See Bernoulli numbers in Wikipedia for details on Bernoulli numbers.

They may have been selected as an example because:

• They cannot be enumerated using the Difference Engine
• The values seem to be fairly random
• The values require a machine that can handle very large numbers
• In approximations of trigonometric functions, they allow less calculating steps than other approximations known at that time. Note that Tchebysheff's Approximation was published several years later (1859).
• They are a sequence where the n-th number can be calculated from the previous numbers

Note that variants for the indexing and sign for Bernoulli numbers were proposed since that time; the later forms allow shorter writing of the formulas.

In Note G, equation 8 shows the recursion formula used by AAL:

`0 = -1/2 * (2n-1)/(2n+1) + B_1 <sup>((1))</sup> /(2*3*4)) + ... + B_(2n-1)`

and in terms of the number to be calculated:

`-B_(2n-1) = -1/2 * (2n-1)/(2n+1) + B_1 ((2n)/2) + B_3 ((2n*(2n-1)*(2n-2)) /(2*3*4)) + B_5 ((2n*(2n-1)... (2n-4)) /(2*3*4*5*6)) +...`

Using `A_i` as abbreviation for the factors gives

`B_(2n-1) = A_0(n) + A_1(n) * B_1 + A_3(n) * B_3 + ... + A_5(n) * B_(2n-3)`

In the text, AAL wrotes: A₁, A₃&c. being … functions of n …, but does not use a corresponding notation in the equation, which was done in the above equation.

Thus, the coefficients – which are not given in Note G, are:

`A_0(n) = 1/2 * (2n-1)/(2n+1)`

`A_1(n) = -(2n) / 2`

`A_3(n) = A_1(n) * (2n-1)/3 * (2n-2)/4`

`A_5(n) = A_3(n) * (2n-3)/5 * (2n-4)/6`

This makes clear, that for `A_5` and the following the calculation steps are structurally equal and thus can be calculated within a loop.

Note the signs in contrast to Note G.

(to be continued)

`A_0(1) = 1/2 * 1/3 = 1/6`

`B_1 = A_0(1) = 1/6`

`A_0(2) = 3/10`

`A_1(2) = -2`

`B_3 = A_0(2) + A_1(2) * B_1 = 3/10 - 2/6 = (9 - 10)/30 = - 1/30`

`A_0(3) = 5/14`

`A_1(3) = -3`

`A_3(3) = -3 * 5/3 * 4/4 = -5`

`B_5 = 5/14 - 3/6 + 5/30 = (5 - 7)/14 + 1/6 = -1/7 + 1/6 = 1/42`

`A_0(4) = 7/18`

`A_1(4) = -4`

`A_3(4) = -4 * 7/3 * 6/4 = -14`

`A_5(4) = -14 * 5/5 * 4/6 = - 28/3`

`B_7 = 7/18 - 4/6 + 14/30 - 28/(3*42) = (7 - 12)/18 + 7/15 - 14/63 = -5/18 + 7/15 - 14/63 = (-25 + 28)/90 + 14/63 = 17/90 -14/63 = (119 - 140 ) / (7*9*10) = - 21/(7*9*10) = - 1/30`