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%I #18 Oct 21 2019 02:20:05
%S 0,5,0,6,11,11,24,10,29,0,40,20,34,12,42,19,0,17,35,26,27,17,19,0,65,
%T 30,56,57,65,16,0,32,26,35,71,53,18,0,82,42,61,46,39,44,0,26,80,48,71,
%U 40,77,0,43,18,81,16,36,48,0,72,77,25,53,37,59,0,83,25,37,23,62,43,0
%N Let m(n,k) be the sequence defined by m(n,1)=1, m(n,2)=n, m(n,k+2)=(m(n,k+1)+m(n,k))/2 if m(n,k+1)+m(n,k) is even, m(n,k+2)=m(n,k+1)-m(n,k) otherwise. Sequence gives the (experimentally unique) solution to m(n,x)=0 (for k > this solution m(n,k) is constant = +1 or -1 depending on n) or a(n)=0 if there is no solution (in this case the cycle (1,3,2,-1,-3,-2) is reached).
%C Sequence presents similarities with the 3x+1 problem but seems less "random". Hence this sequence presents regularities depending curiously on the number 7. If n==3 (mod 7) there is no solution to m(n,x)=0. If n==0,1,2,4,5 or 6 (mod 7) there is always a unique solution to m(n,x)=0. It seems also that lim_{n->infinity} a(n)/n = 0 (a(10^10)=493) and asymptotically, Sum_{i=1..n} a(i) ~ C*n*(log(n))^2 with C=1.7....
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%F Values of m(7, k) for k = 1..24: 1, 7, 4, -3, -7, -5, -6, -1, 5, 2, -3, -5, -4, 1, 5, 3, 4, 1, -3, -1, -2, -1, 1, 0, hence a(7)=24. For k > 22, m(7, k) = -1.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 11 2002
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