%I #57 May 21 2022 14:49:22
%S 676,69696,94249,698896,5221225,6948496,522808225,617323716,942060249,
%T 10662526601,637832238736,1086078706801,1230127210321,1615108015161,
%U 4051154511504,5265533355625,9420645460249,123862676268321,144678292876441,165551171155561,900075181570009
%N Perfect powers m^k, k >= 2 that are palindromes while m is not a palindrome.
%C Seems to be the "converse" of A348320.
%C The first nine terms are the first nine palindromic squares of sporadic type (A059745). Then, a(10) = 10662526601 = 2201^3 is the only known palindromic cube whose root is not palindromic (see comments in A002780 and Penguin reference).
%C The first square that is not in A059745 is a(13) = 1230127210321 = 1109111^2 = A060087(1)^2 since it is a palindromic square that is not of sporadic type, but with an asymmetric root. Indeed, all the squares of terms in A060087 are terms of this sequence (see Keith link).
%C Also, all the squares of terms in A251673 are terms of this sequence.
%C G. J. Simmons conjectured there are no palindromes of form n^k for k >= 5 (and n > 1) (see Simmons link p. 98), according to this conjecture, we have 2 <= k <= 4.
%D David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Revised Edition), Penguin Books, 1997, entry 10662526601, page 188.
%H Michael Keith, <a href="/A002778/a002778_1.pdf">Classification and enumeration of palindromic squares</a>, J. Rec. Math., Vol. 22, No. 2 (1990), pp. 124-132. [Annotated scanned copy]. See foot of page 130.
%H Gustavus J. Simmons, <a href="/A002778/a002778_2.pdf">Palindromic Powers</a>, J. Rec. Math., Vol. 3, No. 2 (1970), pp. 93-98 [Annotated scanned copy]
%e 676 = 26^2, 10662526601 = 2201^3, 12120030703002121 = 110091011^2 are terms.
%t seq[max_] := Module[{m = Floor@Sqrt[max], s = {}, n, p}, Do[If[PalindromeQ[k], Continue[]]; n = Floor@Log[k, max]; Do[If[PalindromeQ[(p = k^j)], AppendTo[s, p]], {j, 2, n}], {k, 1, m}]; Union[s]]; seq[10^10] (* _Amiram Eldar_, Oct 12 2021 *)
%o (Python)
%o def ispal(n): s = str(n); return s == s[::-1]
%o def aupto(limit):
%o aset, m, mm = set(), 10, 100
%o while mm <= limit:
%o if not ispal(m):
%o mk = mm
%o while mk <= limit:
%o if ispal(mk): aset.add(mk)
%o mk *= m
%o mm += 2*m + 1
%o m += 1
%o return sorted(aset)
%o print(aupto(10**13)) # _Michael S. Branicky_, Oct 12 2021
%o (PARI) ispal(x) = my(d=digits(x)); d == Vecrev(d); \\ A002113
%o lista(nn) = {my(list = List()); for (k=2, sqrtint(nn), if (!ispal(k), my(q = k^2); until (q > nn, if (ispal(q), listput(list, q)); q *= k;););); vecsort(list,,8);} \\ _Michel Marcus_, Oct 20 2021
%Y Cf. A002113, A002780, A028818, A060087, A251673, A348320.
%Y Cf. A059745 (a subsequence).
%Y Subsequence of A001597 and of A075786.
%K nonn,base
%O 1,1
%A _Bernard Schott_, Oct 12 2021
%E a(18)-a(21) from _Amiram Eldar_, Oct 12 2021