# Number Theory/Egreteau's nabla triangle

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Egreteau's ${\displaystyle \nabla}$ triangle is a different expansion of the binomial coefficients, which allows their use for negative ${\displaystyle -}$ and positive numbers at the same time; together with the subtraction rule, this in return allows for the creation of completely new ${\displaystyle p-}$nomial triangles. The ${\displaystyle \nabla}$ indicates that negative binomial coefficients ${\displaystyle -}$ and the subtraction rule is employed. In contrary to Egreteau's ${\displaystyle -}$ and Pascal's triangle, this binomial triangle starts with a positive ${\displaystyle -}$ and a negative entry.

## Definition

The entry in the ${\displaystyle n-}$th row and ${\displaystyle k-}$th column of Egreteau's ${\displaystyle \nabla}$ triangle is denoted by:

${\displaystyle \displaystyle\binom{-n}{-k}_{\nabla}}$

We set:

${\displaystyle \begin{array}{lll} \displaystyle\binom{0}{0}_{\nabla} := 0\text{,} & \displaystyle\binom{-n}{0}_{\nabla} := -1\text{,} & \displaystyle\binom{-n}{-n}_{\nabla} := 1 \\ \end{array}}$

for ${\displaystyle n \geq 1}$. Further, Egreteau's ${\displaystyle \nabla}$ triangle fulfills the subtraction rule:

${\displaystyle \displaystyle\binom{-n - 1}{-k - 1}_{\nabla} = \displaystyle\binom{-n}{-k}_{\nabla} - \displaystyle\binom{-n}{-k - 1}_{\nabla}}$

for all ${\displaystyle k}$, ${\displaystyle n \geq 2}$. With the stated binomial coefficients the triangle can be visualized as follows:

${\displaystyle \begin{array}{ccccccccc} \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & \\ & \displaystyle\binom{-3}{0}_{\nabla} & & \displaystyle\binom{-3}{-1}_{\nabla} & & \displaystyle\binom{-3}{-2}_{\nabla} & & \displaystyle\binom{-3}{-3}_{\nabla} & \\ & & & & & & & & \\ & & \displaystyle\binom{-2}{0}_{\nabla} & & \displaystyle\binom{-2}{-1}_{\nabla} & & \displaystyle\binom{-2}{-2}_{\nabla} & &\\ & & & & & & & & \\ & & & \displaystyle\binom{-1}{0}_{\nabla} & & \displaystyle\binom{-1}{-1}_{\nabla} & & & \\ & & & & & & & & \\ & & & & \displaystyle\binom{0}{0}_{\nabla} & & & & \\ \end{array}}$

which continues with ${\displaystyle n = 4}$, ${\displaystyle 5}$, ${\displaystyle \dots}$; equivalent to the binomial coefficients are their values, which are as follows:

${\displaystyle \begin{array}{ccccccccccc} \dots & & \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & & & \\ & -1 & & -2 & & \hspace{8pt}4 & & -4 & & 1 & \\ & & & & & & & & & & \\ & & -1& & \hspace{8pt}1 & & -3 & & 1 & &\\ & & & & & & & & & & \\ & & & -1 & & -2 & & \hspace{8pt}1 & & &\\ & & & & & & & & & & \\ & & & & -1 & & 1 & & & & \\ & & & & & & & & & & \\ & & & & & \hspace{8pt}0 & & & & &\\ \end{array}}$

## Characteristics

With the established binomial coefficients and rule, Egreteau's ${\displaystyle \nabla}$ triangle has some analogous characteristics as those triangles we have already encountered. The most obvious one is:

${\displaystyle \displaystyle\binom{-n}{-1}_{\nabla} = \begin{cases} -2 & \ \textrm{if } \vert -n \vert \textrm{ is even}\\ 1 & \ \textrm{if } \vert -n \vert \textrm{ is odd}\\ \end{cases} \\ }$

for ${\displaystyle n \geq 2}$ and:

${\displaystyle \displaystyle\binom{-n}{-n + 1}_{\nabla} = -n}$

for ${\displaystyle n \geq 1}$. This yields:

${\displaystyle \displaystyle\binom{-n}{-2}_{\nabla} = \begin{cases} -3 \bigg( \displaystyle\frac{\vert -n \vert - 1}{2} \bigg) & \ \textrm{if } \vert -n \vert \textrm{ is odd}\\ \vert -n \vert + \bigg(\displaystyle\frac{\vert -n \vert}{2} - 2\bigg) & \ \textrm{if } \vert -n \vert \textrm{ is even}\\ \end{cases} \\ }$

for ${\displaystyle n \geq 2}$.

## Theorem

Let ${\displaystyle 2 \leq n \in \mathbb{N}}$. Then we have:

${\displaystyle \displaystyle\sum\limits_{k = 0}^{n}~\displaystyle\binom{-n}{-k}_{\nabla} = -2}$

For ${\displaystyle n = 0}$ and ${\displaystyle n = 1}$ the sum equals obviously to zero.

#### Proof

This is a direct consequence of:

${\displaystyle \displaystyle\binom{-n - 1}{-k - 1}_{\nabla} = \displaystyle\binom{-n}{-k}_{\nabla} - \displaystyle\binom{-n}{-k - 1}_{\nabla}}$

which creates an alternating pattern of the sign within the triangle. Those values then annihilate each other. Therefore by induction follows automatically the made statement.

## Theorem

Let ${\displaystyle 2 \leq n \in \mathbb{N}}$ and ${\displaystyle \Delta}$ be the difference operator. Then we have:

${\displaystyle \displaystyle\Delta_{k = 0}^{n} \displaystyle\binom{-n}{-k}_{\nabla} = 0}$

#### Proof

This is a direct consequence of:

${\displaystyle \displaystyle\binom{-n - 1}{-k - 1}_{\nabla} = \displaystyle\binom{-n}{-k}_{\nabla} - \displaystyle\binom{-n}{-k - 1}_{\nabla}}$

By induction follows automatically the made statement.

## Remark: representation of Egreteau's ${\displaystyle \nabla}$ triangle

All ${\displaystyle p-}$nomial triangles refer to polynomials. For Egreteau's ${\displaystyle \nabla}$ triangle we have:

${\displaystyle \begin{array}{ll} -a + b & \\ & \\ -a^2 - 2ab + b^2 & \\ & \\ -a^3 + a^2b - 3ab^2 + b^3 & \\ & \\ -a^4 - 2a^3b +4a^2b^2 - 4ab^3 + b^4 & \\ & \\ \hspace{50pt} \dots & \\ \end{array}}$