A126760 a(0) = 0, a(2n) = a(n), a(3n) = a(n), a(6n+1) = 2n + 1, a(6n+5) = 2n + 2.

0, 1, 1, 1, 1, 2, 1, 3, 1, 1, 2, 4, 1, 5, 3, 2, 1, 6, 1, 7, 2, 3, 4, 8, 1, 9, 5, 1, 3, 10, 2, 11, 1, 4, 6, 12, 1, 13, 7, 5, 2, 14, 3, 15, 4, 2, 8, 16, 1, 17, 9, 6, 5, 18, 1, 19, 3, 7, 10, 20, 2, 21, 11, 3, 1, 22, 4, 23, 6, 8, 12, 24, 1, 25, 13, 9, 7, 26, 5, 27, 2, 1, 14, 28, 3, 29, 15, 10, 4, 30, 2

Offset: 0

Author

N. J. A. Sloane, Feb 19 2007

Keywords

• nonn

Comments

For further information see A126759, which provided the original motivation for this sequence.

From Antti Karttunen, Jan 28 2015: (Start)

The odd bisection of the sequence gives A253887, and the even bisection gives the sequence itself.

A254048 gives the sequence obtained when this sequence is restricted to A007494 (numbers congruent to 0 or 2 mod 3).

For all odd numbers k present in square array A135765, a(k) = the column index of k in that array. (End)

A322026 and this sequence (without the initial zero) are ordinal transforms of each other. - Antti Karttunen, Feb 09 2019

Also ordinal transform of A065331 (after the initial 0). - Antti Karttunen, Sep 08 2024

Links

• Antti Karttunen, Table of n, a(n) for n = 0..19683

Crossrefs

One less than A126759.

Cf. A003586, A003602, A007310, A064989, A249746, A253887, A254048, A273669, A322317, A323881 (Dirichlet inverse), A323882.

Cf. A347233 (Möbius transform) and also A349390, A349393, A349395 for other Dirichlet convolutions.

Ordinal transform of A065331 and of A322026 (after the initial 0).

Related arrays: A135765, A254102.

Programs

Mathematica

f[n_] := Block[{a}, a[0] = 0; a[1] = a[2] = a[3] = 1; a[x_] := Which[EvenQ@ x, a[x/2], Mod[x, 3] == 0, a[x/3], Mod[x, 6] == 1, 2 (x - 1)/6 + 1, Mod[x, 6] == 5, 2 (x - 5)/6 + 2]; Table[a@ i, {i, 0, n}]] (* Michael De Vlieger, Feb 03 2015 *)

PARI

A126760(n)={n&&n\=3^valuation(n,3)<M. F. Hasler, Jan 19 2016

Formula

a(n) = A126759(n)-1. [The original definition.]

From Antti Karttunen, Jan 28 2015: (Start)

a(0) = 0, a(2n) = a(n), a(3n) = a(n), a(6n+1) = 2n + 1, a(6n+5) = 2n + 2.

Or with the last clause represented in another way:

a(0) = 0, a(2n) = a(n), a(3n) = a(n), a(6n+1) = 2n + 1, a(6n-1) = 2n.

Other identities. For all n >= 1:

a(n) = A253887(A003602(n)).

a(6n-3) = a(4n-2) = a(2n-1) = A253887(n).

(End)

a(n) = A249746(A003602(A064989(n))). - Antti Karttunen, Feb 04 2015

a(n) = A323882(4*n). - Antti Karttunen, Apr 18 2022

Extensions

Name replaced with an independent recurrence and the old description moved to the Formula section - Antti Karttunen, Jan 28 2015