A355616 a(n) is the number of distinct lengths between consecutive points of the Farey sequence of order n.
1, 1, 2, 3, 5, 6, 9, 11, 14, 15, 21, 23, 29, 31, 34, 38, 48, 49, 59, 63, 67, 71, 83, 86, 97, 100, 110, 115, 132, 133, 150, 158, 165, 169, 182, 187, 208, 213, 222, 228, 252, 254, 280, 287, 297, 304, 331, 337, 362, 367, 379, 387, 418, 423, 437, 450, 464, 472, 509, 513, 548, 556, 573, 589, 608, 611, 652, 665, 681, 685
Offset: 1
Keywords
Examples
For n=5, the Farey sequence (completely reduced fractions) is [0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1]. The distinct lengths between consecutive points are {1/5, 1/20, 1/12, 1/15, 1/10} so a(5) = 5.
Programs
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Mathematica
a[n_] := FareySequence[n] // Differences // Union // Length; Table[a[n], {n, 1, 70}] (* Jean-François Alcover, Jul 16 2022 *)
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PARI
vp(n) = my(list = List()); for (k=1, n, for (i=0, k, listput(list, i/k))); vecsort(list,,8); a(n) = my(v=vp(n)); #Set(vector(#v-1, k, abs(v[k+1]-v[k]))); \\ Michel Marcus, Jul 10 2022
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Python
from fractions import Fraction from itertools import chain def compute(n): marks = [[(a, b) for a in range(0, b + 1)] for b in range(1, n + 1)] marks = sorted(set([Fraction(a, b) for a, b in chain(*marks)])) dist = [(y - x) for x, y in zip(marks, marks[1:])] return len(set(dist))
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