2026-01-05
In this manuscript we derive formula for multifold sums of powers using Newton’s formula and central differences.
In this manuscript we derive formula for multifold sums of powers using Newton’s formula and central differences.
The idea to derive sums of powers using difference operator and Newton’s series is quite generic, thus, formulas for sums of powers using forward and backward differences can be found in the works (Kolosov 2026a, 2026b).
We define the following recurrence for multifold sums of powers, which we utilize throughout the paper. \[\begin{align*} \Sigma^{0}\,{n}^{m} &= n^m \\ \Sigma^{1}\,{n}^{m} &= \Sigma^{0}\,{1}^{m} + \Sigma^{0}\,{2}^{m} + \cdots + \Sigma^{0}\,{n}^{m} \\ \Sigma^{r+1}\,{n}^{m} &= \Sigma^{r}\,{1}^{m} + \Sigma^{r}\,{2}^{m} + \cdots + \Sigma^{r}\,{n}^{m} \end{align*}\] Consider the Newton’s formula (Newton, Isaac and Chittenden, N.W. 1850) in central differences
Proposition 1 (Newton’s series in central differences). \[\begin{align*} f(x) &= \sum_{k=0}^{\infty} \frac{x^{[k]}}{k!} \delta^{k} f(0) \end{align*}\] where \(\delta^{k} f(0)= \sum_{j=0}^{k} (-1)^{j} \binom{k}{j} f\left(\frac{k}{2} - j\right)\) is central finite difference in zero, and \(x^{[k]} = n \left( n + \frac{k}{2} -1 \right)\left( n + \frac{k}{2} -2 \right) \cdots \left( n - \frac{k}{2} +1 \right)\) is central factorial.
Lemma 2 (Central factorial). \[\begin{align*} n^{[k]} = n \left( n + \frac{k}{2} -1 \right)\left( n + \frac{k}{2} -2 \right) \cdots \left( n - \frac{k}{2} +1 \right) = n \prod_{j=1}^{k-1} \left( n + \frac{k}{2} -j \right) \end{align*}\]
We observe that central factorials are closely related to falling factorials \(\left(x\right)_{n} = x(x-1)(x-2)(x-3)\cdots(x-n+1)=\prod_{k=0}^{n-1}(x-k)\). Therefore, \[\begin{align*} n^{[k]} = n \left(n+\frac{k}{2}-1\right)_{k-1} \end{align*}\] To derive formula for multifold sums of powers, we follow the strategy to express the Newton’s formula [eq ref] in terms of binomial coefficients, then to reach closed forms of column sum of binomial coefficients by means of hockey stick identity. Therefore,
Proposition 3. For \(k \geq 1\) \[\begin{align*} \frac{n^{[k]}}{k!} &= \frac{n}{k!}\left(n+\frac{k}{2}-1\right)_{k-1} = \frac{n}{k (k-1)!} \left(n+\frac{k}{2}-1\right)_{k-1} = \frac{n}{k} \binom{n+\frac{k}{2}-1}{k-1} \end{align*}\]
Proof. The identity above is true because \(\frac{\left(x\right)_{n}}{n!} = \binom{x}{n}\). ◻
Which yields Newton’s formula for powers, in terms of central differences. For positive integers \(n \geq 1\) and \(m \geq 1\) \[\begin{align*} n^m = \sum_{k=1}^{m} \frac{n}{k} \binom{n+\frac{k}{2}-1}{k-1} \delta^{k} 0^m \end{align*}\]
Although based on Newton’s interpolation series, the formula above begins the summation from \(k=1\) to avoid division by zero in \(\frac{n}{k}\). It is a valid trick, because the central difference \(\delta^{k} 0^n\) is zero for all \(n \geq 1\) and \(k=0\).
By factoring out and simplifying the term \(n\), we get \[\begin{align*} n^{m-1} = \sum_{k=1}^{m} \frac{1}{k} \binom{n+\frac{k}{2}-1}{k-1} \delta^{k} 0^m \end{align*}\] We may observe that the operator of central finite difference of power \(\delta^{k} 0^m\) requires the parity of its arguments \(m\) and \(k\) meaning that both \(m\) and \(k\) must be \(m\bmod2 = k\bmod2\) so that \(\delta^{k} 0^m\) is non-zero.
By setting \(m \rightarrow 2m\) we get \[\begin{align*} n^{2m-1} = \sum_{k=1}^{2m} \frac{1}{k} \binom{n+\frac{k}{2}-1}{k-1} \delta^{k} 0^{2m} \end{align*}\] Thus, the central difference \(\delta^{k} 0^{2m}\) for odd \(k\) is zero. Since \(k\) runs over all integers in the range \(0\leq k \leq 2m\), we are able to omit odd values of \(k\) \[\begin{align*} n^{2m-1} = \sum_{k=1}^{m} \frac{1}{2k} \binom{n+k-1}{2k-1} \delta^{2k} 0^{2m} \end{align*}\] Thus, formula for ordinary sums of odd powers is
Proposition 4 (Ordinary sums of powers). \[\begin{align*} \Sigma^{1}\,{n}^{2m-1} = \sum_{k=1}^{m} \frac{1}{2k} \binom{n+k}{2k} \delta^{2k} 0^{2m} \end{align*}\]
Proof. We have \(\Sigma^{1}\,{n}^{2m-1} = \sum_{k=1}^{m} \frac{1}{2k} \delta^{2k} 0^{2m} \sum_{j=1}^{n} \binom{j+k-1}{k-1}\).
By hockey stick identity \(\sum_{j=1}^{n} \binom{j+k-1}{2k-1} = \binom{n+k}{2k}\), thus the statement follows. ◻
Therefore,
Theorem 5 (Multifold sums of odd powers).
\[\begin{align*} \Sigma^{r}\,{n}^{2m-1} = \sum_{k=1}^{m} \frac{1}{2k} \binom{n+k-1+r}{2k-1+r} \delta^{2k} 0^{2m}. \end{align*}\]
Proof. We have \(\Sigma^{1}\,{n}^{2m-1} = \sum_{k=1}^{m} \frac{1}{2k} \delta^{2k} 0^{2m} \sum_{j=1}^{n} \binom{j+k-1}{2k-1}\).
By hockey stick identity \(\sum_{j=1}^{n} \binom{j+k-1}{k-1} = \binom{n+k}{2k}\). By induction the claim follows. ◻
The formula for multifold sums of odd powers can be rewritten in
terms of central factorial numbers of the second kind \(T(n,k)\), because \[\begin{align*}
k! T(n, k) = \delta^k 0^n
\end{align*}\] The identity above is well discussed in (Steffensen,
Johan Frederik 1927; Carlitz and Riordan 1963; Riordan 1968). The
non-zero central factorial numbers \(T(2m,2k)\) are registered as A008957 in the
OEIS (Sloane, Neil
J.A. and others 2003).
Hence,
Proposition 6 (Multifold sums of odd powers in central factorial numbers).
\[\begin{align*} \Sigma^{r}\,{n}^{2m-1} = \sum_{k=1}^{m} (2k-1)! \binom{n+k-1+r}{2k-1+r} T(2m,2k). \end{align*}\]
It is remarkable that exactly this formula can be found in Donald Knuth’s work Johann Faulhaber and sums of powers, see (Knuth, Donald E. 1993).
In this manuscript we derived formula for multifold sums of powers using Newton’s formula and central differences. Additionally, we discussed the connection with classical formulas for sums of powers (Knuth 1993 (Knuth, Donald E. 1993)).
Version: Local-0.1.0
License: This work is licensed under a CC BY 4.0 License.
DOI: https://doi.org/10.5281/zenodo.18096789
Sources: github.com/kolosovpetro/SumsOfPowersViaCentralDifferences
ORCID: 0000-0002-6544-8880
Email: kolosovp94@gmail.com