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The rule for generating the harmonic triangle is that you add
two consecutive entries to give the entry between them in the row
above. Equivalently, from any term, to get the next term, subtract
that term from the corresponding term on the row above. It is still
possible to work downwards row by row because the entry at the left
hand end of the $n$th row is ${1\over n}$.
For example the third row of the harmonic triangle is:
$${1\over 3},\ {1\over 6},\ {1\over 3}$$ and the fourth row is:
$${1\over 4},\ {1\over 12},\ {1\over 12}, \ {1\over 4}$$ which is
given by $${1\over 4},\ \ \left[{1\over 3}{1\over 4}\right] ,\ \
\left[{1\over 6}{1\over 12}\right],\ \ \left[{1\over 3}{1\over
12}\right]$$

\begin{array}{ccccccccccc} & & & & & \frac{1}{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} \\ & & & & & ... & & & & & \end{array} 
The entries in the harmonic triangle are related by a similar rule to the entries in Pascal's triangle and both sets of entries involve the binomial coefficients. We denote the $r$th entry in the $n$th row of Pascal triangle by the binomial coefficient: $${n1\choose r1}= {(n1)!\over (r1)!(nr)!}.$$ For example the sixth row is $${5 \choose 0} = 1 , {5 \choose 1}= 5, {5\choose 2}=10, {5\choose 3}= 10, {5\choose 4}= 5, {5\choose 5} =1 $$  \begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1& & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1& & 4 & & 6 & & 4 & & 1 & \\ 1& & 5& & 10 & & 10 & & 5 & & 1 \\ & & & & & ... & & & & & \end{array} 