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A115789 a(n) = (floor((n+1)*Pi) - floor(n*Pi)) mod 2.

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%I A115789 #13 Apr 13 2024 12:42:09
%S A115789 1,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,
%T A115789 1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,
%U A115789 1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1
%N A115789 a(n) = (floor((n+1)*Pi) - floor(n*Pi)) mod 2.
%C A115789 The arithmetic mean (1/(n+1))*Sum_{k=0..n} a(k) converges to 4 - Pi. What is effectively the same: the Cesaro limit (C1) of a(n) is 4 - Pi.
%F A115789 a(n) = (floor((n+1)*Pi) - floor(n*Pi)) mod 2.
%e A115789 a(6)=1 because 7*Pi=21.99..., 6*Pi=18.84... and so a(6) = (21 - 18) mod 2 = 1;
%e A115789 a(7)=0 because 8*Pi=25.13... and so a(7) = (25 - 21) mod 2 = 0.
%Y A115789 Cf. A000796, A022844, A063438, A115788, A115790.
%K A115789 nonn
%O A115789 0,1
%A A115789 _Hieronymus Fischer_, Jan 31 2006