A375602 Triangle read by rows, where row n is a block of length n*(n^2 + 1)/2, which is a permutation of the numbers of its constituents; see Comments.

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

Offset: 1

Author

Boris Putievskiy, Aug 20 2024

Keywords

• nonn
• tabf

Comments

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

Generalization of the Cantor numbering method for n (n > 1) adjacent diagonals. In this approach, the block number n combines n neighboring diagonals.

Each block is filled sequentially, starting from the top of the leftmost vertical strip and moving downwards and then rightwards to the next strip. In block number n the first (n - 1)*n/2 + 1 strips each have a length of n. The remaining n - 1 strips have lengths that decrease sequentially from n - 1 down to 1. See Example the array of permutations of the triangle read upward antidiagonals.

Examples

			Triangle begins:
     k = 1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
  n=1:   1;
  n=2:   2,  4,  3,  5,  6;
  n=3:   7, 10, 13, 16,  8, 11, 14, 17, 19,  9, 12, 15, 18, 20, 21;
Subtracting (n - 1)*n*(n^2-n+2)/8 from each term in row n is a permutation of 1 .. n(n^2+1)/2:
  1,
  1, 3, 2,  4, 5,
  1, 4, 7, 10, 2, 5, 8, 11, 13, 3, 6, 9, 12, 14, 15,
  ...
The triangle's rows of permutations can be arranged as n successive upward antidiagonals in an array:
   1, 3, 5, 10, 13, 15, ...
   1, 4, 7, 11, 14, ...
   2, 4, 8, 12, ...
   1, 5, 9, ...
   2, 6, ...
   3, ...
		

Links

• Boris Putievskiy, Table of n, a(n) for n = 1..9316
• 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.

Crossrefs

Cf. A002024, A002260, A002817, A006003 (row lengths),

Programs

Mathematica

a[n_]:=Module[{L,Ld,Rd,P,Result},L=Ceiling[(Sqrt[4*Sqrt[8*n+1]-3]-1)/2];
Ld=Ceiling[(Sqrt[8*n+1]-1)/2]; Rd=n-(Ld-1)*Ld/2; P=L*Rd+Ld-L*(L+1)/2-Max[Rd-(L^2-L+2)/2,0]*(Max[Rd-(L^2-L+2)/2,0]+1)/2; Result=P+(L-1)*L*(L^2-L+2)/8; Result]
Nmax= 21; Table[a[n],{n,1,Nmax}]

Formula

Linear sequence:

a(n) = P(n) + (L(n) - 1)*L*(L(n)^2 - L(n) + 2)/8, where L(n) =ceiling((sqrt(4*Sqrt(8*n + 1) - 3) - 1)/2), Ld(n) = ceiling((Sqrt(8*n + 1) - 1)/2), Ld(n) = A002024(n), Rd(n) = n - (Ld(n) - 1)*Ld(n)/2, Rd(n) = A002260(n),

P(n) = L(n)*Rd(n) + Ld(n) - L(n)*(L(n) + 1)/2 - Max[Rd(n) - (L(n)^2 - L(n) + 2)/2, 0]*(Max[Rd(n) - (L(n)^2 - L(n) + 2)/2, 0] + 1)/2.

Triangular array T(n,k) for 1 <= k <= n(n^2+1)/2 (see Example):

T(n,k) = (n - 1)*n*(n^2 - n + 2)/8 + P(n,k), T(n,k) = A002817(n-1) + P(n,k), where P(n,k) = n*Rd(n,k) + Ld(n,k) - n - Max[Rd(n,k) - (n^2 - n + 2)/2, 0]*(Max[Rd(n,k) - (n^2 - n + 2)/2, 0] + 1)/2, where Ld(n,k) = Ceiling[(Sqrt[(n^2 - n + 1)^2 + 8*k] - (n^2 - n + 1))/2].