Download e-book for iPad: Lectures on Integer Partitions by Herbert S. Wilf

By Herbert S. Wilf

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2n. Now b(2n + 1) = b(n), because if we are given a hyperbinary expansion of 2n + 1, the “1” must appear, hence by subtracting 1 from both sides and dividing by 2, we’ll get a hyperbinary representation of n. Conversely, given such an expansion of n, double each part and add a 1 to obtain a representation of 2n + 1. Furthermore, b(2n + 2) = b(n) + b(n + 1), for a hyperbinary expansion of 2n + 2 might have either two 1’s or no 1’s in it. If it has two 1’s, then by deleting them and dividing by 2 we obtain an expansion of n.

To illustrate we write down the lists of properties (diseases): gaps = 0 or 1 11 21 22 32 33 43 parts ≡ 1 or 4 mod 5 2 3 5 7 8 10 It should be quite clear that there is no way to order the properties so that Remmel’s theorem will apply. To see that this does not work by the sieve method notice that the partitions of 4 with exactly one gap of size 0 or 1 are: 22, 1111 and the partitions of 4 with exactly one part size congruent to 0, 2 or 3 mod 5 are: 31, 22, 211 Thus, these two sets of properties are not sieve-equivalent since the numbers of partitions are different.

4, part 1, 2026–2028. [7] — —, A Rogers-Ramanujan bijection, J. Combin. Theory Ser. A 31 (1981), no. 3, 289–339. [8] J. W. L. Glaisher, A theorem in partitions, Messenger of Math. 12 (1883), 158-170. [9] Basil Gordon, Sieve-equivalence and explicit bijections, J. Combin. Theory Ser. A 34 (1983), no. 1, 90–93. [10] Ronald Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison Wesley, Reading, 1989. [11] G. H. Hardy and S. Ramanujan, Asymptotic formulæ in combinatory analysis, Proc.

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Lectures on Integer Partitions by Herbert S. Wilf


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