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

1,2

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.

Links

Table of n, a(n) for n=1..68.

Example

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.

Prog

(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('(?P<zeros>0*)(?P<middle>.*[^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

Crossrefs

A subsequence of A242414.

Cf. also A242418, A085924.

Sequence in context: A072774 A062770 A359889 * A317710 A303554 A325328

Adjacent sequences: A236507 A236508 A236509 * A236511 A236512 A236513

Keyword

nonn,easy

Author

Christian Perfect, Jan 27 2014

Status

approved