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A076539 Numerators a(n) of fractions slowly converging to Pi: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < Pi, then a(n+1) = a(n) + 1, otherwise a(n+1) = a(n).

%I A076539 #19 Jun 25 2022 00:35:10
%S A076539 0,1,2,3,3,4,5,6,6,7,8,9,9,10,11,12,12,13,14,15,15,16,17,18,18,19,20,
%T A076539 21,21,22,23,24,25,25,26,27,28,28,29,30,31,31,32,33,34,34,35,36,37,37,
%U A076539 38,39,40,40,41,42,43,43,44,45,46,47,47,48,49,50,50,51,52,53,53,54,55
%N A076539 Numerators a(n) of fractions slowly converging to Pi: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < Pi, then a(n+1) = a(n) + 1, otherwise a(n+1) = a(n).
%C A076539 a(n) + b(n) = n and as n -> +infinity, a(n)/b(n) converges to Pi. For all n, a(n)/b(n) < Pi.
%F A076539 a(1) = 0, b(n) = n - a(n), if (a(n) + 1)/b(n) < Pi, then a(n+1) = a(n) + 1, otherwise a(n+1) = a(n).
%F A076539 a(n) = floor(n*Pi/(Pi+1)). - _Vladeta Jovovic_, Oct 04 2003
%e A076539 a(7)= 5 so b(7) = 7 - 5 = 2.
%e A076539 a(8) = 6 because (a(7) + 1)/b(7) = 6/2 which is < Pi. So b(8) = 8 - 6 = 2.
%e A076539 a(9) = 6 because (a(8) + 1)/b(8) = 7/2 which is not < Pi.
%t A076539 Array[Floor[# Pi/(Pi + 1)] &, 73] (* _Michael De Vlieger_, Jan 11 2018 *)
%Y A076539 Cf. A074840, A074065, A060143.
%Y A076539 Partial sums of A144609.
%K A076539 easy,frac,nonn
%O A076539 1,3
%A A076539 Robert A. Stump (bee_ess107(AT)msn.com), Oct 18 2002