First we define a dummy program to go on the z s key. The true z s key is supposed to take two numbers from the stack and return one number, so DEL as a dummy definition will make sure the stack comes out right.
2: 4 1: 4 2: 4
1: 2 . 1: 2
. .
4 RET 2 C-x ( DEL C-x ) Z K s RET 2
The last step replaces the 2 that was eaten during the creation of the dummy z s command. Now we move on to the real definition. The recurrence needs to be rewritten slightly, to the form s(n,m) = s(n-1,m-1) - (n-1) s(n-1,m).
(Because this definition is long, it will be repeated in concise form below. You can use M-# m to load it from there.)
2: 4 4: 4 3: 4 2: 4
1: 2 3: 2 2: 2 1: 2
. 2: 4 1: 0 .
1: 2 .
.
C-x ( M-2 RET a = Z [ DEL DEL 1 Z :
4: 4 2: 4 2: 3 4: 3 4: 3 3: 3
3: 2 1: 2 1: 2 3: 2 3: 2 2: 2
2: 2 . . 2: 3 2: 3 1: 3
1: 0 1: 2 1: 1 .
. . .
RET 0 a = Z [ DEL DEL 0 Z : TAB 1 - TAB M-2 RET 1 - z s
(Note that the value 3 that our dummy z s produces is not correct; it is merely a placeholder that will do just as well for now.)
3: 3 4: 3 3: 3 2: 3 1: -6
2: 3 3: 3 2: 3 1: 9 .
1: 2 2: 3 1: 3 .
. 1: 2 .
.
M-TAB M-TAB TAB RET M-TAB z s * -
1: -6 2: 4 1: 11 2: 11
. 1: 2 . 1: 11
. .
Z ] Z ] C-x ) Z K s RET DEL 4 RET 2 z s M-RET k s
Even though the result that we got during the definition was highly bogus, once the definition is complete the z s command gets the right answers.
Here's the full program once again:
C-x ( M-2 RET a =
Z [ DEL DEL 1
Z : RET 0 a =
Z [ DEL DEL 0
Z : TAB 1 - TAB M-2 RET 1 - z s
M-TAB M-TAB TAB RET M-TAB z s * -
Z ]
Z ]
C-x )
You can read this definition using M-# m (read-kbd-macro)
followed by Z K s, without having to make a dummy definition
first, because read-kbd-macro doesn't need to execute the
definition as it reads it in. For this reason, M-# m is often
the easiest way to create recursive programs in Calc.