A072222 - OEIS

%I #23 Aug 20 2023 16:57:21

%S 1,1,0,1,5,4,1,7,6,3,9,8,5,11,10,7,13,12,9,15,14,11,17,16,13,19,18,15,

%T 21,20,17,23,22,19,25,24,21,27,26,23,29,28,25,31,30,27,33,32,29,35,34,

%U 31,37,36,33,39,38,35,41,40,37,43,42,39,45,44,41,47,46,43,49,48,45,51

%N a(n) = (abs(n-1-a(n-2)) mod n) + (abs(n-1-a(n-1)) mod (n-1)), a(0) = 1, a(1) = 1.

%C Proof of the formula from _Ralf Stephan_: If a(n) is in a suitable range, it is possible to omit the abs and the mod function. So for n > 6, a(n) simplifies to a(n) = 2*n-2 - a(n-1) - a(n-2). Substituting a(n-1), we get a(n) = 2*n - 2 - (2*(n-1) - 2 - a(n-2) - a(n-3)) - a(n-2) = a(n-3) + 2. - Lambert Herrgesell (zero815(AT)googlemail.com), Jan 18 2007

%H Clifford A. Pickover, <a href="https://doi.org/10.1016/B978-044450002-1/50022-9">The Crying of Fractal Batrachion</a>, Chapter 19 in Chaos and Fractals, A Computer Graphical Journey, 1998, Pages 127-131

%F For n > 6, a(n) = a(n-3) + 2. - _Ralf Stephan_, May 09 2004

%t f[n_] := f[n] = Mod[ Abs[n - 1 - f[n - 2]], n] + Mod[ Abs[n - 1 - f[n - 1]], n - 1]; f[0] = 1; f[1] = 1; Table[ f[n], {n, 0, 75}]

%K nonn,easy

%O 0,5

%A _Roger L. Bagula_, Jul 04 2002

%E Edited by _Robert G. Wilson v_, Jul 15 2002