Has anyone noticed this pattern?

Question

I've been messing around a bit and I noticed a curious pattern when it comes to progressions of powers.

Let's take the progression of consecutive integers:

$1,2,3,4,5,6,7,...$

Obviously it's an arithmetic progression with a common distance of 1. And obviously 1 = 1!

So what if we take the progression of the squares of the above sequence?

$1^2,2^2,3^2,4^2,5^2,6^2,7^2,.. = 1, 4, 9, 16, 25, 36, 49,... $

we notice that the distances between the numbers are

$3,5,7,9,11,13,...$

which is an arithmetic progression with a common distance of 2. Take note that 2 = 2!

Now if we do the following for the cubes:

$1^3,2^3,3^3,4^3,5^3,6^3,7^3,...=1, 8, 27, 64, 125, 216, 343,...$

we notice that the distances between the numbers are

$7, 19, 37, 61, 91, 127,...$

At the first look there doesn't seem to be a pattern here however if we take the distances of the distances we get:

$12, 18, 24, 30, 36,...$

which as before we notice that it is an arithmetic progression with a common distance of 6. Take note that 6 = 3!

Now let's take the progression of the numbers raised in the power of 4:

$1^4,2^4,3^4,4^4,5^4,6^4,7^4,...=1, 16, 81, 256, 625, 1296, 2401,...$

As before we take the distances:

$ 15, 65, 175, 369, 671, 1105,...$

then the distances of the distances:

$ 50, 110, 194, 302, 434,... $

and finally the distances of the distances of the distances:

$ 60, 84, 108, 132,...$

We notice again that it's an arithmetic progression with a common distance of 24. Take note that 24 = 4!

I am not sure if this pattern has been observed but it looks promising, especially for series formulas.

Answer 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?

Answer 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.