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A378684
a(n) = A378200(A378200(n)).
6
1, 3, 5, 4, 2, 6, 8, 14, 10, 12, 11, 9, 13, 7, 15, 17, 27, 19, 25, 21, 23, 22, 20, 24, 18, 26, 16, 28, 30, 44, 32, 42, 34, 40, 36, 38, 37, 35, 39, 33, 41, 31, 43, 29, 45, 47, 65, 49, 63, 51, 61, 53, 59, 55, 57, 56, 54, 58, 52, 60, 50, 62, 48, 64, 46, 66, 68, 90, 70, 88, 72, 86, 74, 84, 76, 82, 78, 80, 79, 77, 81, 75, 83, 73, 85, 71, 87, 69, 89, 67
OFFSET
1,2
COMMENTS
The sequence can be regarded as a triangular array read by rows. Each row is a permutation of a block of consecutive numbers; the blocks are disjoint and every positive number belongs to some block. The length of row n is 4n-3 = A016813(n+1), n > 0.
The sequence can also be regarded as a table read by upward antidiagonals. For n>1 row n joins two consecutive antidiagonals.
The sequence is an intra-block permutation of the positive integers.
Generalization of Cantor numbering method.
The sequence A378200 generates the cyclic group C6 under composition. The elements of C6 are the successive compositions of A378200 with itself: A378684(n) = A378200(A378200(n)) = A378200(n)^2, A378762(n) = A378200(n)^3, A379342(n) = A378200(n)^4, A378705(n) = A378200(n)^5. The identity element is A000027(n) = A378200(n)^6. - Boris Putievskiy, Jan 10 2025
FORMULA
Linear sequence: (a(1),a(2), ... a(A000384(n+1)) is permutation of the positive integers from 1 to A000384(n+1). (a(1),a(2), ... a(A000384(n+1)) = (A378200(1), A378200(2), ... A378200(A000384(n+1)))^2.
Triangular array T(n,k) for 1 <= k <= 4n - 3 (see Example): T(n,k) = A000384(n-1) + P(n,k), P(n, k) = k + 1 if k < m(n) and k mod 2 = 1, P(n, k) = 2*m(n) - k if k < m(n) and k mod 2 = 0, P(n, k) = k if k >= m(n) and k mod 2 = 1, P(n, k) = 2*m(n) - k - 1 if k >= m(n) and k mod 2 = 0, where m(n) = 2n - 1.
EXAMPLE
Triangle array begins:
k= 1 2 3 4 5 6 7 8 9
n=1: 1;
n=2: 3, 5, 4, 2, 6;
n=3: 8, 14, 10, 12, 11, 9, 13, 7, 15;
(1,3,5, ...7,15) = (A378200(1), A378200(2), A378200(3), ... A378200(14), A378200(15))^2.
For n > 1, each row of triangle array joins two consecutive upward antidiagonals in the table:
1, 5, 6, 12, 15, ...
3, 2, 10, 7, 21, ...
4, 14, 13, 25, 26, ...
8, 9, 19, 18, 34, ...
11, 27, 24, 42, 41, ...
...
Subtracting (n-1)*(2*n-3) from each term in row n produces a permutation of numbers from 1 to 4*n-3:
1;
2, 4, 3, 1, 5;
2, 8, 4, 6, 5, 3, 7, 1, 9;
MATHEMATICA
P[n_, k_]:=Module[{m=2*n-1}, If[k<m, If[OddQ[k], k+1, 2*m-k], If[OddQ[k], k, 2*m-k-1]]]
Nmax=3; Flatten[Table[P[n, k]+(n-1)*(2*n-3), {n, 1, Nmax}, {k, 1, 4 n-3}]]
KEYWORD
nonn,tabf
AUTHOR
Boris Putievskiy, Dec 03 2024
STATUS
approved