A363321 Number of fractions of the Farey sequence of order n, F_n, that coincide with those of the sequence of the #{F_n} equally distributed fractions between 0 and 1.

2, 3, 3, 5, 5, 7, 5, 3, 9, 7, 9, 3, 3, 7, 17, 17, 15, 11, 13, 11, 19, 15, 5, 25, 21, 5, 11, 17, 25, 3, 7, 13, 5, 29, 27, 41, 35, 33, 7, 17, 7, 3, 5, 3, 3, 23, 17, 5, 19, 15, 25, 9, 35, 47, 29, 5, 31, 3, 7, 27, 9, 5, 5, 61, 5, 9, 23, 41, 51, 15, 29, 3, 9, 23, 31, 3, 7, 33, 3, 3

Offset: 1

Author

Andres Cicuttin, May 27 2023

Keywords

• nonn

Comments

The conjectured formula below suggests that as the value of n increases, the proportion of terms in the Farey sequence F_n that align with the #F_n rationals, evenly distributed between 0 and 1, tends to decrease.

Examples

			For n = 5, we have the Farey sequence F_5 = {0, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1} with 11 terms, and the corresponding sequence S_5 = {0, 1/10, 1/5, 3/10, 2/5, 1/2, 3/5, 7/10, 4/5, 9/10, 1} consisting of the 11 equidistant fractions {x/10} with 0 <= x <= 10. Since there are 5 fractions (0, 2/5, 1/2, 3/5, 1) in the same positions in both sequences, F_5 and S_5, then a(5) = 5.
		

Links

• Eric Weisstein's World of Mathematics, Farey Sequence.
• Wikipedia, Farey Sequence.

Crossrefs

Cf. A005728, A262669, A262670, A363300.

Programs

Mathematica

a[n_]:= Module[{len, fn, sn},
fn = FareySequence[n];
len = Length[fn];
sn = Range[0, len - 1]/(len - 1);
Count[fn - sn, 0]];
Table[a[j], {j, 1, 80}]

Formula

Conjecture: lim_{n->infinity} a(n)/A005728(n) = 0.