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A243973
a(n) is the repeating digit in 2^A243972(n).
0
2, 5, 4, 8, 9, 1, 8, 9, 1, 8, 3, 5, 1, 6, 7, 3, 1, 4, 7, 5, 1, 2, 8, 0, 7, 7, 8, 1, 4, 1, 8, 5, 7, 5, 9, 4, 8, 1, 6, 7, 3, 5, 0, 4, 9, 8, 6, 7, 4, 1, 6, 3, 0, 2, 9, 8, 4, 5, 8, 9, 2, 3, 0, 3, 3, 5, 1, 6, 3, 4, 6, 6, 8, 6, 4, 8, 4, 6, 5, 7, 4, 1, 1, 7, 1, 1, 9, 3, 2, 5, 2, 2, 3
OFFSET
1,1
COMMENTS
In case of a tie, choose the smaller integer.
EXAMPLE
2^A243972(4) = 2^23 = 8388608 contains four 8's. Thus A243972(4) = 23 and here, since 8 is the repeating digit, a(4) = 8.
PROG
(Python)
def b():
n = 1
k = 1
while k < 50000:
st = str(2**k)
if len(st) >= n:
for a in range(10):
count = 0
for i in range(len(st)):
if st[i] == str(a):
count += 1
if count == n:
print(a, end=', ')
n += 1
k = 0
break
k += 1
else:
k += 1
b()
CROSSREFS
Cf. A243972.
Sequence in context: A187012 A134079 A033686 * A286015 A183542 A328203
KEYWORD
nonn,base
AUTHOR
Derek Orr, Jun 16 2014
STATUS
approved