A368485 -id:A368485 - cp's OEIS frontend

A369475 Lexicographically earliest infinite sequence such that, from all indices n with the same a(n) value, the terms reached by a single jump are all distinct, where jumps are allowed from location i to i+-a(i).

1, 2, 2, 3, 4, 1, 5, 3, 2, 5, 6, 1, 7, 4, 6, 3, 1, 8, 8, 2, 5, 7, 3, 5, 6, 9, 1, 10, 11, 1, 12, 3, 2, 3, 10, 4, 13, 1, 14, 6, 2, 3, 9, 5, 15, 7, 2, 9, 13, 7, 5, 4, 4, 4, 6, 10, 12, 11, 9, 2, 10, 16, 1, 15, 3, 4, 5, 17, 1, 18, 9, 12, 3, 6, 5, 19, 1, 20, 9, 15

Offset: 1

Neal Gersh Tolunsky, Jan 23 2024

Consider each index i as a location from which one can jump a(i) terms forwards or backwards. From all indices with the same a(n) value, every jump is to a distinct term.

Another way to define the sequence is to consider every possible ordered pair of values of the form (origin value, destination value)--every such ordered pair is distinct.

			a(5)=4 because:
a(5) cannot be 1 because then we would have two jumps from a term with the same value 2, both landing on the value 1--ordered pair (2,1) twice:
  1, 2, 2, 3, 1
        2---->1
  1<----2
a(5) cannot be 2 because we would have two jumps from the same a(n) value 2 to the same value 2--ordered pair (2,2) twice:
  1, 2, 2, 3, 2
        2---->2
        2<----2
a(5) cannot be 3 because we would have two jumps from the same a(n) value 2 to the same a(n) value 3--ordered pair (2,3) twice:
  1, 2, 2, 3, 3
     2---->3
        2---->3
a(5) can be 4 without contradiction.

Pontus von Brömssen, Table of n, a(n) for n = 1..10000
Pontus von Brömssen, Plot of (n,a(n)) for n = 1..1000000.
Pontus von Brömssen, Plot of (n, log(a(n))/log(n)) for n = 2..1000000.

Cf. A368485, A367849, A367832, A367467.

More terms from Pontus von Brömssen, Jan 24 2024

A369852 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).

Original entry on oeis.org

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

Offset: 1

Neal Gersh Tolunsky, Feb 06 2024

nonn

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.

			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

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Cf. A369475, A368485, A367849, A367832, A367467.

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 *)

More terms from Giorgos Kalogeropoulos, Feb 28 2024

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A369475 Lexicographically earliest infinite sequence such that, from all indices n with the same a(n) value, the terms reached by a single jump are all distinct, where jumps are allowed from location i to i+-a(i).

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