A117970 -id:A117970 - cp's OEIS frontend

A117971 One-based position of the first 2 from the least significant digit in the ternary expansion of 2^n, or 0 if there are no 2's present.

0, 1, 0, 1, 2, 1, 4, 1, 0, 1, 2, 1, 3, 1, 3, 1, 2, 1, 5, 1, 8, 1, 2, 1, 11, 1, 11, 1, 2, 1, 3, 1, 3, 1, 2, 1, 4, 1, 4, 1, 2, 1, 5, 1, 4, 1, 2, 1, 3, 1, 3, 1, 2, 1, 6, 1, 8, 1, 2, 1, 4, 1, 7, 1, 2, 1, 3, 1, 3, 1, 2, 1, 12, 1, 7, 1, 2, 1, 6, 1, 10, 1, 2, 1, 3, 1, 3, 1, 2, 1, 4, 1, 4, 1, 2, 1, 6, 1, 4, 1, 2, 1, 3, 1

Offset: 0

Eric W. Weisstein, Apr 06 2006

a(0), a(2) and a(8) are the conjectured to be the only terms equal to 0.

			For n=0, 2^0 = 1 is also "1" in base-3, thus there are no 2-digits present, and therefore a(0) = 0.
For n=4, 2^4 = 16, which in base-3 is "121" as 1*(3^2) + 2*3 + 1 = 16, so the rightmost 2 occurs at two steps from the end, therefore a(4) = 2.
For n=5, 2^5 = 32, which in base-3 is "1012" as 1*(3^3) + 1*3 + 2*1 = 32, so the rightmost 2 occurs as the least significant digit (which is the position 1), therefore a(5) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..16384
Eric Weisstein's World of Mathematics, Ternary

Cf. A004642 (ternary expansion of 2^n), A007089. See also A117970.

pf2[n_]:=Module[{p=Position[Reverse[IntegerDigits[2^n,3]],2,{1},1]},If[p=={},0,p]]; Flatten[Array[pf2,110]] (* Harvey P. Dale, Nov 30 2013 *)

A117971(n) = { my(n=(2^n),i=1); while(n, if(2==(n%3),return(i)); i++; n \= 3); (0); }; \\ Antti Karttunen, Mar 30 2021

Edited by Charles R Greathouse IV, Aug 05 2010

Term a(0) = 0 prepended, examples added and the definition clarified by Antti Karttunen, Mar 30 2021

A270026 a(n) is the smallest b for which the base-b representation of n contains at least one 0 (or 0 if no such base exists).

Original entry on oeis.org

0, 2, 3, 2, 2, 2, 7, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Nathan Fox, Mar 08 2016

For n > 1, a(n)=2 whenever n+1 is not a power of 2.
It is conjectured that the only terms that are neither 2 nor 3 in this sequence are a(7) and a(32767), which are 7 and 5 respectively.
a(n) > 0 for n >= 2 since the base-n representation of n is 10.

Nathan Fox, Table of n, a(n) for n = 1..40000
Eric Weisstein's World of Mathematics, Ternary

Cf. A117970, A216194.

Table[SelectFirst[Range[2, 1200], DigitCount[n, #, 0] > 0 &], {n, 2, 120}] (* Michael De Vlieger, Mar 09 2016, Version 10 *)

a(n) = if (n==1, 0, my(b=2); while(vecmin(digits(n, b)), b++); b); \\ Michel Marcus, Mar 09 2016

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

Original entry on oeis.org

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

Robert Saye, Feb 25 2022

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).

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

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

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}]

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

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

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

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A117971 One-based position of the first 2 from the least significant digit in the ternary expansion of 2^n, or 0 if there are no 2's present.

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A270026 a(n) is the smallest b for which the base-b representation of n contains at least one 0 (or 0 if no such base exists).

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

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