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A069206
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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).
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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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*(ln(n))^2 with C=1.7....
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LINKS
| Index entries for sequences related to 3x+1 (or Collatz) problem
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FORMULA
| Values of m(7, k) for k =1 to 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
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CROSSREFS
| Sequence in context: A098403 A166126 A200419 * A145091 A091685 A062824
Adjacent sequences: A069203 A069204 A069205 * A069207 A069208 A069209
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 11 2002
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