Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below/Lemma 2

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Theorem

Consider the Leibniz harmonic triangle:

$\begin{array}{r|rrrrrr}

n & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & \frac 1 1 \\ 1 & \frac 1 2 & \frac 1 2 \\ 2 & \frac 1 3 & \frac 1 6 & \frac 1 3 \\ 3 & \frac 1 4 & \frac 1 {12} & \frac 1 {12} & \frac 1 4 \\ 4 & \frac 1 5 & \frac 1 {20} & \frac 1 {30} & \frac 1 {20} & \frac 1 5 \\ 5 & \frac 1 6 & \frac 1 {30} & \frac 1 {60} & \frac 1 {60} & \frac 1 {30} & \frac 1 6 \\ \end{array}$


Let $\tuple {n, m}$ be the element in the $n$th row and $m$th column.

Then:

$\ds \forall r \in \N_{>0}: \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$

That is, each number in the Leibniz harmonic triangle is equal to the sum of the number below it, $\paren {r - 1}$ numbers diagonally below that number, and the number to the right of the last number.

This is similar to Rising Sum of Binomial Coefficients (Hockey-stick Identity) for Pascal's triangle.


Proof

Proof by induction:

For all $r \in \N_{>0}$, let $\map P r$ be the proposition:

$\ds \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$


Basis for the Induction

$\map P 1$ is:

$\tuple {n, m} = \tuple {n + 1, m + 1} + \tuple {n + 1, m}$

which is shown in Lemma 1.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P s$ is true, where $s \ge 1$, then it logically follows that $\map P {s + 1}$ is true.


So this is our induction hypothesis:

$\ds \tuple {n, m} = \tuple {n + s, m + s} + \sum_{k \mathop = 1}^s \tuple {n + k, m + k - 1}$


Then we need to show:

$\ds \tuple {n, m} = \tuple {n + s + 1, m + s + 1} + \sum_{k \mathop = 1}^{s + 1} \tuple {n + k, m + k - 1}$


Induction Step

This is our induction step:

\(\ds \) \(\) \(\ds \tuple {n + s + 1, m + s + 1} + \sum_{k \mathop = 1}^{s + 1} \tuple {n + k, m + k - 1}\)
\(\ds \) \(=\) \(\ds \tuple {n + s + 1, m + s + 1} + \tuple {n + s + 1, m + s + 1 - 1} + \sum_{k \mathop = 1}^s \tuple {n + k, m + k - 1}\)
\(\ds \) \(=\) \(\ds \tuple {n + s, m + s} + \sum_{k \mathop = 1}^s \tuple {n + k, m + k - 1}\) Lemma 1
\(\ds \) \(=\) \(\ds \tuple {n, m}\) Induction Hypothesis

So $\map P s \implies \map P {s + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall r \in \N_{>0}: \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$

$\blacksquare$

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