A348516 a(n) is the least positive integer k such that the base 3 representation of n^k contains equally many 1's and 2's, or 0 if no k with this property exists.

1, 0, 7, 0, 16, 1, 7, 1, 22, 0, 16, 1, 16, 6, 2, 1, 8, 6, 7, 1, 4, 1, 66, 9, 22, 3, 2, 0, 15, 1, 16, 2, 32, 1, 6, 9, 16, 2, 11, 6, 19, 13, 2, 13, 1, 1, 10, 22, 8, 2, 1, 6, 1, 159, 7, 1, 20, 1, 3, 6, 4, 2, 15, 1, 11, 3, 66, 6, 1, 9, 1, 6, 22, 2, 4, 3, 1, 2, 2, 2, 6

Offset: 0

Author

Dimiter Skordev, Oct 21 2021

Keywords

• nonn
• base

Comments

a(3*n) = a(n) for any positive integer n because multiplication by 3 does not change the counts of the digits 1 and 2 in the base 3 representation. Hence a(n) reaches any of its values at infinitely many n.

There are infinitely many n with a(n) = 1 that are not divisible by 3, e.g. the numbers of the form (3^m + 2)(3^(m-1) + 3^(m-2) + ... + 3 + 1), m = 1, 2, 3, ...

Of course, a(n^a(n)) = 1 whenever a(n) > 0. More generally, if a(n) = p*q, where p and q are positive integers, then a(n^p) = q (hence any positive divisor of a nonzero term of the sequence is a term too). If a(n) = 0 then a(n^p) = 0 for any positive integer p.

In the absence of a proof that a(n) = 0 only for the numbers n which are powers of 3, it would be desirable to have at least an algorithm whose application to any concrete n answers the question whether a(n) = 0.

Except for the case when the number a(n) is 0, it is the least positive integer k such that n^k is a term of the sequence A039001.

Problem: Are there positive integers not occurring in the sequence a(1),a(2),a(3),...?

Examples

			a(2) = 7 because the base 3 representations of 2^1, 2^2, 2^3, 2^4, 2^5, 2^6 and 2^7 are 2, 11, 22, 121, 1012, 2101 and 11202 respectively.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..19683

Crossrefs

Cf. A039001.

Programs

Mathematica

Array[If[IntegerQ@ Log[3, #], 0, Block[{k = 1}, While[Unequal @@ Most@ DigitCount[#^k, 3], k++]; k]] &, 72] (* Michael De Vlieger, Oct 21 2021 *)

PARI

isp3(n) = my(q); isprimepower(n,&q) && (q==3);
isok(k, n) = my(d=digits(n^k, 3)); #select(x->(x==1), d) == #select(x->(x==2), d);
a(n) = if ((n==1) || isp3(n), return (0)); my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Oct 22 2021

Python

h=[0,1,-1]
def d(x):
    y,d=x,0
    while y>0: d,y=d+h[y%3],y//3
    return d
def a(n):
    v,a,x=n,0,1
    while v%3==0: v=v//3
    if v>1:
        while d(x)!=0: a,x=a+1,v*x
    return a

Python

from gmpy2 import digits
def A348516(n):
    k, s = 1, digits(n,3).rstrip('0')
    if s == '1' or s == '': return 1-len(s)
    m = int(s,3)
    mk = m
    while s.count('1') != s.count('2'): k += 1; mk *= m; s = digits(mk,3)
    return k # Chai Wah Wu, Nov 11 2021

Extensions

a(0) from Michel Marcus, Nov 11 2021