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

1,1

Comments

Row 3 of the array in A007062.

Links

Table of n, a(n) for n=1..72.

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

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)

Example

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, ...

Mathematica

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

Crossrefs

Cf. A007062.

Sequence in context: A046573 A006287 A087225 * A343404 A220182 A076064

Adjacent sequences: A366616 A366617 A366618 * A366620 A366621 A366622

Keyword

nonn,easy

Author

Wesley Ivan Hurt, Oct 14 2023

Status

approved