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13, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143, 1872, 1898, 1911, 1924, 1937, 1950, 1963, 1976, 1989, 2002, 2015, 3744, 3757, 3783, 3796, 3809, 3822, 3835, 3848, 3861, 3874, 3887, 5616, 5629, 5642, 5668, 5681, 5694, 5707, 5720, 5733, 5746, 5759, 7488, 7501
(Python)
from sympy.ntheory import digits
from itertools import groupby
def ok(n):
return all(len(list(g))==2 for k, g in groupby(digits(n, 12)[1:]))
print(list(filter(ok, range(1, 7502)))) # Michael S. Branicky, Apr 27 2021
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Every run Numbers each of whose runs of digits of n in base 12 has length 2.
a(n) = 13*A043316(n) (= 13*n for n < 12). - M. F. Hasler, Feb 02 2014
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Vincenzo Librandi, <a href="/A033010/b033010.txt">Table of n, a(n) for n = 1..1400</a>
Select[Range[10000], Union[Length/@Split[IntegerDigits[#, 12]]]=={2}&] (* Vincenzo Librandi, Feb 05 2014 *)
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