A351927 Smallest positive integer k such that 2^k has no '0' in the last n digits of its ternary expansion.

1, 2, 4, 10, 15, 15, 15, 15, 15, 15, 50, 50, 101, 101, 101, 101, 143, 143, 143, 143, 143, 143, 143, 143, 143, 1916, 1916, 1916, 1916, 1916, 1916, 82286, 1134022, 1639828, 3483159, 3483159, 3483159, 3917963, 3917963, 3917963, 4729774, 4729774, 9827775, 9827775, 43622201, 43622201, 43622201

Offset: 1

Author

Robert Saye, Feb 25 2022

Keywords

• nonn
• base

Comments

The powers of two are required to have at least n ternary digits, i.e., 2^k >= 3^(n-1).

Sloane (1973) conjectured that every power 2^n with n > 15 has a '0' somewhere in its ternary expansion (see A102483 and A346497).

Links

• Robert I. Saye, On two conjectures concerning the ternary digits of powers of two, arXiv:2202.13256 [math.NT], 2022.

Crossrefs

Cf. A004642, A117970, A102483, A346497, A351928.

Programs

Mathematica

smallest[n_] := Module[{k}, k = Max[1, Ceiling[(n - 1) Log[2, 3]]];  While[MemberQ[Take[IntegerDigits[2^k, 3], -n], 0], ++k]; k]; Table[smallest[n], {n, 1, 20}]

PARI

a(n) = my(k=1); while(!vecmin(Vec(Vecrev(digits(2^k,3)), n)), k++); k; \\ Michel Marcus, Feb 26 2022

Python

from sympy.ntheory.digits import digits
def a(n, startk=1):
    k = max(startk, len(bin(3**(n-1))[2:]))
    pow2 = 2**k
    while 0 in digits(pow2, 3)[-n:]:
        k += 1
        pow2 *= 2
    return k
an = 0
for n in range(1, 32):
    an = a(n, an)
    print(an, end=", ") # Michael S. Branicky, Mar 10 2022

Python

from itertools import count
def A351927(n):
    kmax, m = 3**n, (3**(n-1)).bit_length()
    k2 = pow(2,m,kmax)
    for k in count(m):
        a = k2
        if 3*a >= kmax:
            while a > 0:
                a, b = divmod(a,3)
                if b == 0:
                    break
            else:
                return k
        k2 = 2*k2 % kmax # Chai Wah Wu, Mar 19 2022