A085924 -id:A085924 - cp's OEIS frontend

A103653 Members of A085924 that are not members of A072774.

90, 126, 198, 234, 270, 300, 306, 342, 350, 378, 414, 522, 525, 550, 558, 588, 594, 650, 666, 702, 738, 774, 810, 825, 846, 850, 918, 950, 954, 975, 980, 1026, 1062, 1078, 1098, 1134, 1150, 1206, 1242, 1274, 1275, 1278, 1314, 1422, 1425, 1450, 1452, 1494

Offset: 1

David Wasserman, Feb 11 2005

All terms shown have exactly 3 distinct prime factors. a(101) = 2940 =2^2*3*5*7^2 is the first member with more than 3. a(107) = 3072 = 2^10*3 is the first member with less than 3.

			90 = 2^1*3^2*5^1 is a member because the concatenation of the exponents is 121.

A242414 Numbers whose prime factorization viewed as a tuple of nonzero powers is palindromic.

Original entry on oeis.org

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, 42, 43, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95, 97, 100

Offset: 1

Antti Karttunen, May 30 2014

nonn

The fixed points of permutation A069799.
Differs from its subsequence, A072774, Powers of squarefree numbers, for the first time at n=68, as here a(68) = 90 is included, as 90 = p_1^1 * p_2^2 * p_3^1 has a palindromic tuple of exponents, even although not all of them are identical.
Differs from its another subsequence, A236510, in that, although numbers like 42 = 2^1 * 3^1 * 5^0 * 7^1, with a non-palindromic exponent-tuple (1,1,0,1) are excluded from A236510, it is included in this sequence, because here only the nonzero exponents are considered, and (1,1,1) is a palindrome.
Differs from A085924 in that as that sequence is subtly base-dependent, it excludes 1024 (= 2^10), as then the only exponent present, 10, and thus also its concatenation, "10", is not a palindrome when viewed in decimal base. On the contrary, here a(691) = 1024.

			As 1 has an empty factorization, (), which also is a palindrome, 1 is present.
As 42 = 2 * 3 * 7 = p_1^1 * p_2^1 * p_4^1, and (1,1,1) is palindrome, 42 is present.
As 90 = 2 * 9 * 5 = p_1^1 * p_2^2 * p_3^1, and (1,2,1) is palindrome, 90 is present.
Any prime power (A000961) is present, as such numbers have a factorization p^e (e >= 1), and any singleton sequence (e) by itself forms a palindrome.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Fixed points of A069799.
Complement: A242416.
A000961, A072774 and A236510 are subsequences.
Cf. A242418, A085924.

Select[Range[100], PalindromeQ[FactorInteger[#][[All, 2]]]&] (* Jean-François Alcover, Feb 09 2025 *)

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.

Original entry on oeis.org

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

Christian Perfect, Jan 27 2014

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.

			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.

A subsequence of A242414.

Cf. also A242418, A085924.

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

A383106 Numbers k such that A382883(k) != 0.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 100

Offset: 1

Peter Luschny, Apr 16 2025

nonn

See the comments in A382883.

Cf. A382883, A382943 (complement), A383016, A383017.

Cf. A005117 and similars: A062770, A072774, A085924, A242414, A304449.

V[n_, e_] := If[e == 1, 1, IntegerExponent[n, e]]; f[n_] := f[n] = -DivisorSum[n, V[n, #] * f[#] &, # < n &]; f[1] = 1; Select[Range[100], f[#] != 0 &] (* Amiram Eldar, Apr 29 2025 *)

def A383106List(upto): return [n for n in srange(1, upto) if A382883(n) != 0]

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A103653 Members of A085924 that are not members of A072774.

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A242414 Numbers whose prime factorization viewed as a tuple of nonzero powers is palindromic.

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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.

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A383106 Numbers k such that A382883(k) != 0.

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