A374447 Inverse permutation to A373498.

2, 1, 3, 9, 4, 7, 6, 8, 5, 10, 20, 11, 18, 13, 16, 15, 17, 12, 19, 14, 21, 35, 22, 33, 24, 31, 26, 29, 28, 30, 23, 32, 25, 34, 27, 36, 54, 37, 52, 39, 50, 41, 48, 43, 46, 45, 47, 38, 49, 40, 51, 42, 53, 44, 55

Offset

1,1

Comments

Triangle read by rows where row n is a block of length 4*n-1 which is a permutation of the numbers of its constituents.

Generalization of the Cantor numbering method for two adjacent diagonals. A pair of neighboring diagonals are combined into one block.

The sequence is an intra-block permutation of positive integers.

The sequence A373498 generates the cyclic group C6 under composition. The elements of C6 are the successive compositions of A373498 with itself: A374494 = A373498(A373498) = A373498^2, A370655 = A373498^3, A374531 = A373498^4, A374447 = A373498^5. The identity element is A000027 = A373498^6. - Boris Putievskiy, Aug 03 2024

Links

Boris Putievskiy, Table of n, a(n) for n = 1..9870

Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.

Index entries for sequences that are permutations of the natural numbers

Formula

Linear sequence:

a(n) = P(n) + (L(n)-1)*(2*L(n)-1), where L(n) = ceiling((sqrt(8*n+1)-1)/4),

L(n) = A204164(n), P(n) = 4*L(n) - R(n) - 1, if R(n) <= 2*L(n) and R(n) mod 2 = 1, P(n) = R(n) - 1, if R(n) <= 2*L(n) and R(n) mod 2 = 0, P(n) = R(n), if R(n) > 2*L(n) and R(n) mod 2 = 1, P(n) = - 2*L(n) + R(n), if R(n) > 2*L(n) and R(n) mod 2 = 0.

a(n) = A373498(A373498(A373498(A373498(A373498(n))))).

Triangular array T(n,k) for 1 <= k <= 4*n-1 (see Example):

T(n,k) = (n-1)*(2*n-1) + P(n,k), where P(n,k) = 4*n - k - 1, if k <= 2*n and k mod 2 = 1, P(n,k) = k-1, if k <= 2*n and k mod 2 = 0, P(n,k) = k, if k > 2*n and k mod 2 = 1, P(n,k) = -2*n + k, if k > 2*n and k mod 2 = 0.

Example

Triangle begins:
     k = 1   2   3   4   5   6   7   8   9  10  11
  n=1:   2,  1,  3;
  n=2:   9,  4,  7,  6,  8,  5, 10;
  n=3:  20, 11, 18, 13, 16, 15, 17, 12, 19, 14, 21;
The triangle's rows can be arranged as two successive upward antidiagonals in an array:
    2,  3,  7, 10, 16, 21, ...
    1,  4,  5, 13, 14, 26, ...
    9,  8, 18, 19, 31, 34, ...
    6, 11, 12, 24, 25, 41, ...
   20, 17, 33, 32, 50, 51, ...
   15, 22, 23, 39, 40. 60, ...
Subtracting (n-1)*(2*n-1) from each term in row n is a permutation of 1 .. 4*n-1:
    2,1,3,
    6,1,4,3,5,2,7,
   10,1,8,3,6,5,7,2,9,4,11
   ...
The inverse permutation of each permutation in example A373498 is equal to the corresponding permutation above:
   (2,1,3)^(-1) = (2,1,3),
   (2,6,4,3,5,1,7)^(-1) = (6,1,4,3,5,2,7),
   (2,8,4,10,6,5,7,3,9,1,11)^(-1) = (10,1,8,3,6,5,7,2,9,4,11).
The 5th power of each permutation in example A373498 is equal to the corresponding permutation above:
   (2,1,3)^5 = (2,1,3),
   (2,6,4,3,5,1,7)^5 = (6,1,4,3,5,2,7),
   (2,8,4,10,6,5,7,3,9,1,11)^5 = (10,1,8,3,6,5,7,2,9,4,11).

Mathematica

Nmax=21;
a[n_]:=Module[{L, R, P, Result}, L=Ceiling[(Sqrt[8*n+1]-1)/4];
R=n-(L-1)*(2*L-1); P=Which[R<=2*L&&Mod[R, 2]==1, 4*L-R-1, R<=2*L&&Mod[R, 2]==0, R-1, R>2*L&&Mod[R, 2]==1, R, R>2*L&&Mod[R, 2]==0, -2*L+R];
Result=P+(L-1)*(2*L-1);
Result]
Table[a[n], {n, 1, Nmax}]
Nmax=21;
a[n_]:=Module[{L, R, P, Result}, L=Ceiling[(Sqrt[8*n+1]-1)/4];
R=n-(L-1)*(2*L-1); P=Which[R<=2*L-1&&Mod[R, 2]==1, R+1, R<=2*L-1&&Mod[R, 2]==0, R+2*L, R>2*L-1&&Mod[R, 2]==1, R, R>2*L-1&&Mod[R, 2]==0, 4*L-1-R];
Result=P+(L-1)*(2*L-1);
Result]
Table[a[n], {n, 1, Nmax}] (* A373498 *)
Table[a[a[a[a[a[n]]]]], {n, 1, Nmax}] (* this sequence *)

Crossrefs

Cf. A000027, A004767 (row lengths), A204164, A370655, A373498, A374494, A374531.

Sequence in context: A099644 A369446 A058113 * A249456 A234746 A359648

Adjacent sequences: A374444 A374445 A374446 * A374448 A374449 A374450

Keyword

nonn,tabf

Author

Boris Putievskiy, Jul 08 2024

Status

approved