A291926 a(n) is the smallest integer k>=0 such that 2^k contains every digit in base n, or 0 if no such integer exists.

1, 5, 0, 18, 25, 20, 0, 61, 68, 64, 72, 103, 110, 134, 0, 138, 141, 140, 141, 172, 191, 228, 225, 244, 306, 281, 272, 339, 384, 412, 0, 390, 421, 372, 472, 395, 441, 486, 495, 473, 566, 576, 629, 735, 626, 661, 706, 707, 741, 825, 782, 751, 811, 924, 930, 908, 927, 975, 1049, 934, 1018, 1070, 0

Offset: 2

Author

Ely Golden, Sep 05 2017

Keywords

• nonn
• base

Comments

a(n) >= ceiling(log_2(n)*(n-1)), whenever a(n)>0. This is because in order for an integer to have n digits in base n it must have at least a magnitude of n-1 in base n.

a(n) = 0 at all powers of 2 (except 2 itself). This is because powers of 2 in power-of-2 bases can only have 2 distinct digits. Is a(n) equal to 0 for any other values of n?

It seems that the base n representation of 2^(2*n^2) contains all n digits whenever n is not a power of 2. A proof of this would yield a negative answer to the above question. In the absence of a negative answer to this question, at least an algorithm would be desirable whose application to any concrete value of n solves the problem whether a(n)=0 for this n (for instance, if a(n)Dimiter Skordev, Aug 18 2021

Examples

			a(3) = 5, since 2^5 is the smallest power of 2 which contains every digit in base 3: Namely, 2^5 is 1012 in base 3, whereas the previous powers are 1, 2, 11, 22, and 121, respectively, none of which contain all possible base-3 digits.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 2..512 (n = 2..256 from Ely Golden).

Crossrefs

Cf. A090493.

Programs

Mathematica

TakeWhile[#, # > -1 &] &@ Table[If[And[IntegerQ@ #, # > 1] &@ Log2@ n, 0, SelectFirst[Range[2^11], Times @@ DigitCount[2^#, n] > 0 &]] /. k_ /; MissingQ@ k -> -1, {n, 2, 64}] (* Michael De Vlieger, Sep 05 2017 *)

PARI

a(n) = {if (n==2, return (1)); if (ispower(n,,&k) && (k==2), return (0)); k = 1; while (#Set(digits(2^k, n)) != n, k++); k;} \\ Michel Marcus, Sep 06 2017

Python

def floorLog(b,n):
    x=-1
    while(n>0):
        x+=1
        n//=b
    return x
def distinctDigits(n,b):
    li=[]
    while(n>0):
        li.append(n%b)
        n//=b
    li=list(set(li))
    li.sort()
    return li
def iroot(k,n):
    u, s = n, n+1
    while u < s:
        s = u
        t = (k-1) * s + n // (s**(k-1))
        u = t // k
    return s
def perfectPower(n):
    if(n==1): return 0
    x=1
    for i in range(2,floorLog(2,n)+1):
        if(iroot(i,n)**i==n): x=i
    return x
def leastPandigital(b,n):
    if(n<=1 or b<=1): return 0
    if(n==2): return 2 if (b==(1<
				

Python

from sympy.ntheory.digits import digits
def a(n):
    b = bin(n)[2:]
    if b.strip('0') == '1': return int(n == 2)
    k = (len(b)-1)*(n-1)
    while len(set(digits(2**k, n)[1:])) != n: k += 1
    return k
print([a(n) for n in range(2, 65)]) # Michael S. Branicky, Oct 07 2021