A075786 Palindromic perfect powers.

1, 4, 8, 9, 121, 343, 484, 676, 1331, 10201, 12321, 14641, 40804, 44944, 69696, 94249, 698896, 1002001, 1030301, 1234321, 1367631, 4008004, 5221225, 6948496, 100020001, 102030201, 104060401, 121242121, 123454321, 125686521, 400080004, 404090404, 522808225, 617323716, 942060249

Offset: 1

Author

Zak Seidov, Oct 10 2002

Keywords

• nonn
• base

Comments

Up to 10^12, there are only 43 perfect powers which are palindromic.

The sequence is infinite, for instance it contains (10^k+1)^2. - Emmanuel Vantieghem, Sep 29 2017

Conjecture: there are no palindromic perfect powers with prime exponent > 3. - Chai Wah Wu, Aug 26 2021

Examples

			343 = 7^3 is a term as it is a palindrome and a perfect power. - _David A. Corneth_, Mar 23 2021
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..2001 (n = 1..176 from Michael S. Branicky, n = 177..504 from David A. Corneth)

Crossrefs

Cf. A001597, A002113, A076443.

Programs

Mathematica

a = {}; Do[q = IntegerDigits[n]; p = FromDigits[ Join[ q, Reverse[ Drop[q, -1]]]]; If[ Apply[ GCD, Last[ Transpose[ FactorInteger[p]]]] > 1, a = Append[a, p]]; p = FromDigits[ Join[ q, Reverse[q]]]; If[ Apply[ GCD, Last[ Transpose[ FactorInteger[p]]]] > 1, a = Append[a, p]], {n, 1, 10^5}]

Python

from math import isqrt
def ispal(n): s = str(n); return s == s[::-1]
def athrough(digits):
  found, limit = {1}, 10**digits
  for k in range(2, isqrt(limit) + 1):
    kpow = k*k
    while kpow < limit:
      if ispal(kpow): found.add(kpow)
      kpow *= k
  return sorted(found)
print(athrough(9)) # Michael S. Branicky, Mar 23 2021

Extensions

Edited and extended by Robert G. Wilson v, Oct 11 2002

More terms from David A. Corneth, Mar 24 2021

b-file corrected and extended by Chai Wah Wu, Aug 26 2021