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A255541 a(n) = 1+Sum_{k=1..2^n-1} A000010(k).

%I A255541 #41 Aug 12 2025 20:30:59
%S A255541 1,2,5,19,73,309,1229,4959,19821,79597,318453,1274563,5097973,
%T A255541 20397515,81591147,326371001,1305482159,5222040189,20888133573,
%U A255541 83552798667,334211074959,1336845501841,5347382348679,21389531880435,85558125961121,342232529890275,1368930120480617,5475720508827645,21902882035220391,87611528574186091,350446114129452131,1401784457568941917,5607137830212707769
%N A255541 a(n) = 1+Sum_{k=1..2^n-1} A000010(k).
%C A255541 Number of fractions in Farey series of order 2^n-1.
%H A255541 D. T. Ashley, J. P. DeVoe, K. Perttunen, C. Pratt, and A. Zhigljavsky, <a href="http://esrg.sourceforge.net/docs/paper_brap_detailed.pdf">On Best Rational Approximations Using Large Integers</a>
%H A255541 Wikipedia, <a href="http://en.wikipedia.org/wiki/Farey_sequence">Farey Sequence</a>
%F A255541 a(n) ~ (2^n-1)^2 / Pi.
%F A255541 a(n) = 2+A015614(2^n-1).
%F A255541 a(n) = A005728 (2^n-1). - _Michel Marcus_, Feb 27 2015
%F A255541 a(n) = (3+A018805(2^n-1))/2. - _Colin Linzer_, Aug 06 2025
%e A255541 For each n, measure the size of the set of reduced fractions with a denominator less than 2^n:
%e A255541 a(0) = 1 since the set of reduced fractions with denominator less than 2^0 = 1 is {0}.
%e A255541 a(1) = 2 since the set of reduced fractions with denominator less than 2^1 = 2 is {0, 1}.
%e A255541 a(2) = 5 since the set of reduced fractions with denominator less than 2^2 = 4 is {0, 1/3, 1/2, 2/3, 1}.
%e A255541 a(3) = 19 since the set of reduced fractions with denominator less than 2^3 = 8 is {0, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1}.
%t A255541 k = s = 1; lst = {}; Do[While[k < 2^n, s = s + EulerPhi@ k; k++]; AppendTo[lst, s], {n, 0, 26}]; lst
%t A255541 a[n_] := 1 + (1/2) Sum[ MoebiusMu[k]*Floor[n/k]*Floor[1 + n/k], {k, n}]; Array[a, 27, 0]
%Y A255541 Cf. A007305, A007306, A000010, A049643, A006842/A006843 (Farey fractions).
%K A255541 nonn
%O A255541 0,2
%A A255541 _Robert G. Wilson v_, Feb 24 2015