login
A235713
Least number k such that 2^k begins with exactly n identical digits.
1
1, 25, 143, 1598, 10627, 194220, 26399, 37811757, 15689797, 1609719151, 42001126081, 42116737194, 2277292670319, 8475536580225, 57483036385216, 1185808703658960, 2800250032195382, 203292337502775829, 294180235677139843, 28666496154250702728
OFFSET
1,2
COMMENTS
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.
a(8) > 3*10^7, a(9) = 15689797 (repeating digit is 3). - Lars Blomberg, Jul 02 2014
LINKS
EXAMPLE
2^25 = 33554432 begins with two identical digits ('33'). Thus a(2) = 25.
PROG
(Python)
def b(n):
..for k in range(1, 2*10**5):
....st = str(2**k)
....count = 0
....if len(st) >= n:
......for i in range(len(st)):
........if st[i] == st[0]:
..........count += 1
........else:
..........break
......if count == n:
........return k
n = 1
while n < 10:
..print(b(n), end=', ')
..n += 1
CROSSREFS
Sequence in context: A147360 A147223 A184057 * A072471 A017042 A100255
KEYWORD
nonn,base
AUTHOR
Derek Orr, Jun 13 2014
EXTENSIONS
a(8)-a(12) from Hiroaki Yamanouchi, Jul 13 2014
a(13)-a(20) from Hiroaki Yamanouchi, May 31 2015
STATUS
approved