A373498 a(a(a(n))) = A370655(n).
2, 1, 3, 5, 9, 7, 6, 8, 4, 10, 12, 18, 14, 20, 16, 15, 17, 13, 19, 11, 21, 23, 31, 25, 33, 27, 35, 29, 28, 30, 26, 32, 24, 34, 22, 36, 38, 48, 40, 50, 42, 52, 44, 54, 46, 45, 47, 43, 49, 41, 51, 39, 53, 37, 55
Offset: 1
Examples
Triangle begins: k = 1 2 3 4 5 6 7 8 9 10 11 n=1: 2, 1, 3; n=2: 5, 9, 7, 6, 8, 4, 10; n=3: 12, 18, 14, 20, 16, 15, 17, 13, 19, 11, 21; The triangle's rows can be arranged as two successive upward antidiagonals in an array: 2, 3, 7, 10, 16, 21, ... 1, 9, 4, 20, 11, 35, ... 5, 8, 14, 19, 27, 34, ... 6, 18, 13, 33, 24, 52, ... 12, 17, 25, 32, 42, 51, ... 15, 31, 26, 50, 41, 73, ... Subtracting (n-1)*(2*n-1) from each term in row n is a permutation of 1 .. 4*n-1: 2,1,3, 2,6,4,3,5,1,7, 2,8,4,10,6,5,7,3,9,1,11, ... The 3rd power of each permutation is equal to the corresponding permutation in example A370655: (2,1,3)^3 = (2,1,3), (2,6,4,3,5,1,7)^3 = (1,2,4,3,5,6,7), (2,8,4,10,6,5,7,3,9,1,11)^3 = (3,4,1,2,6,5,7,10,9,8,11).
Links
- Boris Putievskiy, Table of n, a(n) for n = 1..9870 _Boris Putievskiy_, Aug 02 2024
- 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
- _Boris Putievskiy_, Aug 02 2024
Programs
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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-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}] Table[a[a[a[n]]],{n,1,Nmax}] (* A370655 *)
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), R(n) = n - (L(n)-1)*(2*L(n)-1),
P(n) = R(n) + 1 if R(n) <= 2*L(n)-1 and R(n) mod 2 = 1, P(n) = 2*L(n) + R(n) if R(n) <= 2*L(n)-1 and R(n) mod 2 = 0, P(n) = R(n) if R(n) > 2*L(n)-1 and R(n) mod 2 = 1, P(n) = 4*L(n) - R(n) - 1 if R(n) > 2*L(n)-1 and R(n) mod 2 = 0.
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) = k + 1 if k <= 2*n-1 and k mod 2 = 1, P(n,k) = 2*n + k if k <= 2*n-1 and k mod 2 = 0,
P(n,k) = k if k > 2*n-1 and k mod 2 = 1, P(n,k) = 4*n - k - 1 if k > 2*n-1 and k mod 2 = 0.
Comments