A366619 Group the natural numbers into blocks of size 2: [1,2], [3,4], ... and reverse the order of the numbers within each block. Then group into blocks of size 3 and reverse the order in each block.

4, 1, 2, 5, 6, 3, 10, 7, 8, 11, 12, 9, 16, 13, 14, 17, 18, 15, 22, 19, 20, 23, 24, 21, 28, 25, 26, 29, 30, 27, 34, 31, 32, 35, 36, 33, 40, 37, 38, 41, 42, 39, 46, 43, 44, 47, 48, 45, 52, 49, 50, 53, 54, 51, 58, 55, 56, 59, 60, 57, 64, 61, 62, 65, 66, 63, 70, 67, 68, 71, 72, 69

Offset: 1

Author

Wesley Ivan Hurt, Oct 14 2023

Keywords

• nonn
• easy

Comments

Row 3 of the array in A007062.

Examples

			Group natural numbers into blocks of size 2: [1, 2], [3, 4], [5, 6], ...
Reverse the order in each block: [2, 1], [4, 3], [6, 5], ...
Group the remaining sequence into blocks of size 3: [2, 1, 4], [3, 6, 5], ...
Reverse the order in each block to get a(n): 4, 1, 2, 5, 6, 3, ...
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,1,1,0,-1).

Crossrefs

Cf. A007062.

Programs

Mathematica

Table[1 + Mod[n, 2] + 2 Floor[3 Floor[(n - 1)/3]/2 + Mod[-n, 3]/2], {n, 100}]

Formula

a(n) = 1 + (n mod 2) + 2*floor(3*floor((n - 1)/3)/2 + (-n mod 3)/2).

G.f.: x*(4 + x - 2*x^2 + 3*x^4)/((1 - x)^2*(1 + x)*(1 + x + x^2)). - Stefano Spezia, Oct 14 2023

a(n+6) = a(n) + 6. - Joerg Arndt, Oct 15 2023

From Wesley Ivan Hurt, Oct 15 2023: (Start)

a(n) = n - (-1)^n + 2*cos(2*(n - 1)*Pi/3) + 2*sin(2*(n - 1)*Pi/3)/sqrt(3).

a(n) = a(n-2) + a(n-3) - a(n-5) for n >= 6. (End)