 By René Schoof

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Extra resources for Algebra 2. The symmetric groups Sn

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Are the pentagonal numbers. Algebraic and Geometric Methods in Enumerative Combinatorics 19 9. (Partitions into distinct parts) Let DistPartition be the family of partitions into distinct parts, weighted by xn yk where n is the sum of the parts and k is the number of parts. Then DistPartition ∼ = {1, 1} × {2, 2} × · · · , where wt(i) = xi y and wt(i) = 1 for i = 1, 2, . , so the corresponding generating function is DistPartition(x, y) = (1 + xy)(1 + x2 y)(1 + x3 y) · · · 10. (Partitions into restricted parts) It is clear how to adapt the previous generating functions to partitions where the parts are restricted.

Hk are half-pyramids, which are pyramids containing no bricks to the left of the bottom brick. This decomposition is reversible. To recover x, we drop hk , hk−1 , . . , h1 , p from the top in that order; each piece is dropped in its correct horizontal position, and some of its bricks may get stuck on the previous pieces. This shows that the corresponding combinatorial classes satisfy X∼ = P × Seq(H). 9. Each new half-pyramid is obtained by pushing up the leftmost remaining brick (which is not necessarily in the bottom row), dragging with it all the bricks that it encounters along the way.

If there are k summands equal to 2 there must be n − 2k summands equal to 1, and there are n−2k+k = n−k ways of ordering the summands. Therefore k k n/2 an = ∑ k=0 n−k n n−1 n−2 = + + +··· . 1) This is a pretty good answer. It is certainly an explicit formula, and it may be used to compute an directly for small values of n. It does have two drawbacks. ” In practice, it is also not as useful as it seems; after computing a few examples, we will soon notice that computing binomial coefficients is a non-trivial task.