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A243973 a(n) is the repeating digit in 2^A243972(n). 0

%I #5 Jun 21 2014 22:38:04

%S 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,

%T 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,

%U 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

%N a(n) is the repeating digit in 2^A243972(n).

%C In case of a tie, choose the smaller integer.

%e 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.

%o (Python)

%o def b():

%o ..n = 1

%o ..k = 1

%o ..while k < 50000:

%o ....st = str(2**k)

%o ....if len(st) >= n:

%o ......for a in range(10):

%o ........count = 0

%o ........for i in range(len(st)):

%o ..........if st[i] == str(a):

%o ............count += 1

%o ........if count == n:

%o ..........print(a,end=', ')

%o ..........n += 1

%o ..........k = 0

%o ..........break

%o ......k += 1

%o ....else:

%o ......k += 1

%o b()

%Y Cf. A243972.

%K nonn,base

%O 1,1

%A _Derek Orr_, Jun 16 2014

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Last modified March 29 08:13 EDT 2024. Contains 371265 sequences. (Running on oeis4.)