

A069206


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).


0



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, 30, 56, 57, 65, 16, 0, 32, 26, 35, 71, 53, 18, 0, 82, 42, 61, 46, 39, 44, 0, 26, 80, 48, 71, 40, 77, 0, 43, 18, 81, 16, 36, 48, 0, 72, 77, 25, 53, 37, 59, 0, 83, 25, 37, 23, 62, 43, 0
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OFFSET

1,2


COMMENTS

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....


LINKS

Table of n, a(n) for n=1..73.
Index entries for sequences related to 3x+1 (or Collatz) problem


FORMULA

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.


CROSSREFS

Sequence in context: A320375 A200419 A271522 * A291800 A091685 A062824
Adjacent sequences: A069203 A069204 A069205 * A069207 A069208 A069209


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Apr 11 2002


STATUS

approved



