A Determinant Formula for Sums of Powers of Integers

In this note, we first derive a recursive formula for the sum of powers Sk(n) = 1 k + 2k + · · · + nk, with k and n non-negative integers. We then apply it to establish, via Cramer’s rule, an explicit determinant formula for Sk(n) involving the Bernoulli numbers and the binomial (n+ 12). Evaluating the determinant gives us directly Sk(n) in the form of the so-called Faulhaber polynomial, namely as a sum of even or odd powers of (n+ 12). Furthermore, a connection with Hessenberg matrices is shown. Mathematics Subject Classification: 11B57, 11C08, 11B68


Introduction
For k and n non-negative integers, define the sum of powers and set S k (0) = 0 for all k.In this note (Section 2), we derive the following recursive formula for S k (n): where B j is the j-th Bernoulli number.Next, in Section 3, we apply (1) to obtain, by Cramer's rule, an explicit determinant formula for S k (n) in terms of the Bernoulli numbers and the binomial (n + 1  2 ).Evaluating the corresponding determinant yields S k (n) in the form of a polynomial in n + 1  2 .This type of polynomials are known in the literature as Faulhaber polynomials, after the German mathematician Johann Faulhaber (1580-1635) who expressed for the first time the S k (n)'s as polynomials in S 1 (n) = 1  2 n(n + 1) [6,7,11,1] (see Theorem 4.1).As will be shown in Section 4, the latter is equivalent to the fact that the S k (n)'s are the sums only of even or odd powers of (n + 1  2 ) depending on whether k is odd or even (Theorem 4.2).In Section 5, we define an n×n Hessenberg matrix H n and write down the recurrence relation satisfied by the sequence of determinants {det(H n ), n ≥ 1}.Then, formula (1) is rederived from said recurrence relation.Some concluding remarks are provided in Section 6.
We point out that this note could be of interest for use in introductory courses on number theory or discrete mathematics as enrichment material related to the sums of powers of integers.

Proof of the recursive formula
The proof of formula (1) is based on the following lemma.
Lemma 2.1.For n, k ≥ 1, we have ( Proof.Next, we give the most simple and direct proof of (2) one can conceivably construct.For that, it suffices to display (n + 1)S k−1 (n) as the sum of the n + 1 rows and then note that this sum can be decomposed as the sum of the following two pieces: Clearly, summing the rows of the piece on the left gives n i=1 S k−1 (i).Further, the entries in the i-th column of the piece on the right sum to i k , and hence the sum of all these columns amounts to S k (n).Thus, the lemma follows.
For a historical account of equation ( 2), the reader is referred to Katz's paper [10].It is a well-known result, on the other hand, that S k−1 (n) is expressible as a polynomial in n of degree k with zero constant term: Moreover, the remaining coefficients a k−1,r are connected to the Bernoulli numbers through the relation (see, for example, [13]) Equipped with relations (2), (3), and (4), the proof of (1) is straightforward.
For that, we first evaluate S k−1 (i) for each i = 1, 2, . . ., n.Using (3) we have Then summing terms corresponding to the same coefficient a k−1,r in all these equations yields Substituting this expression into (2) and recalling (4), one finally gets formula (1).

A determinant formula for sums of powers of integers
Unless otherwise stated, in what follows we drop the explicit dependence of S k (n) on n and write this merely as S k .It is easily seen that the well-known formulas for S 1 and S 2 where N = n + 1 2 .Now, by letting k = 3, 4, 5, . . ., p in (1), we obtain the system of equations The sets of equations ( 5) and ( 6) can be combined in matrix form as Then, solving for S p and applying Cramer's rule to the above p × p lower triangular system of linear equations yields By expanding the determinant with respect to the last column, this simplifies to Evaluating the determinant of order p − 1 in (8) gives us S p as a polynomial in N .So, for the first values of p we find and so on.Note that S 1 , S 3 , and S 5 involve only even powers of N , whereas S 2 , S 4 , and S 6 involve only odd ones.As we will see in the next section, this feature holds generally so that S p (p ≥ 1) is expressible as an even or odd polynomial in N depending on whether p is odd or even [8].
On the other hand, it is clear from (8) that the term of maximum degree in N corresponds to the product of the elements on the main diagonal of the determinant.The term of maximum degree of S p will be thus given by Indeed, the simple formula proves a fairly good approximation to the sums of powers [3].
Unfortunately, the determination of the trailing coefficients of S p using ( 8) is considerably more difficult to achieve than it was for the leading coefficient.Upon examination of the above formulas for S 2 , . . ., S 6 , however, one can correctly guess that the next highest term of S p (p > 1) is given by − p 24 N p−1 .A complete characterization of the coefficients of S p expressed as a polynomial in N is given elsewhere using the technique of generating functions [5].Finally, we note that, for any given p, the set of power sums {S 1 , S 2 , . . ., S p } can alternatively be obtained by inverting the square matrix in equation ( 7).

Symmetry and Faulhaber polynomials
An immediate consequence of formula (8) is that, for all p ≥ 1, S p is a multiple of S 1 since the factor N 2 − 1 4 is proportional to S 1 .Furthermore, for odd p ≥ 3, S 2  1 divides S p , the quotient S p /S 2 1 being a polynomial in S 1 .On the other hand, for even p ≥ 2, S 2 divides S p , the quotient S p /S 2 being a polynomial in S 1 .These results were essentially known to Faulhaber by 1615, and can be formally established in the following theorem [1,7,11].Theorem 4.1 (Faulhaber).(i) For p = 3, 5, 7, . . .there exists a polynomial F p in S 1 of degree 1  2 (p − 3) such that S p = S 2 1 F p (S 1 ).(ii) For p = 2, 4, 6, . . .there exists a polynomial F p in S 1 of degree 1 2 (p − 2) such that S p = S 2 F p (S 1 ).Let us now focus on the symmetry properties of S p .For this purpose, it is convenient to extend the definition of S p to view it as a polynomial function of the continuous real variable x.Consider first the case of odd p. Regarding S 1 , from the expression For odd p ≥ 3, from Theorem 4.1 we have that S p (x) = S 1 (x) 2 F p S 1 (x) , and then, noting (9), it follows that So, for odd p ≥ 1 and for all values of x, it is the case that S p x − 1 2 = S p −x− 1 2 revealing the fact that S p (x) is symmetric with respect to reflection in the line x = − 1 2 .This amounts to saying that, for odd p, S p (x) can be expressed as an even polynomial in x + 1 2 .Similarly, from the expression S 2 (x) = 1 3 (2x + 1)S 1 (x), one can easily check that Now, for even p ≥ 2, from Theorem 4.1 we have that S p (x) = S 2 (x)F p S 1 (x) .Then, using ( 9) and ( 10), it is readily seen that, for even p ≥ 2 and for all values of x, S p x − 1 2 = −S p − x − 1 2 .This means that S p (x) is symmetric with respect to reflection in the point (− 1 2 , 0), and thus S p (x) can be expressed as an odd polynomial in x + 1  2 .Therefore, we can state the following version of Faulhaber's theorem.Theorem 4.2 (Faulhaber).(i) For p = 1, 3, 5, . . .there exists an even polynomial F p in N = n + 1 2 of degree p + 1 such that S p = F p (N ).(ii) For p = 2, 4, 6, . . .there exists an odd polynomial F p in N of degree p + 1 such that S p = F p (N ).
We have just shown that Theorem 4.1 implies Theorem 4.2.It can be shown, conversely, that Theorem 4.2 implies Theorem 4.1, so that these are actually equivalent to each other.Interestingly, the implication Theorem 4.2 ⇒ Theorem 4.1 is nicely seen by using the formula (8).For that, let us denote the determinant appearing in (8) as det(H p−1 ) and consider separately the cases of odd and even p: • For odd p, from Theorem 4.2 we have that S p is expressible as an even polynomial in N of degree p + 1.So, from (8) we deduce that det(H p−1 ) must be an even polynomial in N of degree p − 1, and thus Since N 2 = 2S 1 + 1 4 , this can be written as Clearly, for p = 1 we have that a = 0 and b = c = • • • = d = 0. Now the key point to observe is that, for odd p ≥ 3, a = 0.This follows from the fact that for this case, namely for odd p ≥ 3, we have that S p (x) = pS p−1 (x), and then S p (x = 0) = 0 [14].Hence, differentiating (11) with respect to x gives S p (x) , and then S p (x = 0) ∝ a .Thus putting a = 0 in (11) yields the Faulhaber form, . Moreover, it can be proved that b , c , c , . . ., d = 0.
• For even p, from Theorem 4.2 we have that S p is expressible as an odd polynomial in N of degree p+1.Hence, from (8) it follows that det(H p−1 ) must be an odd polynomial in N of degree p − 1, and thus Factorizing N and noting that N 2 − 1 4 N ∝ S 2 , we have Then, as N 2 = 2S 1 + 1 4 , we retrieve the Faulhaber form, . Likewise, it turns out that a , b , c , . . ., d = 0.

A connection with Hessenberg matrices
An n × n matrix H n = (h ij ) is called a (lower) Hessenberg matrix if all entries above the superdiagonal are zero but the matrix is not lower triangular, i.e., , where h i,i+1 = 0 for some i = 1, 2, . . ., n − 1.Consider the sequence of determinants {det(H n ), n ≥ 1} and define det(H 0 ) = 1.Then it can be shown (see, for example, [4]) that {det(H n ), n ≥ 1} satisfies the recurrence relation where Now, it is easy to see that formula (8) can be rewritten as where H p−1 is (whenever p ≥ 3) a lower Hessenberg matrix of order p − 1 with Then, letting n = p − 1 and replacing det(H p−1 ) and det(H r−1 ) by the corresponding expression in equation ( 13), recurrence (12) becomes Multiplying both sides of this equation by 2(−1) p+1 S 1 , we get Thus dividing both sides of this equation by (p + 1)!, and after renaming the index p as k, we come full circle and find again formula (1).

Conclusion
In this note, we have derived a recursive formula for the sums of powers S p (n) = n r=1 r p .Using Cramer's rule, this formula allows us to obtain a determinant formula for S p (n) in terms of the Bernoulli numbers and the binomial (n + 1  2 ).Evaluating the determinant gives us S p (n) as a sum of even or odd powers of (n + 1  2 ).We have also shown the equivalence between the two forms of Faulhaber's polynomials for expressing S p (n).Furthermore, the relationship between the recursion formula (1) for S p (n) and the recurrence (12) for the sequence of determinants of a Hessenberg matrix has been worked out.
It should be mentioned that there are other determinant formulas giving S p (n) as a polynomial in n or in n + 1 [2].Such formulas for S p (n), however, involve determinants of order p + 1, unlike our determinant which is of order p − 1.Moreover, to the best of our knowledge, both the recursive formula (1) and the resulting determinant formula (8) for S p (n) appear to be new.