cp's OEIS Frontend

A073720 Let b(1) = 1, b(k+1) = b(k) - k*trunc(k/b(k)+1), where trunc(x) = floor(x) if x>= 0, trunc(x) = ceiling(x) otherwise. Sequence a(n) gives the successive absolute values taken by b(k).

%I A073720 #8 Jun 17 2013 15:20:59
%S A073720 1,11,58,293,1468,7343,36718,183593,917968,4589843,22949218,114746093,
%T A073720 573730468,2868652343,14343261718,71716308593,358581542968,
%U A073720 1792907714843,8964538574218,44822692871093,224113464355468
%N A073720 Let b(1) = 1, b(k+1) = b(k) - k*trunc(k/b(k)+1), where trunc(x) = floor(x) if x>= 0, trunc(x) = ceiling(x) otherwise. Sequence a(n) gives the successive absolute values taken by b(k).
%F A073720 It appears that for n>1 a(n)=( 47*5^(n-2)-3 )/4 and if 2*a(n-1)+1 < k < 2*a(n)+1, then b(k)= -a(n), if k = 2*a(n)+1 b(k)= +a(n).
%F A073720 Empirical g.f.: -x*(3*x^2-5*x-1) / ((x-1)*(5*x-1)). - _Colin Barker_, Jun 17 2013
%K A073720 nonn
%O A073720 1,2
%A A073720 _Benoit Cloitre_, Aug 30 2002