OFFSET
1,1
COMMENTS
The powers of two are required to have at least n ternary digits, i.e., 2^k >= 3^(n-1).
Erdős (~1978) conjectured that 1, 4, and 256 are the only powers of two whose ternary expansion consists solely of 0's and 1's.
LINKS
Paul Erdős, Some unconventional problems in number theory, Mathematics Magazine, Vol. 52, No. 2 (1979), pp. 67-70.
Robert I. Saye, On two conjectures concerning the ternary digits of powers of two, arXiv:2202.13256 [math.NT], 2022.
MATHEMATICA
smallest[n_] := Module[{k}, k = Max[1, Ceiling[(n - 1) Log[2, 3]]]; While[MemberQ[Take[IntegerDigits[2^k, 3], -n], 2], ++k]; k]; Table[smallest[n], {n, 1, 20}]
PROG
(PARI) a(n) = my(k=max(1, logint(3^(n-1), 2))); while(#select(x->(x==2), 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 2 in digits(pow2, 3)[-n:]:
k += 1
pow2 *= 2
return k
an = 0
for n in range(1, 22):
an = a(n, an)
print(an, end=", ") # Michael S. Branicky, Feb 27 2022
(Python)
from itertools import count
def A351928(n):
kmax, m = 3**n, (3**(n-1)).bit_length()
k2 = pow(2, m, kmax)
for k in count(m):
a = k2
while a > 0:
a, b = divmod(a, 3)
if b == 2:
break
else:
return k
k2 = 2*k2 % kmax # Chai Wah Wu, Mar 19 2022
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Robert Saye, Feb 25 2022
STATUS
approved