Results 221 to 228 of about 3,063 (228)

Bijections for Entringer families

open access: green, 2010
Yoann Gelineau, Heesung Shin, Jiang Zeng
openalex   +1 more source

Identities versus bijections [PDF]

open access: possible, 2004
Consider the infinite product (1+x)(1+x 2)(1+x 3)(1+x 4) · · · and expand it in the usual way into a series \( \sum_{n \geq 0} \) a n x n by grouping together those products that yield the same power x n . By inspection we find for the first terms $$ \label{1} \prod_{k \geq 1} (1+x^{k}) = 1 + x + x^{2} + 2x^{3} + 2x^{4} + 3x^{5} + 4x^{6} + 5x^{7} +
Martin Aigner, Günter M. Ziegler
openaire   +1 more source

Differences of Bijections

The American Mathematical Monthly, 2019
When is a function from an abelian group to itself expressible as a difference of two bijections? Answering this question for finite cyclic groups solves a problem about juggling.
Daniel J. Velleman   +2 more
openaire   +2 more sources

Bijective A-transducers

20th Annual Symposium on Foundations of Computer Science (sfcs 1979), 1979
In this paper we study bijective a-transducers. We derive necessary and sufficient conditions on pairs of regular sets (R,S) such that a bijective a-transducer, mapping R cnto S exists. The results obtained allow the systematic construction of an a-transducer, mapping a set R onto a set S bijectively for surprisingly "different" regular sets R and S.
Hermann A. Maurer, H. Nivat
openaire   +2 more sources

Bijections for the Schröder Numbers

Mathematics Magazine, 2000
(2000). Bijections for the Schroder Numbers. Mathematics Magazine: Vol. 73, No. 5, pp. 369-376.
Robert A. Sulanke, Louis W. Shapiro
openaire   +2 more sources

Geometrical Bijections in Discrete Lattices

Combinatorics, Probability and Computing, 1999
We define uniformly spread sets as point sets in d-dimensional Euclidean space that are wobbling equivalent to the standard lattice ℤd. A linear image ϕ(ℤd) of ℤd is shown to be uniformly spread if and only if det(ϕ) = 1. Explicit geometrical and number-theoretical constructions are given.
Wolfgang Thumser   +3 more
openaire   +3 more sources

An Extension of Franklin’s Bijection

2001
The author gives a purely combinatorial proof of the identity \[ \prod_{n>m}(1-q^n) = \sum_{n=1}^{\infty} (-1)^n \left[ {n+m \atop m} \right] q^{nm+n(3n+1)/2}(1-q^{2n+m+1}), \] which generalizes Franklin's proof for the case \(m=0\).
openaire   +3 more sources

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