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The difference between consecutive members of a difference set $ ({\rm mod}$ $ v)$


Author: E. Spence
Journal: Proc. Amer. Math. Soc. 46 (1974), 150-154
MSC: Primary 05B10
DOI: https://doi.org/10.1090/S0002-9939-1974-0345843-2
MathSciNet review: 0345843
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Abstract: Let $ \mathcal{D} = \{ {d_1},{d_2}, \cdots ,{d_k}\} $ be a difference set $ \pmod v$ so that for any $ d\not\equiv0\pmod v$ there are exactly $ \lambda $ pairs $ ({d_i},{d_j}),{d_i},{d_j}\in \mathcal{D}$ such that $ {d_i} - {d_j} \equiv d\pmod v$. Suppose further that $ 0 \leq {d_1} < {d_2} < \cdots < {d_k} < v$ and write $ {d_{k + 1}} = v+ {d_1}$. The following two results are proved:

(i) $ \Sigma _{i = 1}^k{({d_{i + 1}} - {d_i})^2} = O(({v^2}/k)\log k),$

(ii) $ {\max _{1 \leqslant i \leqslant k}}({d_{i + 1}} - {d_i}) = O(v/\surd k).$


References [Enhancements On Off] (What's this?)

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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1974-0345843-2
Keywords: Difference set $ \pmod v$
Article copyright: © Copyright 1974 American Mathematical Society

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