Leibniz harmonic triangle

The Leibniz harmonic triangle is a triangular arrangement of fractions in which the outermost diagonals consist of the reciprocals of the row numbers and each inner cell is the absolute value of the cell above minus the cell to the left. To put it algebraically, (where $r$ is the number of the row, starting from 1, and $c$ is the column number, never more than r) and $L (r, c) = | L (r - 1, c - 1) - L (r, c - 1)|.$

The first eight rows are:

$\begin{array}{cccccccccccccccccc} & & & & & & & & & 1 & & & & & & & &\\ & & & & & & & & \frac{1}{2} & & \frac{1}{2} & & & & & & &\\ & & & & & & & \frac{1}{3} & & \frac{1}{6} & & \frac{1}{3} & & & & & &\\ & & & & & & \frac{1}{4} & & \frac{1}{12} & & \frac{1}{12} & & \frac{1}{4} & & & & &\\ & & & & & \frac{1}{5} & & \frac{1}{20} & & \frac{1}{30} & & \frac{1}{20} & & \frac{1}{5} & & & &\\ & & & & \frac{1}{6} & & \frac{1}{30} & & \frac{1}{60} & & \frac{1}{60} & & \frac{1}{30} & & \frac{1}{6} & & &\\ & & & \frac{1}{7} & & \frac{1}{42} & & \frac{1}{105} & & \frac{1}{140} & & \frac{1}{105} & & \frac{1}{42} & & \frac{1}{7} & &\\ & & \frac{1}{8} & & \frac{1}{56} & & \frac{1}{168} & & \frac{1}{280} & & \frac{1}{280} & & \frac{1}{168} & & \frac{1}{56} & & \frac{1}{8} &\\ & & & & &\vdots & & & & \vdots & & & & \vdots& & & & \\ \end{array}$

The denominators are listed in (sequence A003506 in OEIS), while the numerators are all 1s.

Whereas each entry in Pascal's triangle is the sum of the two entries in the above row, each entry in the Leibniz triangle is the sum of the two entries in the row below it. For example, in the 5th row, the entry (1/30) is the sum of the two (1/60)s in the 6th row.

Just as Pascal's triangle can be computed by using binomial coefficients, so can Leibniz's: $L(r, c) = \frac{1}{r {r-1 \choose c-1}}$ . Furthermore, the entries of this triangle can be computed from Pascal's, "the terms in each row are the initial term divided by the corresponding Pascal triangle entries."^[1]

This triangle can be used to obtain examples for the Erdős–Straus conjecture when n is divisible by 4.

If one takes the denominators of the nth row and adds them, then the result will equal $n 2^{n - 1}$ . For example, for the 3rd row, we have 3 + 6 + 3 = 12 = 3 × 2².

It is worth noting that $L(r, c) = \int_0^1 \! x ^ {c - 1} (1 - x)^{r-c} \,dx \,.$

References

^ Wells, David (1986). The Penguin Dictionary of Curious and Interesting Numbers, p.98. ISBN 9780140261493.