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

%I A352040 #53 Apr 22 2022 09:58:00
%S A352040 68,88,119,200,209,246,291,318,396,398,443,443,495,586,592,622,646,
%T A352040 707,758,758,813,866,875,903,923,1001,1022,1022,1105,1111,1111,1231,
%U A352040 1243,1245,1327,1342,1419,1453,1453,1454,1534,1536,1537,1626,1676,1699,1699,1763
%N A352040 a(n) is the least number k such that 2^k contains each of the 10 digits at least n times.
%C A352040 a(1)-a(8) were given in the solution to Problem 410 (Crux Mathematicorum, 1979), but a(2) = 170 is wrong.
%C A352040 a(1) was calculated by Rudolph Ondrejka in 1976.
%H A352040 Amiram Eldar, <a href="/A352040/b352040.txt">Table of n, a(n) for n = 1..10000</a>
%H A352040 James Gary Propp, <a href="https://cms.math.ca/publications/crux/issue?volume=5&amp;issue=1">Problem 410</a>, Crux Mathematicorum, Vol. 5, No. 1 (1979), p. 17; <a href="https://cms.math.ca/publications/crux/issue?volume=5&amp;issue=10">Solution to Problem 410</a>, ibid., Vol. 5, No. 10 (1979), pp. 296-299.
%F A352040 Conjecture: a(n) ~ c*n, where c = 10*log_2(10) = 33.21928... .
%F A352040 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
%e A352040 a(1) = 68 since 2^68 = 295147905179352825856 is the least power of 2 that contains all the 10 digits at least once.
%p A352040 a:= proc(n) option remember; local k; for k from a(n-1)
%p A352040       while min((p-> seq(coeff(p, x, j), j=0..9))(add(
%p A352040             x^i, i=convert(2^k, base, 10))))<n do od; k
%p A352040     end: a(0):=0:
%p A352040 seq(a(n), n=1..50);  # _Alois P. Heinz_, Apr 22 2022
%t A352040 s = Table[Min[DigitCount[2^n, 10, Range[0, 9]]], {n, 1, 2500}]; Table[FirstPosition[s, _?(# >= n &)], {n, 1, Max[s]}] // Flatten
%o A352040 (Python)
%o A352040 from sympy import ceiling, log
%o A352040 def A352040(n):
%o A352040     k = 10*n-1+int(ceiling((10*n-1)*log(5,2)))
%o A352040     s = str(c := 2**k)
%o A352040     while any(s.count(d) < n for d in '0123456789'):
%o A352040         c *= 2
%o A352040         k += 1
%o A352040         s = str(c)
%o A352040     return k   # _Chai Wah Wu_, Apr 16 2022
%Y A352040 Cf. A000079, A007377, A020862, A130694, A130696.
%K A352040 nonn,base
%O A352040 1,1
%A A352040 _Amiram Eldar_, Apr 16 2022