Has anyone noticed this pattern?

Solution 1:

HINT:

If the $n$ th $\displaystyle T_n=n^r$

$\displaystyle T_{m+1}-T_m=(m+1)^r-m^r=\sum_{k=0}^{r-1}\binom mkm^k$

Observe that the difference of order $O(r-1)$

If we set $T'_m= T_{m+1}-T_m,$

$T'_{s+1}-T'_s $ will be of order $O(r-2)$ and so on

Reference :

Finite Difference I, II

Finite Sum of Power?

Solution 2:

If you have any polynomial with rational coefficients that takes positive integers to integers, it can be written uniquely in the form $$ f(x) = a_n{x \choose n} + \ldots + a_1 {x \choose 1} + a_0 $$ where if $x$ is not an integer we define $$ {x \choose n} = \frac{x(x-1)\ldots(x-1+n)}{n!} $$ (mnemonic: we are taking the formula for ${x \choose n}$ where $x$ is an integer and cancelling the part of the denominator that won't make sense when $x$ isn't.)

Note that, by considering ${x \choose n}$ itself, we see that there are polynomials with rational coefficients that take positive integers to integers but don't themselves have integer coefficients.

In fact, we can recover $a_n$ as $\Delta^n(0)$ where $\Delta f(m):= f(m+1) - f(m)$ and the $n$ means to iterate the operator $n$ times.

You have applied this construction to the polynomial $x^n$ and recovered that $a_n = n!$ in this case. You can see this without doing any computation by degree considerations.