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 A074065 Numerators a(n) of fractions slowly converging to sqrt(3): let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < sqrt(3), then a(n+1) = a(n) + 1, else a(n+1)= a(n). 2

%I

%S 0,1,1,2,3,3,4,5,5,6,6,7,8,8,9,10,10,11,12,12,13,13,14,15,15,16,17,17,

%T 18,19,19,20,20,21,22,22,23,24,24,25,25,26,27,27,28,29,29,30,31,31,32,

%U 32,33,34,34,35,36,36,37,38,38,39,39,40,41,41,42,43,43,44,45,45,46,46

%N Numerators a(n) of fractions slowly converging to sqrt(3): let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < sqrt(3), then a(n+1) = a(n) + 1, else a(n+1)= a(n).

%C a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to sqrt(3). For all n, a(n) / b(n) < sqrt(3).

%F a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < sqrt(3), then a(n+1) = a(n) + 1, else a(n+1) = a(n).

%F a(n) = floor(n*(3-sqrt(3))/2). - _Vladeta Jovovic_, Oct 04 2003

%e a(6)= 3 so b(6) = 6 - 3 = 3. a(7) = 4 because (a(6) + 1) / b(6) = 4/3 which is < sqrt(3). So b(7) = 7 - 4 = 3. a(8) = 5 because (a(7) + 1) / b(7) = 5/3 which is < sqrt(3).

%Y Cf. A074840.

%K easy,frac,nonn

%O 0,4

%A Robert A. Stump (bee_ess107(AT)msn.com), Sep 15 2002

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Last modified March 20 21:49 EDT 2019. Contains 321352 sequences. (Running on oeis4.)