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N < 1 + q1 In (3), note the particular case of t = 0, i.e.
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Using (4) with t = 0,
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So for this range of N, r always equals N and s always equals 0. The two corner points, u1 = qt = 1, uN = qt−1 + rqt = r = N, simply remain 1 and N for all such N. Due to Corollary 5, the gap-length L1 + L2 does not exist because s = 0 = qt − 1.
A simple example of this case is with α = 2∕11 = [0;5,2], where q1 = 5. So for all N < 1 + q1 = 6, the corner points are 1 and N, and only two gap-lengths will exist.
N = qt+1 The below range of N which decides t (from (3)):
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can be split into two subranges:
Refer to relation (4). In (a), (N − qt−1) ranges from qt to at+1qt − 1. So, r would take values 1 to at+1 − 1. For all N in this subrange, the two corner points ((u1,uN) or (uN,u1)) will be: qt and qt−1 + rqt.
In (b), (N −qt−1) ranges from at+1qt to at+1qt + (qt − 1). So, r remains fixed at at+1, and s varies from 0 to qt − 1. The two corner points remain fixed at: qt and (qt−1 + at+1qt = qt+1). It is this qt+1 corner point which continues to remain the corner point when N crosses this range and enters into a new range with “t” as t + 1.
qt increments in N Also observe how the corner points change as N varies in the range for a particular t: qt−1 + qt ≤ N < qt + qt+1.
Suppose t is even (odd case is similar) and so u1 = qt and uN = qt−1 + rqt. For all N in this range, u1 remains fixed, but uN changes whenever r changes. When N starts at qt−1 + qt, r = 1, so uN = qt−1 + qt = N. Then, for every qt increment in N, r increments by 1, and so uN changes too. This also implies simultaneous changes in the gap-lengths due to theorem 2.
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[1] Tony Van Ravenstein. The Three Gap Theorem (Steinhaus Conjecture). J. Austral. Math. Soc. (Series A) 45 (1988), 360-370.
[2] Victor Beresnevich, Nicol Leong. Sums of reciprocals and the three distance theorem. http://eprints.whiterose.ac.uk/121745/ (2017).
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