Here is a suitable set of rules to solve the first part of the problem:

[ seq(n, c) := seq(n/2, c+1) :: n%2 = 0, seq(n, c) := seq(3n+1, c+1) :: n%2 = 1 :: n > 1 ]

Given the initial formula ``seq(6, 0)'`, application of these
rules produces the following sequence of formulas:

seq( 3, 1) seq(10, 2) seq( 5, 3) seq(16, 4) seq( 8, 5) seq( 4, 6) seq( 2, 7) seq( 1, 8)

whereupon neither of the rules match, and rewriting stops.

We can pretty this up a bit with a couple more rules:

[ seq(n) := seq(n, 0), seq(1, c) := c, ... ]

Now, given ``seq(6)'` as the starting configuration, we get 8
as the result.

The change to return a vector is quite simple:

[ seq(n) := seq(n, []) :: integer(n) :: n > 0, seq(1, v) := v | 1, seq(n, v) := seq(n/2, v | n) :: n%2 = 0, seq(n, v) := seq(3n+1, v | n) :: n%2 = 1 ]

Given ``seq(6)'`, the result is ``[6, 3, 10, 5, 16, 8, 4, 2, 1]'`.

Notice that the n > 1 guard is no longer necessary on the last rule since the n = 1 case is now detected by another rule. But a guard has been added to the initial rule to make sure the initial value is suitable before the computation begins.

While still a good idea, this guard is not as vitally important as it
was for the `fib`

function, since calling, say, ``seq(x, [])'`
will not get into an infinite loop. Calc will not be able to prove
the symbol ``x'` is either even or odd, so none of the rules will
apply and the rewrites will stop right away.

Go to the first, previous, next, last section, table of contents.