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A075786
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Palindromic perfect powers.
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7
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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
(list;
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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343 = 7^3 is a term as it is a palindrome and a perfect power. - David A. Corneth, Mar 23 2021
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MATHEMATICA
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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}]
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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b-file corrected and extended by Chai Wah Wu, Aug 26 2021
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STATUS
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approved
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