A351928 Smallest positive integer k such that 2^k has no '2' in the last n digits of its ternary expansion.
2, 2, 6, 8, 8, 8, 20, 24, 24, 24, 72, 186, 186, 332, 332, 1134, 1134, 1134, 1134, 1134, 1134, 25458, 25458, 25458, 25458, 25458, 25458, 159140, 249968, 249968, 249968, 249968, 249968, 249968, 249968, 249968, 9076914, 9076914, 9076914, 9076914, 9076914, 9076914, 90062678
Offset: 1
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.
Programs
-
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}] -
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
Comments