A352040 a(n) is the least number k such that 2^k contains each of the 10 digits at least n times.

68, 88, 119, 200, 209, 246, 291, 318, 396, 398, 443, 443, 495, 586, 592, 622, 646, 707, 758, 758, 813, 866, 875, 903, 923, 1001, 1022, 1022, 1105, 1111, 1111, 1231, 1243, 1245, 1327, 1342, 1419, 1453, 1453, 1454, 1534, 1536, 1537, 1626, 1676, 1699, 1699, 1763

Offset: 1

Author

Amiram Eldar, Apr 16 2022

Keywords

• nonn
• base

Comments

a(1)-a(8) were given in the solution to Problem 410 (Crux Mathematicorum, 1979), but a(2) = 170 is wrong.

a(1) was calculated by Rudolph Ondrejka in 1976.

Examples

			a(1) = 68 since 2^68 = 295147905179352825856 is the least power of 2 that contains all the 10 digits at least once.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• James Gary Propp, Problem 410, Crux Mathematicorum, Vol. 5, No. 1 (1979), p. 17; Solution to Problem 410, ibid., Vol. 5, No. 10 (1979), pp. 296-299.

Crossrefs

Cf. A000079, A007377, A020862, A130694, A130696.

Programs

Maple

a:= proc(n) option remember; local k; for k from a(n-1)
      while min((p-> seq(coeff(p, x, j), j=0..9))(add(
            x^i, i=convert(2^k, base, 10))))Alois P. Heinz, Apr 22 2022

Mathematica

s = Table[Min[DigitCount[2^n, 10, Range[0, 9]]], {n, 1, 2500}]; Table[FirstPosition[s, _?(# >= n &)], {n, 1, Max[s]}] // Flatten

Python

from sympy import ceiling, log
def A352040(n):
    k = 10*n-1+int(ceiling((10*n-1)*log(5,2)))
    s = str(c := 2**k)
    while any(s.count(d) < n for d in '0123456789'):
        c *= 2
        k += 1
        s = str(c)
    return k   # Chai Wah Wu, Apr 16 2022

Formula

Conjecture: a(n) ~ c*n, where c = 10*log_2(10) = 33.21928... .

a(n) >= (10n-1)*log_2(10), i.e., c = 10*log_2(10) is a lower bound on the asymptotic growth rate. - Chai Wah Wu, Apr 16 2022