OFFSET
1,1
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.
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.
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
EXAMPLE
a(1) = 68 since 2^68 = 295147905179352825856 is the least power of 2 that contains all the 10 digits at least once.
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))))<n do od; k
end: a(0):=0:
seq(a(n), n=1..50); # 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
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Amiram Eldar, Apr 16 2022
STATUS
approved