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A076087 a(n)= 7*n - 3*sum(i=1,n,b(i)) (see comment for b(i) definition). 0

%I #5 Mar 30 2012 18:39:09

%S 4,5,6,1,-4,-9,-8,-4,-3,1,-4,0,4,-1,-6,-5,-1,-6,-11,-10,-9,-5,-1,0,-5,

%T -4,0,4,8,12,13,8,3,4,5,9,13,17,18,13,14,9,4,5,6,7,2,-3,1,5,9,4,-1,0,

%U 1,2,-3,-2,-7,-6,-5,-10,-15,-11,-7,-3,1,-4,-9,-14,-10,-9,-8,-7,-12,-11,-10,-15,-20,-16,-15,-20,-25,-21,-17,-13,-9

%N a(n)= 7*n - 3*sum(i=1,n,b(i)) (see comment for b(i) definition).

%C Recalling the Collatz map (cf. A006370 ) : x->x/2 if x is even; x->3x+1 if x is odd, let C_m(n) denotes the image of n after m iterations. Then b(n)= lim k -> infinity C_3k(n) (from the Collatz conjecture C_3k(n) is constant =1,2 or 4 for k large enough). Curiously the graph for a(n) presents "regularities" around zero and a pattern coming bigger and bigger. Compared with a random sequence of form : 7*n-3*sum(k=1,n,r(k)) where r(k) takes random values among (1;2;4).

%e since 3->10->5->16->8->4->2->1 etc. C_6(3)=2 and then for any k>=2 C_3k(3)=2, hence b(3)=2.

%K sign

%O 1,1

%A _Benoit Cloitre_, Oct 30 2002

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