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%I #9 Sep 28 2014 16:05:18
%S 1,7,28,106,391,992,1178,7255,15975,67143,333212,333212,1641257
%N Least number k > 0 such that 2^k contains an n-digit long substring of the infinite string "0123456789012345678901234567890123456....".
%C By A238448, a(10) <= 244178.
%e 2^7 = 128 contains the 2-digit substring "12". Thus a(2) = 7.
%o (Python)
%o def a(n):
%o ..for k in range(1,10**5):
%o ....for i in range(10):
%o ......s = ''
%o ......for j in range(i,i+n):
%o ........dig=j%10
%o ........s+=str(dig)
%o ......if str(2**k).find(s) > -1:
%o ........return k
%o n=1
%o while n < 10:
%o ..print(a(n))
%o ..n+=1
%Y Cf. A238448.
%K nonn,more,hard,base
%O 1,2
%A _Derek Orr_, May 31 2014
%E a(10)-a(13) from _Hiroaki Yamanouchi_, Sep 26 2014
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