*This page is an appendix to "Irish Logarithms Part 2"*

The Jacobi index Ind(*z*) of an integer *z* can be given by

*g*and

*p*for which this equation applies and for which Ind(

*z*) is unique for each value of

*z*in a selected set of numbers. If

*p*is prime and the set of

*z*'s consists of all integers between 1 and

*p*− 1, then there is a

*g*for which each

*z*has an unique index between 0 and

*p*− 2. If so, then:

Ind(*z*_{1}×*z*_{2 }) = Ind(*z*_{1}) + Ind(*z*_{2 }) (2)

So using Ind(), we can multiply by adding.^{[1]}

For a calculator like Verea's, we only need the indices of the 36 integers in the simple multiplication table.

So for a calculator the set of *z*'s differs from the set of all integer between 1 and *p *− 1: there are "gaps" in the collection and the collection does not end with a prime ( *p *− 1 = 81).

Because in the calculator only numbers less than 10 are multiplied, equation (2) does not have to apply for all *z*_{1} and *z*_{2}, but only for *z*_{1}, *z*_{2} <10. The resulting index Ind(*z*_{1}×*z*_{2 }) must be unique for all unique simple products.

So we do not have to apply the Jacobi indices literally for the multiplication table of a calculating machine. We can still try using equation (1) to generate indices that meet our goal. There is no guarantee that the indices that we find are the **smallest **numbers causing Ind(*z*_{1} × *z*_{2}) = Ind(*z*_{1}) + Ind(*z*_{2}). Our final goal is to minimize the largest index.

Using equation (1) with *p* = 11 and *g* = 2, I found two sets of indices with the largest index less than 100. The following table shows the indices for the integers <10:

z |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

Ind(z) |
0 | 1 | 18 | 2 | 44 | 19 | 7 | 4 | 36 |

Ind(z) |
0 | 1 | 8 | 2 | 44 | 9 | 27 | 4 | 16 |

Other choices of *p*
and *g*
may provide better (i.e. lower) indices.

The indices of Schumacher's slide rule are generated with
*p* = 101 and
*g* = 2.