This number triangle is made up of the denominators of the Leibniz Harmonic Triangle. From the earlier post about that number triangle, you can see that there are two ways of generating the Harmonic Denominator Triangle:

1. take the reciprocal of the entries in the Leibniz Harmonic Triangle; or

2. multiply the entries of Pascal's Triangle by n+d (here we are using the 'triangular number style' indexing of Pascal's Triangle, rather than the usual 'binomial coefficient style' indexing).

The entries of the Harmonic Denominator Triangle ($g^d_n$) are given by:

\[g^d_n = \frac{1}{h^d_n}\]

and

\[g^d_n = (n+d)t^d_n\]

Where $h^d_n$ are the entries in the Leibniz Harmonic Triangle and $t^d_n$ are the entries in Pascal's Triangle. The connection to the triangular numbers gives us the general formula:

\[g^d_n = \frac{n(n+1)\cdots(n+d)}{d!}\]

I am looking forward to learning more about this number triangle (and the Leibniz Harmonic Triangle too). If you make use of OEIS, you'll see that the The HDT contains a lot of well-known sequences.

A quick look at the rows and columns and you'll find the a bunch of well-known sequences: A005430, A002457, A002378, A027480, and A033488, to name a first few.

The whole triangle has an entry in OEIS: A003506. One of the comments in the OEIS entry for the HDT points out a neat way to express the relationship between Pascal's Triangle and the Harmonic Denominator Triangle. The entries of the

*k-*th row of the HDT are the coefficients of the first derivative of a polynomial whose coefficients are the entries of the (

*k*+1)-th row of Pascal's Triangle. More specifically, the coefficients of $(x+1)^{k+1}$ give you the

*k*+1 row of Pascal's Triangle, while the coefficients of $\frac{d}{dx}(x+1)^{k+1}$ give you the

*k-*th row of the Harmonic Denominator Triangle.