A348429 - OEIS

%I #41 Oct 21 2021 10:40:58

%S 1,4,8,9,121,343,484,1331,10201,12321,14641,40804,44944,1002001,

%T 1030301,1234321,1367631,4008004,100020001,102030201,104060401,

%U 121242121,123454321,125686521,400080004,404090404,1003003001,10000200001,10221412201,12102420121,12345654321,40000800004

%N Perfect powers m^k, m >= 1, k >= 2 such that m and m^k both are palindromes.

%C Complement of A348319 relative to the positive perfect powers A001597.

%C This sequence is infinite since each square (10^m+1)^2 is a term for m >= 0 and A033934 is a subsequence.

%C Observation: terms always contain an odd number of digits.

%C For k = 2, subsequence of palindromes whose square root is a palindrome is A057136 (see A057135).

%C For k = 3, except for 2201^3 = 10662526601, all known palindromic cubes have a palindromic rootnumber (see A002780 and A002781).

%C For k = 4, all known integers whose fourth power is a palindrome are also palindromes (see A056810 and subsequence A186080).

%C For k >= 5, G. J. Simmons conjectured there are no palindromes of the form m^k for k >= 5 and m > 1 (see Simmons link p. 98); according to this conjecture, all the terms are of the form (palindrome)^k, with 2 <= k <= 4.

%H Michael S. Branicky, <a href="/A348429/b348429.txt">Table of n, a(n) for n = 1..1024</a> (all terms with <= 40 digits)

%H Michael S. Branicky, <a href="/A348429/a348429.txt">Python program</a>

%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 First few terms are equal to 1, 2^2, 2^3, 3^2, 11^2, 7^3, 22^2, 11^3, 101^2, 111^2, 11^4 = 121^2, 202^2, 212^2, 1001^2, 101^3, 1111^2, 111^3.

%t Block[{n = 10^6, nn, s}, s = Select[Range[2, n], PalindromeQ]; nn = Max[s]^2; {1}~Join~Union@ Reap[Table[Do[If[PalindromeQ[m^k], Sow[m^k]], {k, 2, Log[m, nn]}], {m, s}]][[-1, -1]]] (* _Michael De Vlieger_, Oct 18 2021 *)

%o (Python) # see link for faster version

%o def ispal(n): s = str(n); return s == s[::-1]

%o def aupto(limit):

%o     aset, m, mm = {1}, 2, 4

%o     while mm <= limit:

%o         if 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**11)) # _Michael S. Branicky_, Oct 18 2021

%o (PARI) ispal(x) = my(d=digits(x)); d == Vecrev(d); \\ A002113

%o isok(m) = if (m==1, return (1)); my(p); ispal(m) && ispower(m, , &p) && ispal(p); \\ _Michel Marcus_, Oct 19 2021

%o (PARI) ispal(x) = my(d=digits(x)); d == Vecrev(d); \\ A002113

%o lista(nn) = {my(list = List(1)); 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. A001597, A002113, A033934, A056810, A186080, A348319, A348320.

%K nonn,base

%O 1,2

%A _Bernard Schott_, Oct 18 2021