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Introduction

Lets choose a natural number $n$ and generate a sequence $ k_1, k_2, k_3, ... k_n $ where each $k_n$ is replaced using $ f(k_a)= (a+1)^n - a^n $

The sequence then is $ (2^n-1), (3^n-2^n),(4^n-3^n),\dots,((n+1)^n-n^n). $

Let now for example $n=5$

The sequence is then: $31, 211,781,2101,4651$

Now lets build a number pyramid where the item in the next row is made by subtracting the two numbers behind it (basicly like when using a number pyramid to find the next number in a sequence if possible):

$$ (31) (211) (781) (2101) (4651) $$

$$ (180) (570) (1320) (2550) $$

$$ (390) (750) (1230) $$

$$ (360) (480) $$

$$ (120) $$

By expanding our pyramid with the 6th element in the first/top/longest row; its $9031$ and by climbing upwards the same way it gives us another $120$ at the last/bottom row which tells us that the pyramid is complete since the last row starts repeating.

The 120, or the last element (the one that completes the pyramid where $n=5$) is also $5!$

In general, for any natural $n$, we can generate a pyramid like this with $n$ rows and $n$ elements in the first row, and the $n!$ in the last row alone.

From this I concluded that $ n!=(n+1)^n-nn^n+\binom n2(n-1)^n-\dots $ which I was wondering about in my previous question: Expressing Factorials with Binomial Coefficients

I then have concluded that for all pyramids in general (where $n\ge2$ so it can have at least 2 rows) that the element $m$ in the row before the last one can be calculated using $$(\frac{n+(2m-1)}{2})n!$$

I then have concluded that the first element in the row before the one that is before the last one (or simply third from the bottom) can be calculated with $$ \frac{(3n^2+n+2)}{24}n! $$

And these 2 conclusions have been made by experimenting and comparing different pyramids.

Question

Instead of trying to find connections between the elements and writing formulas and expressions for elements in each row like I did with the previous two conclusions, how would one "create" a formula, something like $F(x,y,n)=...$ to calculate the $x$th element in the $y$th row for a pyramid $n$ of this type ($n$ rows) ?

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Vepir
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1 Answers1

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Using more readable notation: $u_{n}(k)=(k+1)^{n}-k^{n}$ for $k=1,\ldots, n$

Now $u_{n-1}(k)=u_{n}(k+1)-u_{n}(k)=(k+2)^{n}-2(k+1)^{n}+k^{n}$ for $k=1,\ldots, n-1$

$u_{n-2}(k)=(k+3)^{n}-3(k+2)^{n}+3(k+1)^{n}-k^{n}$ for $k=1,\ldots, n-2$

For $j=0,\ldots, n-1$,

$u_{n-j}(k)=(k+j+1)^{n}-\binom{j+1}{1}(k+j)^{n}+\binom{j+1}{2}(k+j-1)^{n}-\ldots+(-1)^{j+1}k^{n}$

where $k=1,\ldots, n-j$

What I've observed:

$u_{1}(1)=n!$ for $n\geq 1$

$u_{2}(k)=n!\left( k+\frac{n-1}{2} \right)$ for $n\geq 2$

$u_{3}(k)=n!\left( \frac{k^{2}}{2}+\frac{n-2}{2}k+\frac{(n-2)(3n-5)}{24} \right)$ for $n\geq 3$

$u_{4}(k)=n!\left( \frac{k^{3}}{6}+\frac{n-3}{4}k^{2}+\frac{(n-3)(3n-8)}{24}k+\frac{(n-2)(n-3)^{2}}{48} \right)$ for $n\geq 4$

Ng Chung Tak
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