the left-hand side, is replaced by the 'non-central' x + r, namely The If now the powers (x + r) n of the left-hand side of (19) are replaced by the noncentral re-scaled falling factorials (sx + r) n , we obtain some new coefficients in the linear combination of the right-hand side of (19).We have The numbers C r,s (n, k) are called non-central generalized factorial coefficients (or Gould-Hopper numbers).Observe that (20) share with (4) the fact that both formulas express certain falling factorials (those of the corresponding left-hand sides) as linear combinations of some other falling factorials (those of the corresponding right-hand sides).So we can see the non-central generalized factorial coefficients C r,s (n, k) of ( 20) as the objects playing a similar role to the generalized Stirling numbers of the second kind S r,s (p, k) of (4).Some examples are the following We have the explicit formula [8, p. 335, ex. 35 A connection of C r,s (n, k) with S r (n, k) is the following where lim s→∞ (rs the electronic journal of combinatorics 21(1) (2014), #P1.10 (Observe that we have also a relation between the Blasiak's generalized Stirling numbers of the second kind and some generalized Lah numbers: see (8).) Formula ( 20) can be written as which gives us the binomial coefficient sn+r p written as a linear combination of the binomial coefficients n k , k = 0, 1, . . ., p. (Again we see that (24) has the same flavor of (15).)Note that ( 14) contains (15) (with integer p > 1) which is not contained in (24).On the other hand, formula (24) in the case s = 1, r = 0, is not contained in (14).In fact, we have not any interesting intersection in both results, which reminds us that, despite the shared flavor of both objects, they appear in different directions.Some examples from (24) are the following

The main results
In the following proposition we consider a version of (4) (and give a simple induction argument to prove it).This result (25), together with the corresponding formula (28) from proposition 3, are the starting point of the results we will obtain in the remaining sections.
Proposition 2. The following formula holds Proof.By induction on p.For p = 1 formula ( 25) is trivial.Suppose it is true for a given p ∈ N.Then, by using ( 7) and the induction hypothesis we have as expected.
From expression (25), we have at once the corresponding expression of the partial sums of the sequence n r p in terms of binomial coefficients.In fact, by using that we have for l ∈ N n r However, we are interested in expressions of the partial sums n r p * l 1 as linear combinations of the binomials n s , n s+1 , n s+2 ,. . ., since we want to study the number arrays formed by the corresponding coefficients of these expressions.Proposition 3.For k pr − s + 1 we have where the coefficients a (r,s) Proof.We proceed by induction on k.The case k = pr − s + 1 is (25).Suppose the result is valid for a given k pr − s + 1.That is, suppose that By taking the convolution in both sides of (30) with 1 we obtain that Thus, we need to prove that and this is an easy exercise left to the reader.
We form the (r, s, p)-arrays with the coefficients a (r,s) k,j (k for rows and j for columns).The first (pr − s) rows are cancelled, so the first non-cancelled row is the (pr − s + 1)th one, which contains the coefficients of the right-hand side of (25) (corresponding to k = pr − s + 1 of (28)).The first (s − 1) columns are cancelled, and the corresponding columns of the k-th row goes from the s-th column up to the (k − p (r − s) + s − 1)th column (with zeros in the subsequent places).Since the coefficients a (r,s) k,j have the Pascal's triangle property a (r,s) k+1,j+1 , we can also form the (r, s, p)-arrays beginning with the first non-cancelled row, and filling-out the array with the Pascal's triangle recurrence.
For example, in the case r = s (formula ( 18)), if we set k = s (p − 1) + 2, we can write for sums of powers of integers.)In the case r = s (18), if we set p = 1 we obtain that (for k 1) which gives us the partial sums of the sequence n s in terms of the binomials n j+s , j = 0, 1, . . ., k − 1.For example, the electronic journal of combinatorics 21(1) (2014), #P1.10 , and so on.That is, the (s, s, 1)-array is a "right-shifted Pascal's triangle", with its first column (the constant sequence 1) in the s-th column of the array.
In the case r = 2s we can use (8) to write (28) as or where k 2ps − s + 1.Note then that the (2s, s, p)-array is a "piece of Pascal's triangle multiplied by (sp)! (s!) p ": the first non-cancelled row of the array (the (2ps − s + 1)-th row) contains the s (p − 1)-th row of Pascal's triangle (multiplied by the factor (sp)! (s!) p ); the first non-cancelled column of the array (the s-th column) contains the first column of Pascal's triangle (multiplied by the factor (sp)! (s!) p ).That is, the s-th column of the array is the constant sequence (sp)!(s!) p .The subsequent rows and columns of the array are the corresponding subsequent rows and columns of Pascal's triangle multiplied by (sp)! (s!) p .See for example the (2, 1, 3)-array below, which is a Pascal's triangle beginning in row s (p − 1) = 2 and multiplied by (sp)! (s!) p = 6, and the (4, 2, 3)-array which is a Pascal's triangle beginning in row s (p − 1) = 4 and multiplied by (sp)! (s!) p = 90.Some (r, s, p)-arrays are the following:     (23).The coefficients come from k-th line of the (r,s,p)array.
3 Some properties (rows, columns, falling diagonals) From the well-known properties of rows, columns and falling diagonals in Pascal´s triangle, we state in this section the corresponding generalizations of them for our (r, s, p)-arrays.
Most of the proofs of these facts are easy exercises left to the reader.We begin by considering the rows of (r, s, p)-arrays.

Rows
We mention (leaving the proofs to the reader) the generalization of two famous properties of rows of Pascal's triangle: or more explicitly 2. Making the row k pr − s + 1 a single number N k (the elements of the row being the digits, and carrying over when appear elements with more than one digit), this is equal to 11 k−pr+s−1 N pr−s+1 , where N pr−s+1 is the single number corresponding to row k = pr − s + 1.
In A more interesting question arises when one considers the sequence of alternating sums of rows of a (r, s, p)-array.In the case r = s = 1, we proved that this sequence is ((−1) p , 0, 0, 0, . ..) (see formula (26) in [24]).In general, for a (r, s, p)-array, we have only some partial answers, as we show next.
First to all note that for k > pr − s + 1, the alternating row sums of the (r, s, p)-array are equal to 0 (easy proof left to the reader).The alternating sum of the elements of row k = pr − s + 1 is equal to α r,s,p = 1 (s!) p ps j=s (−1) j j!S r,s (p, j) . (38) Thus, (ignoring the cancelled rows) the alternating sums of rows in a (r, s, p)-array, are sequences of the form (α r,s,p , 0, 0, 0, . ..).Let us see some particular cases in which the term α r,s,p has a simple closed formula.(39) In fact, proceeding by induction on p, we see at once that this formula is trivial if p = 1.If it is true for a given p ∈ N, then (according to the recurrence ( 7)) But we have s i=0 (−1) i s i j−i+s s = 1 (see identity (53) below, or identity (1.83) in [14]).Thus, by using the induction hypothesis we have as expected.That is, the alternating sum of rows in the (s, s, p)-arrays is the sequence ((−1) ps , 0, 0, 0, . ..).
Other interesting particular case is when s = 1.We claim that The case p = 1 is obvious.If this formula is true for a given p ∈ N, we have (by using the recurrence (7)) (−1) j j!S r,1 (p, j) as expected.Some examples (within the particular case s = 1 we are considering) are the following: • If r = 1 we have α 1,1,p = (−1) p (as we obtained before).

Columns
According to (29), the s-th column (first non-cancelled column) in the (r, s, p)-array is the (constant) sequence a (r,s) For j s, the j-th column is If j ps, we can shift the corresponding sequence in order to show only the non-zero terms: For j > ps, we have the hockey-stick property (proof left to the reader) In general we have for j > ps and t, m ∈ N, (45)

Falling Diagonals
We consider first falling diagonals of the (r, s, p)-array, beginning in row k pr − s + 1 and column j = s.According to (29) these diagonals are the sequences (where n ∈ N) the electronic journal of combinatorics 21(1) (2014), #P1.10 In particular, observe that the falling diagonal beginning in row k = pr + 1 is (47) That is, the falling diagonal beginning in row k = pr + 1 and column j = s in the (r, s, p)-array is, according to (25), the sequence (the original sequence involved in the array).For example, let us consider the (3, 2, 3)array: the starting point of this array is the sequence n+1 n∈N .When writing this sequence in terms of the binomials n 2 , n 3 , . . ., n 6 , we obtain the coefficients 18, 126, 288, 270, 90 (which form the "first" -non-cancelled-row of the array).Then we form the (3, 2, 3)-array by using the Pascal's triangle recurrence.In this array, the falling diagonal beginning in row k = pr + 1 = 10 and column j = s = 2 is precisely the original sequence n+1 2 n+2 2 n+3 2 n∈N = (18, 180, 900, 3150, . ..).From the falling diagonal beginning in row k = pr +1, the subsequent diagonals can be obtained as the partial sums of the previous diagonal.That is, we have the hockey-stick property (proof left to the reader) where l ∈ N. In fact, observe that the falling diagonal beginning in row k = pr (and column j = s) is and then the falling diagonal beginning in row k = pr + 1 and column j = s, that is, the sequence p j=1 n+s−1+(j−1)(r−s) s , can also be obtained as the partial sums of the falling the electronic journal of combinatorics 21(1) (2014), #P1.10 diagonal beginning in the previous row (the one beginning in row k = pr and column j = s).
The case r = s of (49) give us the "hyper-sums of powers of binomial coefficients", namely Some examples from (49) are given in the following table: (2, 2, 2) Table 3: Hockey-stick property on falling diagonals of a (r,s,p)-array.Now let us consider the falling diagonals beginning in row k = pr − s + 1 and column j, where s < j ps.According to (29) these are the sequences (where n ∈ N) From ( 51) we see at once that the falling diagonal beginning in row k = pr − s + 1 and column j = sp is the constant sequence (ps)!(s!) p .It turns out that the considered falling diagonals can be described in a different way.Proposition 4. For l = s, s + 1, . . ., sp, and n ∈ N we have the identity Proof.By induction on n.For n = 1 we have to prove that (for s l sp) the electronic journal of combinatorics 21(1) (2014), #P1.10 But this is an easy exercise left to the reader (by using the explicit formula (3) for S r,s (p, l)).If ( 52) is true for all positive integers n, where n ∈ N is given, then as wanted.
The case p = 1 of (52) says that for any n, s ∈ N we have Also, if we set l = sp in (52) (and use that the corresponding falling diagonal is the constant sequence (sp)!(s!) p ), we obtain the following identity valid for any n ∈ N sp i=0 The case r = s of ( 54) is One more particular case of ( 52) is when r = 2s.By using (8) we can write (52) as

Rising diagonals
In this section, related to rising diagonals of the (r, s, p)-arrays, Fibonacci and Lucas numbers will appear in a natural way.Recall that the Fibonacci numbers sequence F n and Lucas numbers sequence L n satisfy the same second-order recurrence a n+2 = a n+1 +a n , with initial conditions F 0 = 0 and F 1 = 1 in the Fibonacci case, and L 0 = 2 and L 1 = 1 in the Lucas case.We will use without further comments the Binet's formulas for F n and L n , namely √ 5 (observe that α + β = 1; we will use this in the proof of lemma 6).
We consider the polynomials Proposition 5.The following formula holds Proof.By induction on p.For p = 1 it is easy to see that both sides of (58) are equal to r!.Let us assume that ( 58) is true for a given p ∈ N.Then, by using (7) we have that

Some further simplifications give us
as wanted (we used (7) in the last step) .
The desired formulas (59) and (60) are obtained from (61) and the hypotheses made on r and p in each case.

Lemma 7. (a)
The following identity holds (b) Let r, l ∈ N be given, such that l r.Then the electronic journal of combinatorics 21(1) (2014), #P1.10 Proof.(a) Some simple manipulations on the left hand side of (62) give us as wanted.
(b) We begin with the left-hand side of (63) and Binet's formulas to write as wanted.
In the following proposition we will use the identities which are consequences of the index-reduction formula the electronic journal of combinatorics 21(1) (2014), #P1.10 r is even or r and p are odd r is odd and p is even Proof.Let us prove first that if r is even or r and p are odd the formula holds for k 2r (p − 1).We proceed by induction on k.For k = 2r (p − 1) we have to prove that .
That is, according to (62) we have to prove that We begin with the left-hand side of (69) to write where we used (63) in the last step.Now use ( 64) and ( 59) to obtain finally that if r is even or r and p are odd we have as expected.The rest of the induction argument is an easy exercise (left to the reader) by using the recurrence of Fibonacci numbers (see Proposition 13 in [24]).Now we prove that if r is odd and p is even we have .
For k = 2r (p − 1) we have to prove (according to (62)) that if r is odd and p is even: We have where we used that r is odd and (63).Now use ( 65) and (60) to conclude finally that for r odd and p even we have as wanted.Again, the rest of the induction argument is an easy exercise left to the reader.
According to (29), the sum of the elements of the rising diagonal beginning in row k 2r (p − 1) and column j = s is Thus, in proposition 8 we have proved that when r is even or r and p are odd, the sum (72) is equal to and that when r is odd and p is even, this sum is equal to called "Stirling-Bernoulli transform of Fibonacci numbers" ( [25, A050946]).This is the sequence that appears in [24] and the result is that (for k 2 (p − 1)) the sum of rising diagonals is equal to the Stirling-Bernoulli transform (75) times the Fibonacci number F k−p+1 when p is odd, and times the Lucas number L k−p+1 when p is even.For r > 1 we have the following generalizations of the sequence (75), which we call "r-Stirling-Bernoulli transform of Fibonacci numbers": 1.If r is even, the r-Stirling-Bernoulli transform of Fibonacci numbers is the p-sequence For example, if r = 2 the sequence is (1, 7, 115, 3499, 170611, . ..) and if r = 4 the sequence is (1, 91, 54091, 116359591, . ..).In this case the result on rising diagonals is the following: In a (r, r, p)-array, where r is even, the sum of the elements of the rising diagonals (beginning in row k 2r (p − 1) and column r) is equal to the r-Stirling-Bernoulli transform of Fibonacci numbers (76) times the Fibonacci number F k−r(p−1) .
2. If r is odd, we have two versions of r-Stirling-Bernoulli transforms of Fibonacci numbers, depending on the parity of p: (b) For p even the sequence is and the result is the following: In a (r, r, p)-array, where r is odd and p is even, the sum of the elements of the rising diagonals (beginning in row k 2r (p − 1) and column r) is equal to the r-Stirling-Bernoulli transform of Fibonacci numbers (78) times the Lucas number L k−r(p−1) .
In  4: Sums of rising diagonals in a (r,r,p)-array, when r is even, or r and p are odd.
Remark 9. We believe that for the general case r > s, the sums of rising diagonals in a (r, s, p)-arrays are also related with Fibonacci and/or Lucas numbers.Note that the sum of the elements of the rising diagonal beginning in column s and row k (s + r)p − 2s is given by In the case r = 2s it is not difficult to see that the sum (79) is equal to However, formula (80) is not an interesting result, since we can write it (by using ( 8)) as and one can see at once that (81) is consequence of the following two easy to prove identities where m is a non-negative integer and r ∈ Z.We obtained explicit values of the sum (79) for some concrete values of r, s, p, k in the remaining cases (when 1 s < r, r = 2s) and observed that Fibonacci and/or Lucas numbers appear in these sums.In table 6 we have some examples.
We mention that we were not able to propose and prove the results (of the last column of Table 6) corresponding to this general situation.We believe that they are not as nice and simple as those we showed in this section for the case r = s.

Some additional results
By equating coefficients of similar powers of x in both sides of (58) we see that In this section we want to obtain some consequences of this identity.First of all, observe that ( 15) and (83) give us that which can be written as After a simplification procedure (that we describe next), using Pascal's triangle recurrence, the right-hand side of (85) can be written in a simpler form, in which we need only 'the half' of the number of coefficients involved in (85).To prove this fact is one of the goals of this section, and it is contained in corollary 11.
The mentioned simplification procedure is as follows.
Step 1 Write the last (the k = sp-th) term of the right-hand side of (85), namely the term (sp)! (s!) p S s,s (p, sp) n+sp−s sp , as Step 2 Write together the first summand of (86), namely , to get the new k = (sp − 1)-th summand of (85) as Step Now we take (89) as the starting point and repeat Procedure, beginning with the last term of the right-hand side of (89) and going backwards, up to the third term (that is, we leave the first two summands of the right-hand side of (89) as they appear in this expression, namely (−1) sp (s!) p By using Gould's identity (3.50) in [13], we see that which can be written as Beginning with the left-hand side of (100) and using (102) we get sp j=s (−1) j j!S s,s (p, j) (easy proof, left to the reader), together with the hypothesis that s is even or s and p are odd, give us from (105) the end of the proof of (100).
(b) The case t = 0 of (99) can be written as sp j=s (−1) j j!S s,s (p, j) = s!S s,s (p, s) .
This is a direct consequence of (58) (with x = 0) and the fact that p is even.So let us consider the cases when t 1.We can write (99) as From Gould's identity (3.50) in [13] we see that which can be written as

x r d s dx s p =
x p(r−s) ps j=s S r,s (p, j) x j d j dx j .

(− 1 )
i+r+1 (i − r)!S r,r (p + 1, i) F i−r is the following: In a (r, r, p)-array, where r and p are odd, the sum of the elements of the rising diagonals (beginning in row k 2r (p − 1) and column r) is equal to the r-Stirling-Bernoulli transform of Fibonacci numbers (77) times the Fibonacci number F k−r(p−1) .

Table 1 :
Examples of sequence
tables 4 and 5 we have some more examples.

Table 5 :
Sums of rising diagonals in a (r,r,p)-array, when r is odd and p is even.
If s is odd and p is even, then for any integer l 0 we have Some examples from (113) and (114) are the following (compare them with similar expressions shown in Table1):