COVERING THE INTEGERS BY ARITHMETIC SEQUENCES .2.

Authors
Citation
Zw. Sun, COVERING THE INTEGERS BY ARITHMETIC SEQUENCES .2., Transactions of the American Mathematical Society, 348(11), 1996, pp. 4279-4320
Citations number
29
Categorie Soggetti
Mathematics, General",Mathematics
ISSN journal
00029947
Volume
348
Issue
11
Year of publication
1996
Pages
4279 - 4320
Database
ISI
SICI code
0002-9947(1996)348:11<4279:CTIBAS>2.0.ZU;2-C
Abstract
Let A = {a(s) + n(s)Z}(k)(s=1) (n(1) less than or equal to ...less tha n or equal to n(k)) be a system of arithmetic sequences where a(1),... , a(k), is an element of Z and n(1),..., n(k), is an element of Z(+). For m is an element of Z(+) system A will be called an (exact) m-cover of Z if every integer is covered by A at least (exactly) m times. In this paper we reveal further connections between the common difference s in an (exact) m-cover of Z and Egyptian fractions. Here are some typ ical results for those m-covers A of Z: (a) For any m(1),..., m(k) is an element of Z(+) there are at least m positive integers in the form Sigma(s is an element of I)m(s)/(n), where I subset of or equal to {1, ..., k}. (b) When n(k-l) < n(k-l+1) =...= n(k) (0 < l < k), either l g reater than or equal to n(k)/n(k-l) or Sigma(s=1)(k-l) 1/n(s) greater than or equal to m, and for each positive integer lambda < n(k)/n(k-l) the binomial coefficient ((l)(lambda)) can be written as the sum of s ome denominators > 1 of the rationals Sigma(s is an element of I)1/n(s ) - lambda/n(k), I not subset of (1,..., k) if A forms an exact m-cove r of Z. (c) If (a(s) + n(s)Z) [GRAPHICS] is not an m-cover of Z, then Sigma(s is an element of I)1/n(s), I subset of or equal to {1,..., k} \ {t} have at least n(t) distinct fractional parts and for each r = 0, ..., n(t) - 1 there exist I-1, I-2 subset of or equal to {1,..., k} \ {t} such that r/n(t) = Sigma(s is an element of I1) 1/n(s) - Sigma(s i s an element of I2) 1/n(s) (mod 1). If A forms an exact m-cover of Z w ith m = 1 or n(1) <...< n(k-l) < n(k-l+1) =...= n(k), (l > O) then for every t = 1,..., k and r = 0, 1,..., n(t) - 1 there is an I subset of or equal to (1,..., k) such that Sigma(s is an element of I)1/n(s) = r/n(t) (mod 1).