A236510 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.

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, 32, 33, 34, 35, 36, 37, 38, 39, 41, 43, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 64, 65, 67, 69, 71, 73, 74, 77, 79, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95

Offset: 1

Author

Christian Perfect, Jan 27 2014

Keywords

• nonn
• easy

Comments

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.

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.

Examples

			14 is a member as 14 = 2^1 * 3^0 * 5^0 * 7^1, and (1,0,0,1) is a palindrome.
42 is not a member as 42 = 2^1 * 3^1 * 5^0 * 7^1, and (1,1,0,1) is not a palindrome.
		

Crossrefs

A subsequence of A242414.

Cf. also A242418, A085924.

Programs

Python

import re
   
def factorize(n):
   for prime in primes:
      power = 0
      while n%prime==0:
         n /= prime
         power += 1
      yield power
   
re_zeros = re.compile('(?P0*)(?P.*[^0])(?P=zeros)')
   
is_palindrome = lambda s: s==s[::-1]
   
def has_palindromic_factorization(n):
   if n==1:
      return True
   s = ''.join(str(x) for x in factorize(n))
   try:
      middle = re_zeros.match(s).group('middle')
      if is_palindrome(middle):
         return True
   except AttributeError:
      return False
   
a = has_palindromic_factorization