A261691 Change of base from fractional base 3/2 to base 3.

0, 1, 2, 6, 7, 8, 21, 22, 23, 63, 64, 65, 69, 70, 71, 192, 193, 194, 207, 208, 209, 213, 214, 215, 579, 580, 581, 621, 622, 623, 627, 628, 629, 642, 643, 644, 1737, 1738, 1739, 1743, 1744, 1745, 1866, 1867, 1868, 1881, 1882, 1883, 1887, 1888, 1889, 1929, 1930

Offset

0,3

Comments

To obtain a(n), we interpret A024629(n) as a base 3 representation (instead of base 3/2). More precisely, if A024629(n) = A007089(m), then a(n) = m.

The digits used in fractional base 3/2 are 0,1, and 2, which are the same as the digits used in base 3.

Links

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Formula

For n = Sum_{i=0..m}c_i*(3/2)^i with each c_i in {0,1,2}, a(n) = Sum_{i=0..m}c_i*3^i.

From Rémy Sigrist, Apr 06 2021: (Start)

Apparently:

- a(3*n) = a(3*n-1) + A003462(1+A087088(n)) for any n > 0,

- a(3*n+1) = a(3*n) + 1 for any n >= 0,

- a(3*n+2) = a(3*n+1) + 1 for any n >= 0,

(End)

Example

The base 3/2 representation of 7 is (2,1,1); i.e., 7 = 2*(3/2)^2 + 1*(3/2) + 1. Since 2*(3^2) + 1*3 + 1*1 = 22, we have a(7) = 22.

Prog

(Sage)
def changebase(n):
    L=[n]
    i=1
    while L[i-1]>2:
        x=L[i-1]
        L[i-1]=x.mod(3)
        L.append(2*floor(x/3))
        i+=1
    return sum([L[i]*3^i for i in [0..len(L)-1]])
[changebase(n) for n in [0..100]]
(PARI) a(n) = { my (v=0, t=1); while (n, v+=t*(n%3); n=(n\3)*2; t*=3); v } \\ Rémy Sigrist, Apr 06 2021

Crossrefs

Cf. A024629, A007089.

Cf. A005836, A023717, A000695, A037453, A037454, A037455, A037456, A037314.

Cf. A003462, A087088.

Sequence in context: A258826 A157671 A196747 * A351088 A296443 A102046

Adjacent sequences: A261688 A261689 A261690 * A261692 A261693 A261694

Keyword

nonn,base

Author

Tom Edgar, Aug 28 2015

Status

approved