

A339624


Perfect powers p^k, k >= 2 of palindromic primes p when p^k is not a palindrome.


1



16, 25, 27, 32, 49, 64, 81, 125, 128, 243, 256, 512, 625, 729, 1024, 2048, 2187, 2401, 3125, 4096, 6561, 8192, 15625, 16384, 16807, 17161, 19683, 22801, 32761, 32768, 36481, 59049, 65536, 78125, 97969, 117649, 124609, 131072, 139129, 146689, 161051, 177147, 262144
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

Equivalently: numbers m with only one prime factor such that the LCM of their palindromic divisors is neither 1 nor m: subsequence of A334392.
G. J. Simmons conjectured there are no palindromes of form n^k for k >= 5 (and n > 1) (see Simmons p. 98). According to this conjecture, these perfect powers are terms: {2^k, k>=4}, {3^k, k>=3}, {5^k, k>=2}, {7^k, k=2 and k>=4}, {11^k, k>=5}, {101^k, k>= 5}, {131^k, k>=2}, ...
From a(1) = 16 to a(17) = 2187, the data is the same as A056781(10) until A056781(26), then a(18) = 2401 and A056781(27) = 4096.


REFERENCES

Murray S. Klamkin, Problems in applied mathematics: selections from SIAM review, (1990), p. 520.


LINKS

Gustavus J. Simmons, Palindromic Powers, J. Rec. Math., 3 (No. 2, 1970), 9398 [Annotated scanned copy].


EXAMPLE

5^2 = 25, 2^6 = 64, 3^4 = 81 are terms.
7^2 = 49 is a term, 7^3 = 343 is not a term, and 7^4 = 2401 is a term.
101^2 = 10201 and 11^4 = 14641 are not terms.


MATHEMATICA

q[n_] := Module[{f = FactorInteger[n]}, Length[f] == 1 && f[[1, 2]] > 1 && PalindromeQ[f[[1, 1]]]]; Select[Range[10^5], !PalindromeQ[#] && q[#] &] (* Amiram Eldar, Dec 10 2020 *)


PROG

(PARI) ispal(n) = my(d=digits(n)); Vecrev(d)==d;
isok(k) = my(p); isprimepower(k, &p) && isprime(p) && ispal(p) &&!ispal(k); \\ Michel Marcus, Dec 10 2020


CROSSREFS

Subsequences: A000079 \ {1,2,4,8}, A000244 \ {1,3,9}, A000351 \ {1,5}, A000420 \ {1,7,343}, A001020 \ {1,11,121,1331,14641}, A096884 \ {1,101, 10201, 1030301, 104060401}.


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



