By THE MATHEMATICAL ASSOCIATION OF AMERICA

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**Additional resources for American Mathematical Monthly, volume 105, number 2, February 1998**

**Sample text**

When m = 1, there are no bricks, and the empty stack in sf maps to the sequence 0 in Qp. For m > 1, we consider three types of stacks and the corresponding sequences, showing first that f restricts as desired. If A E S::, does not cover the first space of the base, then f(A) consists of 0 followed by f(A ' ), where A' is obtained from A by deleting the first space of the base. By the induction hypothesis, f(A' ) E Q::'-l' so f(A) E Q::, and f(A) is q-dominating. If A E S::, covers the first space, let m' = k(q + 1) be the step on which the outline of A first returns to the base (in the top example of Figure 5, k = 2 and m ' = 6).

Figure 4. Flowchart for the determination of Hurwitz systems from D4 • Question. Does D4 act freely, orientation-reversing, on T 5? Step 1. The 8 elements of D4 are generated and stored as integer 3-by-3 matrices. Representation of the generators by integer matrices is essential. Approximations associated with storage of real values and roundoff errors that occur in arithmetic operations with real values would lead to unreliable results. We let a=[~o ~ =~l' b=[~ ~ ~l 1 -1 1 0 0 serve as the generating matrices.

By working up from the leaves, combining q + 1 subtrees at each non-leaf vertex, we see that each tree corresponds to a product of qn + 1 elements in order using a non-associative q + 1-ary operator. Conversely, from such a product we obtain the corresponding tree. Sands [14] was interested in counting the ways to form such a product. In order to count the products, we convert the tree to a sequence of objects and markers. We do this by traversing the tree; beginning at the root, walk around the tree, keeping our left hand on it, until we have traversed every edge twice.

### American Mathematical Monthly, volume 105, number 2, February 1998 by THE MATHEMATICAL ASSOCIATION OF AMERICA

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