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
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 0..19683
Crossrefs
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(6n-3) = a(4n-2) = a(2n-1) = A253887(n).
(End)
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
Comments