A080413 Take the rightmost three binary digits of n (for n<4 padded with leading zeros) and rotate left 1 digit.

0, 2, 4, 6, 1, 3, 5, 7, 8, 10, 12, 14, 9, 11, 13, 15, 16, 18, 20, 22, 17, 19, 21, 23, 24, 26, 28, 30, 25, 27, 29, 31, 32, 34, 36, 38, 33, 35, 37, 39, 40, 42, 44, 46, 41, 43, 45, 47, 48, 50, 52, 54, 49, 51, 53, 55, 56, 58, 60, 62, 57, 59, 61, 63, 64, 66, 68, 70, 65, 67, 69, 71, 72

Offset

0,2

Links

Table of n, a(n) for n=0..72.

Index entries for sequences that are permutations of the natural numbers.

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

Formula

For n>7: a(n) = 8*floor(n/8) + a(n mod 8).

A permutation of natural numbers with inverse = A080414: A080414(a(n))=n, a(A080414(n))=n.

a(a(n))=A080414(n), A080414(A080414(n))=a(n), a(a(a(n)))=n.

From Elmo R. Oliveira, May 11 2026: (Start)

a(n) = a(n-1) + a(n-8) - a(n-9).

G.f.: (2*x*((1+x^4)*(1+x+x^2)) - x^4*(5-x^4)) / ((1+x)*(1+x^2)*(1+x^4)*(x-1)^2). (End)

Example

a(2)=a('010')='100'=4; a(3)=a('011')='110'=6; a(4)=a('100')='001'=1; a(5)=a('101')='011'=3;
a(20)=a('10'100')='10'001'=17; a(21)=a('10'101')='10'011'=19.

Mathematica

LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 2, 4, 6, 1, 3, 5, 7, 8}, 73] (* Georg Fischer, Jul 03 2025 *)

Prog

(Python)
def A080413(n): return ((n&3)<<1)+bool(n&4)+(n&-8) # Chai Wah Wu, Jan 21 2023

Crossrefs

Cf. A004442, A007088, A064429, A080412, A080414.

Sequence in context: A125944 A077069 A261596 * A004517 A254350 A357049

Adjacent sequences: A080410 A080411 A080412 * A080414 A080415 A080416

Keyword

nonn,easy

Author

Reinhard Zumkeller, Feb 17 2003

Status

approved