After seeing the neat little identity $(n+3)^2(n+2)^2(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form $\sum_{k=0}^a\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\epsilon_k=\pm1$. I discovered a relatively simple algorithm to generate these patterns: simply take $(n+2^p1)^p$ and subtract $(n+2^p2)^p$ (using $n\to n1$) to get a polynomial of degree $p1$. Take this difference and subtract $(n+2^p3)^p(n+2^p4)^p$ (using $n\to n2$) to get a polynomial of degree $p2$. Repeat this process until $n^p$ is reach. The first few examples of this are $$ \begin{align} n^0&=1\\ (n+1)^1n^1&=1\\ (n+3)^2(n+2)^2(n+1)^2+n^2&=4\\ (n+7)^3(n+6)^3(n+5)^3+(n+4)^3(n+3)^3+(n+2)^3+(n+1)^3n^3&=48 \end{align} $$ Upon doing this for the next several powers and checking OEIS, it would appear the constant corresponding to the power $p$ is $$\large C_p=2^{\frac{p(p1)}2} p!$$ However, this is an observation only, and I have no idea how to go about proving this. The only thing I notice is that $\frac{p(p1)}{2}=\sum\limits_{k=1}^{p1}k$, but I don't know how to use this fact. Does any one know how to prove this observation?

1$f(n) = \displaystyle\sum_{k = 0}^{a}\epsilon_k(n+k)^p$ is a polynomial with degree at most $p$. So if you can show that $f(n) = C$ for $p+1$ distinct integers $n$, then $f(n)$ must be constant. – JimmyK4542 Dec 24 '14 at 07:21

I already know that it's constant, the nature of the algorithm described above reduces the degree of the polynomial by one each time, until it reaches degree 0. I'm interested in the explicit form of the constant. – Pauly B Dec 24 '14 at 07:39

The fifth power identity can have the small sum, $$\sum_{n=1}^{168}\pm (x+n)^5=480$$ given in [this post](http://math.stackexchange.com/questions/1084103/). – Tito Piezas III Dec 30 '14 at 05:21

Suggestion: Wilson's theory in number theory for congruent modulos. – McTaffy Jul 24 '17 at 15:36
2 Answers
Let $X = \mathbb{R}^{\mathbb{Z}}$ be the space of real valued sequences defined over $\mathbb{Z}$. Let $R : X \to X$ be the operator on $X$ replacing the terms of a sequence by those on their right. More precisely,
$$X \ni (\ell_n)_{n\in\mathbb{Z}} \quad\mapsto\quad ( (R\ell)_n = \ell_{n+1} )_{n\in\mathbb{Z}} \in X$$ The identities you have can be rewritten as
$$\begin{array}{rcl} (R  1)n^1 &=& 1\\ (R^21)(R1) n^2 &=& 4\\ (R^4  1)(R^21)(R1) n^3 &=& 48\\ &\vdots&\\ (R^{2^{p1}}1)(R^{2^{p2}}1)\cdots(R^{2^0}1) n^{p} &\stackrel{?}{=}& C_p = ???\tag{*1} \end{array}$$
Notice for any polynomial $f(n)$ of degree $p$ and leading coefficient A,
$$f(n) = A n^p + A' n^{p1} + ( \text{something of degree }< p1 )$$ We have
$$\begin{align} (R^{2^{p1}}  1) f(n) &= A\left((n+2^{p1})^p  n^p \right) + A'\left((n+2^{p1})^{p1}  n^{p1}\right) + \cdots\\ &= A p2^{p1} n^{p1} + ( \text{mess with degree }< p1 )\\ \end{align} $$ This means $(R^{2^{p1}}1)f(n)$ will be a polynomial with degree $p1$ and leading coefficient $A \cdot p 2^{p1}$. Repeat applying this to the last equation in $(*1)$ $p$ times, we find the LHS of the last equation is equal to a polynomial with degree $0$. i.e. $C_p$ is indeed a constant. Furthermore, $$C_p = 1 (p 2^{p1} )((p1) 2^{p2}) \cdots (2 \cdot 2^{1}) (1 \cdot 2^{0}) = p! 2^{(p1)+(p2)+\cdots + 1} = p!2^{\frac{p(p1)}{2}}$$
 116,756
 6
 165
 318

1@rpm Yes. this is essentially a generalized finite forward difference. – achille hui Dec 24 '14 at 08:25

Note that since $R$ and $1$ commute,
$$
R^{2^k}1=\left[\sum\limits_{j=0}^{2^k1}R^j\right](R1)\tag{1}
$$
Therefore,
$$
\begin{align}
\prod_{k=0}^{n1}\left(R^{2^k}1\right)x^n
&=\left[\prod_{k=0}^{n1}\sum_{j=0}^{2^k1}R^j\right](R1)^nx^n\tag{2a}\\
&=\left[\prod_{k=0}^{n1}\sum_{j=0}^{2^k1}R^j\right]n!\tag{2b}\\
&=\left[\prod_{k=0}^{n1}2^k\right]n!\tag{2c}\\[6pt]
&=2^{n(n1)/2}\,n!\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $(1)$
$\text{(2b)}$: the $n^\text{th}$ forward difference of $x^n$ is $n!$
$\text{(2c)}$: on a constant sequence, $R^j=1$
$\text{(2d)}$: $\sum\limits_{k=0}^{n1}k=n(n1)/2$


@PaulyB: $(R1)^nx^n$ is just the [$n^\text{th}$ forward difference of $x^n$](http://math.stackexchange.com/a/838992), which is $n!$. – robjohn Dec 24 '14 at 17:25

@PaulyB: I've added an explanation to my answer to help clarify things. – robjohn Dec 24 '14 at 17:39