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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.

Newb
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Veritas
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    Have you studied calculus? If so, you might find it interesting to look at the first derivative of $y=x$, the second derivative of $y=x^2$, the third derivative of $y=x^3$, etc. – G Tony Jacobs Jan 30 '14 at 17:54
  • See this. – David Mitra Jan 30 '14 at 17:57
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    This mechanism, of taking "distances" between consecutive values, is useful in generic Integer-Valued Polynomials as well. It even has its own operator, $\Delta$ for general operations. You might consider your particular method to be $p_n(x)=x^n, \Deltap_n=p_n(x+1)-p_n(x)$. – abiessu Jan 30 '14 at 18:02
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    @veritas i don't see why you feel stupid. You noticed a cool pattern and Tony told you why it was related to one you might already know. You can't immediately deduce your formula from Tony's or vice versa. – hunter Jan 30 '14 at 18:06
  • cough factorials? Say, if only we could do this for arbitrary exponents to generalize the factorial. Oh wait, we can and indeed all of this happens to be the motivation to the gamma function! $$n!=\int_0^\infty x^ne^{-x}dx$$ – Simply Beautiful Art Nov 29 '16 at 01:19

2 Answers2

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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?

Community
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lab bhattacharjee
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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.

hunter
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