A236510 - OEIS

%I #18 Jun 08 2017 16:33:45

%S 1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,17,19,21,22,23,25,26,27,29,30,31,

%T 32,33,34,35,36,37,38,39,41,43,46,47,49,51,53,55,57,58,59,61,62,64,65,

%U 67,69,71,73,74,77,79,81,82,83,85,86,87,89,90,91,93,94,95

%N Numbers whose prime factorization viewed as a tuple of powers is palindromic, when viewed from the least to the largest prime present, including also any zero-exponents for the intermediate primes.

%C Compute the prime factorization of n = product(p_i^r_i). If the tuple (r_1,...) is a palindrome (excluding leading or trailing zeros, but including any possible intermediate zeros), n belongs to the sequence.

%C 42 is the first element of A242414 not in this sequence, as 42 = 2^1 * 3^1 * 5^0 * 7^1, and (1,1,0,1) is not a palindrome, although (1,1,1) is.

%e 14 is a member as 14 = 2^1 * 3^0 * 5^0 * 7^1, and (1,0,0,1) is a palindrome.

%e 42 is not a member as 42 = 2^1 * 3^1 * 5^0 * 7^1, and (1,1,0,1) is not a palindrome.

%o (Python)

%o import re

%o ...

%o def factorize(n):

%o ...for prime in primes:

%o ......power = 0

%o ......while n%prime==0:

%o .........n /= prime

%o .........power += 1

%o ......yield power

%o ...

%o re_zeros = re.compile('(?P<zeros>0*)(?P<middle>.*[^0])(?P=zeros)')

%o ...

%o is_palindrome = lambda s: s==s[::-1]

%o ...

%o def has_palindromic_factorization(n):

%o ...if n==1:

%o ......return True

%o ...s = ''.join(str(x) for x in factorize(n))

%o ...try:

%o ......middle = re_zeros.match(s).group('middle')

%o ......if is_palindrome(middle):

%o .........return True

%o ...except AttributeError:

%o ......return False

%o ...

%o a = has_palindromic_factorization

%Y A subsequence of A242414.

%Y Cf. also A242418, A085924.

%K nonn,easy

%O 1,2

%A _Christian Perfect_, Jan 27 2014