cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A367832 Array T(n, k) read by ascending antidiagonals is a dispersion based on A367467. Column 1 lists the numbers which cannot be represented by A367467(m) + m. For k >= 1, T(n, k+1) = A367467(T(n, k)) + T(n, k).

Original entry on oeis.org

1, 4, 2, 7, 6, 3, 11, 9, 10, 5, 14, 12, 15, 17, 8, 18, 16, 20, 25, 29, 13, 21, 19, 27, 34, 42, 49, 22, 24, 23, 32, 46, 58, 71, 83, 37, 28, 26, 39, 54, 78, 99, 121, 141, 63, 31, 30, 44, 66, 92, 133, 169, 206, 240, 107, 35, 33, 51, 75, 112, 157, 227, 288, 351, 409, 182, 38, 36, 56, 87, 128, 191, 268
Offset: 1

Views

Author

Thomas Scheuerle, Dec 02 2023

Keywords

Comments

This sequence is a permutation of the positive integers.
The array T(n, k+1) - T(n, k) for k > 1 is also a permutation of the positive integers.
Columns k > 2 together consist of all the numbers from A003152. These are all the positive numbers of the form floor(m*(1+1/sqrt(2))).
In column 2 are all the numbers from A184119. These are all the numbers of the form floor((2+sqrt(2))*m - sqrt(2)/2).
Column 2 together with the columns k > 2 are all the numbers from A087057; these are all the numbers of the form ceiling(m*sqrt(2)). Together with column 1, which consists of all the numbers from A083051, they cover all positive integers.
An alternative definition that allows this array to be obtained without using A367467:
Take for T(n, 1) and T(n, 2) the first and the second number which do not appear in any row r < n. Complete all rows by the recurrence T(n, k) = floor(T(n, k-1)*(1 + 1/sqrt(2))). Start in the first row with T(1, 1) = 1 and T(1, 2) = 2.
Let Q(n, k) = T(n, k+2) - T(n, k+1) for k > 0. Let b(m) be the row n where the integer m is found in Q. Then we will obtain for (b(n)) the sequence: 1, 1, 1, 2, 1, 3, 2, 4, 1, 3, 5, 2, 6, 4, 1, ... . If we were to remove the first occurrence of each number in this sequence, we would get the same sequence again, hence (b(n)) is a fractal sequence.

Examples

			Array T(n, k) begins:
   1,  2,  3,  5,   8,  13,  22,  37,  63,  107, ...
   4,  6, 10, 17,  29,  49,  83, 141, 240,  409, ...
   7,  9, 15, 25,  42,  71, 121, 206, 351,  599, ...
  11, 12, 20, 34,  56,  99, 169, 288, 491,  839, ...
  14, 16, 27, 46,  78, 133, 227, 387, 660, 1126, ...
  18, 19, 32, 54,  92, 157, 268, 457, 780, 1331, ...
  21, 23, 39, 66, 112, 191, 326, 556, 949, 1620, ...
  ...
		

References

  • Clark Kimberling, Fractal sequences and interspersions, Ars Combinatoria 45 (1997) 157-168.

Crossrefs

Cf. A083050 (a closely related dispersion).

Formula

T(1, k) = A293078(k).
T(n, 1) = A083051(n-1).
T(n, 2) = A184119(n).
Conjectured: T(n, 3) = A328987(n-1).
T(1, k) = 2*T(1, k-1) - T(1, k-2) + floor(T(1, k-2)/2), for k > 2.
T(n, k+1) = floor(T(n, k)*(1+1/sqrt(2))) for k > 1.
T(n, k+1) = A367467(T(n, k)) + T(n, k).

Extensions

Edited by Peter Munn, Dec 11 2023

A367849 Lexicographically earliest infinite sequence of positive integers such that for each n, the values in a path of locations starting from any i=n are all distinct, where jumps are allowed from location i to i+a(i).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 2, 5, 3, 1, 1, 4, 6, 1, 1, 5, 1, 1, 7, 1, 6, 1, 1, 2, 1, 8, 7, 1, 1, 2, 1, 3, 1, 9, 4, 1, 1, 2, 5, 3, 1, 1, 10, 6, 1, 1, 2, 1, 3, 7, 1, 4, 11, 1, 1, 5, 8, 1, 1, 2, 6, 3, 1, 12, 9, 1, 7, 1, 1, 2, 1, 3, 1, 10, 4, 13, 1, 1, 5, 1, 1, 2, 1, 11, 1, 1, 2, 1, 14, 1, 1, 2, 1, 3
Offset: 1

Views

Author

Neal Gersh Tolunsky, Dec 08 2023

Keywords

Comments

Consider each index i as a location from which one can jump a(i) terms forward. No starting index can reach the same value more than once by forward jumps.
The value a(i) at the starting index is not part of the path (and thus allows a(2)=1).
The indices of first occurrences are given by A133263 (essentially triangular numbers + 2).
Changing the definition so that jumps are allowed only from location i to i-a(i) gives A002260.

Examples

			We can see, for example, that the terms reachable by jumping forward continuously from i=1 are all distinct (and in this case are just the positive integers):
  1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 2, 5
  *->1->2---->3------->4---------->5
Beginning at i=9 and jumping forward continuously, we get the sequence 1,2,3,4,5,6,7,9 (in which all terms are likewise distinct).
		

Crossrefs

Cf. A367467, A367832, A133263 (index of first occurrences), A362248.

Programs

  • MATLAB
    function a = A367849( max_n )
        a = zeros(1,max_n); j = find(a == 0,1);
        while ~isempty(j)
            a(j) = 1; k = 1;
            if j+k < max_n
                while a(j+k) == 0
                    a(j+k) = k;
                    if j+2*k-1 < max_n
                        j = j+(k-1); k = k+1;
                    else
                        break;
                    end
                end
            end
            j = find(a == 0,1);
        end
    end % Thomas Scheuerle, Dec 09 2023

Formula

a(A133263(n)) = n + 1.

A368485 Lexicographically earliest infinite sequence of positive integers such that for n > 1, a(n - a(n)) is distinct for all indices n with the same a(n) value.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 3, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 4, 4, 5, 6, 1, 2, 3, 4, 5, 5, 5, 6, 7, 1, 2, 3, 4, 5, 6, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 6, 6, 7, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 7, 7, 7, 8, 8, 9, 10, 11
Offset: 1

Views

Author

Neal Gersh Tolunsky, Dec 28 2023

Keywords

Comments

Consider each index i as a location from which one can jump a(i) terms backwards. Any two distinct indices m and k, where a(m)=a(k), will jump to distinct values. In other words, every 1 will jump back to a distinct value, every 2, 3, etc.

Examples

			We can see, for example, that the values reached by jumping backwards once from each 3 in the sequence are all distinct:
  1, 1, 2, 1, 2, 3, 1, 2, 3, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3
        2<=======*
                 3<=======*
                    1<=======*  4<=======*     5<=======*
		

Crossrefs

Programs

  • PARI
    See Links section.

Extensions

More terms from Rémy Sigrist, Jan 15 2024

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).

Original entry on oeis.org

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

Views

Author

Neal Gersh Tolunsky, Jan 23 2024

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Extensions

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

Views

Author

Neal Gersh Tolunsky, Feb 06 2024

Keywords

Comments

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.

Examples

			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
		

Crossrefs

Programs

  • Mathematica
    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 *)

Extensions

More terms from Giorgos Kalogeropoulos, Feb 28 2024
Showing 1-5 of 5 results.