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A369852
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a(1)=1, a(2)=2; thereafter, any two indices n with different a(n) values reach distinct values by a single jump, where jumps are allowed from location i to i+a(i).
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1
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1, 2, 2, 3, 1, 2, 4, 1, 5, 2, 6, 1, 2, 7, 1, 2, 8, 1, 5, 2, 9, 1, 5, 7, 10, 1, 2, 11, 3, 12, 9, 4, 1, 13, 14, 15, 1, 5, 16, 12, 3, 17, 7, 4, 1, 18, 19, 6, 20, 21, 22, 23, 3, 8, 24, 4, 1, 5, 25, 26, 4, 10, 7, 27, 15, 28, 1, 13, 29, 30, 31, 32, 33, 2, 34, 1, 5, 5
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OFFSET
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1,2
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COMMENTS
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Consider each index i as a location from which one can jump a(i) terms forward. To find a(n) we have to check 2 conditions:
1. The value a(n) can be reached in one jump by at most one distinct value.
2. Location n reaches a location in one jump that is not reached in one jump from a location before n.
Another way to view the sequence is to consider the sets of values that can be reached from each distinct integer by a single jump forward (values reached by 1s in the sequence, values reached by 2s, 3s etc.): all of these sets are disjoint.
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LINKS
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EXAMPLE
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a(4)=3 because:
a(4) cannot be 1 because then we would have two distinct values (a(3)=2, a(4)=1) that reach the same future value a(5)=x:
1, 2, 2, 1, x
2---->x
1->x
a(4) cannot be 2 because then we would have two distinct values (a(1)=1, a(2)=2) reach the same value 2:
1, 2, 2, 2
1->2
2---->2
a(4) can be 3 without contradiction since there is only one distinct value that can reach the value 3 (a(2)=2):
1, 2, 2, 3
2---->3
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MATHEMATICA
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lst={1, 2}; Do[z=1; Quiet@While[l=Join[lst, {z}]; Union[Length@*Union/@ GatherBy[Select[Table[{l[[k]], l[[l[[k]]+k]]}, {k, Length@l}], IntegerQ@Last@#&], Last]]!={1}||
MemberQ[Table[l[[k]]+k, {k, Length@l-1}], Length@l+Last@l], z++]; AppendTo[lst, z], {i, 89}]; lst (* Giorgos Kalogeropoulos, Feb 29 2024 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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