A265667 Permutation of nonnegative integers: a(n) = n + floor(n/3)*(-1)^(n mod 3).

0, 1, 2, 4, 3, 6, 8, 5, 10, 12, 7, 14, 16, 9, 18, 20, 11, 22, 24, 13, 26, 28, 15, 30, 32, 17, 34, 36, 19, 38, 40, 21, 42, 44, 23, 46, 48, 25, 50, 52, 27, 54, 56, 29, 58, 60, 31, 62, 64, 33, 66, 68, 35, 70, 72, 37, 74, 76, 39, 78, 80, 41, 82, 84, 43, 86, 88, 45

Offset: 0

Author

Bruno Berselli, Dec 12 2015 - based on an idea by Paul Curtz

Keywords

• nonn
• easy

Comments

The inverse permutation is given by P(n) = A006368(n-1) + 1, for n >= 1, and P(0) = 0. - Wolfdieter Lang, Sep 21 2021

This permutation is given by A006369(n-1) + 1, with A006369(-1) = -1. Observed by Kevin Ryde. - Wolfdieter Lang, Sep 22 2021

Examples

			-------------------------------------------------------------------------
0, 1, 2, 3,  4, 5, 6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, ...
+  +  +  +   +  +  +   +   +   +   +   +   +   +   +   +   +   +   +
0, 0, 0, 1, -1, 1, 2, -2,  2,  3, -3,  3,  4, -4,  4,  5, -5,  5,  6, ...
-------------------------------------------------------------------------
0, 1, 2, 4,  3, 6, 8,  5, 10, 12,  7, 14, 16,  9, 18, 20, 11, 22, 24, ...
-------------------------------------------------------------------------
		

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Peter Lynch and Michael Mackey, Parity and Partition of the Rational Numbers, arXiv:2205.00565 [math.NT], 2022. See set F p. 4.
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).
• Index entries for permutations of the positive (or nonnegative) integers.

Crossrefs

Cf. A001477, A006368, A006369, A008738, A032793.

Cf. A064455: n+floor(n/2)*(-1)^(n mod 2).

Cf. A265888: n+floor(n/4)*(-1)^(n mod 4).

Cf. A265734: n+floor(n/5)*(-1)^(n mod 5).

Programs

Magma

[n+Floor(n/3)*(-1)^(n mod 3): n in [0..70]];

Mathematica

Table[n + Floor[n/3] (-1)^Mod[n, 3], {n, 0, 70}]

Sage

[n+floor(n/3)*(-1)^mod(n,3) for n in (0..70)]

Formula

G.f.: x*(1 + 2*x + 4*x^2 + x^3 + 2*x^4) / ((1 - x)^2*(1 + x + x^2)^2).

a(n) = 2*a(n-3) - a(n-6).

a(3*k) = 4*k;

a(3*k+1) = 2*k+1, hence a(3*k+1) = a(3*k)/2 + 1;

a(3*k+2) = 4*k+2, hence a(3*k+2) = 2*a(3*k+1) = a(3*k) + 2.

Sum_{i=0..n} a(i) = A008738(A032793(n+1)).

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 + log(2)/4. - Amiram Eldar, Mar 30 2023