%I #29 Jun 01 2015 14:21:34
%S 1,25,143,1598,10627,194220,26399,37811757,15689797,1609719151,
%T 42001126081,42116737194,2277292670319,8475536580225,57483036385216,
%U 1185808703658960,2800250032195382,203292337502775829,294180235677139843,28666496154250702728
%N Least number k such that 2^k begins with exactly n identical digits.
%C a(8) > 200000. The repeating digits that corresponds to a(n) are {2, 3, 1, 1, 1, 1, 7, 1, 3, 1, 1, 7} respectively.
%C a(8) > 3*10^7, a(9) = 15689797 (repeating digit is 3). - _Lars Blomberg_, Jul 02 2014
%H Hiroaki Yamanouchi, <a href="/A235713/b235713.txt">Table of n, a(n) for n = 1..100</a>
%e 2^25 = 33554432 begins with two identical digits ('33'). Thus a(2) = 25.
%o (Python)
%o def b(n):
%o ..for k in range(1,2*10**5):
%o ....st = str(2**k)
%o ....count = 0
%o ....if len(st) >= n:
%o ......for i in range(len(st)):
%o ........if st[i] == st[0]:
%o ..........count += 1
%o ........else:
%o ..........break
%o ......if count == n:
%o ........return k
%o n = 1
%o while n < 10:
%o ..print(b(n),end=', ')
%o ..n += 1
%K nonn,base
%O 1,2
%A _Derek Orr_, Jun 13 2014
%E a(8)-a(12) from _Hiroaki Yamanouchi_, Jul 13 2014
%E a(13)-a(20) from _Hiroaki Yamanouchi_, May 31 2015
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