A023717 -id:A023717 - cp's OEIS frontend

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

Tom Edgar, Aug 28 2015

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.

			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.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Index entries for sequences related to fractional bases.

Cf. A024629, A007089.
Cf. A005836, A023717, A000695, A037453, A037454, A037455, A037456, A037314.
Cf. A003462, A087088.

a[n_] := a[n] = If[n == 0, 0, 3 * a[2 * Floor[n/3]] + Mod[n, 3]]; Array[a, 100, 0] (* Amiram Eldar, Aug 04 2025 *)

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

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]]

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)

A337143 Numbers k for which there are only 3 bases b (2, k+1 and another one) in which the digits of k contain the digit b-1.

Original entry on oeis.org

5, 6, 8, 9, 12, 16, 18, 28, 37, 81, 85, 88, 130, 150, 262, 810, 1030, 1032, 4132, 9828, 9832, 10662, 10666, 562576, 562578

Offset: 1

François Marques, Sep 14 2020

This sequence is the list of indices k such that A337496(k)=3.
Conjecture: this sequence is finite and full. a(26) > 3.8*10^12 if it exists.
All terms of this sequence increased by 1 are either prime numbers, or prime numbers squared, or 2 times a prime number because if b is a strict divisor of k+1, the digit for the units in the expansion of k in base b is b-1 so it must be 2 or the third base. In fact k+1 could have been equal to 8=2*4 but 7 is not a term of the sequence (7 = 111_2 = 21_3 = 13_4 = 7_8).

			a(7)=18 because there are only 3 bases (2, 19 and 3) which satisfy the condition of the definition (18=200_3) and 18 is the seventh of these numbers.

Cf. Numbers with at least one digit b-1 in base b : A074940 (b=3), A337250 (b=4), A337572 (b=5), A011539 (b=10), A095778 (b=11).
Cf. Numbers with no digit b-1 in base b: A005836 (b=3), A023717 (b=4), A020654 (b=5), A037465 (b=6), A020657 (b=7), A037474 (b=8), A037477 (b=9), A007095 (b=10), A065039 (b=11).
Cf. A337496, A337535, A337536.

cp's OEIS Frontend

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

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