A348320 -id:A348320 - cp's OEIS frontend

A348319 Perfect powers m^k, k >= 2 that are palindromes while m is not a palindrome.

676, 69696, 94249, 698896, 5221225, 6948496, 522808225, 617323716, 942060249, 10662526601, 637832238736, 1086078706801, 1230127210321, 1615108015161, 4051154511504, 5265533355625, 9420645460249, 123862676268321, 144678292876441, 165551171155561, 900075181570009

Offset: 1

Bernard Schott, Oct 12 2021

Seems to be the "converse" of A348320.
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).
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).
Also, all the squares of terms in A251673 are terms of this sequence.
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.

			676 = 26^2, 10662526601 = 2201^3, 12120030703002121 = 110091011^2 are terms.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Revised Edition), Penguin Books, 1997, entry 10662526601, page 188.

Michael Keith, Classification and enumeration of palindromic squares, J. Rec. Math., Vol. 22, No. 2 (1990), pp. 124-132. [Annotated scanned copy]. See foot of page 130.
Gustavus J. Simmons, Palindromic Powers, J. Rec. Math., Vol. 3, No. 2 (1970), pp. 93-98 [Annotated scanned copy]

Cf. A002113, A002780, A028818, A060087, A251673, A348320.
Cf. A059745 (a subsequence).
Subsequence of A001597 and of A075786.

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 *)

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

def ispal(n): s = str(n); return s == s[::-1]
def aupto(limit):
    aset, m, mm = set(), 10, 100
    while mm <= limit:
        if not ispal(m):
            mk = mm
            while mk <= limit:
                if ispal(mk): aset.add(mk)
                mk *= m
        mm += 2*m + 1
        m += 1
    return sorted(aset)
print(aupto(10**13)) # Michael S. Branicky, Oct 12 2021

a(18)-a(21) from Amiram Eldar, Oct 12 2021

A348429 Perfect powers m^k, m >= 1, k >= 2 such that m and m^k both are palindromes.

Original entry on oeis.org

1, 4, 8, 9, 121, 343, 484, 1331, 10201, 12321, 14641, 40804, 44944, 1002001, 1030301, 1234321, 1367631, 4008004, 100020001, 102030201, 104060401, 121242121, 123454321, 125686521, 400080004, 404090404, 1003003001, 10000200001, 10221412201, 12102420121, 12345654321, 40000800004

Offset: 1

Bernard Schott, Oct 18 2021

Complement of A348319 relative to the positive perfect powers A001597.
This sequence is infinite since each square (10^m+1)^2 is a term for m >= 0 and A033934 is a subsequence.
Observation: terms always contain an odd number of digits.
For k = 2, subsequence of palindromes whose square root is a palindrome is A057136 (see A057135).
For k = 3, except for 2201^3 = 10662526601, all known palindromic cubes have a palindromic rootnumber (see A002780 and A002781).
For k = 4, all known integers whose fourth power is a palindrome are also palindromes (see A056810 and subsequence A186080).
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.

			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.

Michael S. Branicky, Table of n, a(n) for n = 1..1024 (all terms with <= 40 digits)
Michael S. Branicky, Python program
Gustavus J. Simmons, Palindromic Powers, J. Rec. Math., Vol. 3, No. 2 (1970), pp. 93-98 [Annotated scanned copy].

Cf. A001597, A002113, A033934, A056810, A186080, A348319, A348320.

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 *)

ispal(x) = my(d=digits(x)); d == Vecrev(d); \\ A002113
isok(m) = if (m==1, return (1)); my(p); ispal(m) && ispower(m, , &p) && ispal(p); \\ Michel Marcus, Oct 19 2021

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

# see link for faster version
def ispal(n): s = str(n); return s == s[::-1]
def aupto(limit):
    aset, m, mm = {1}, 2, 4
    while mm <= limit:
        if ispal(m):
            mk = mm
            while mk <= limit:
                if ispal(mk): aset.add(mk)
                mk *= m
        mm += 2*m + 1
        m += 1
    return sorted(aset)
print(aupto(10**11)) # Michael S. Branicky, Oct 18 2021

cp's OEIS Frontend

A348319 Perfect powers m^k, k >= 2 that are palindromes while m is not a palindrome.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Python

Extensions

A348429 Perfect powers m^k, m >= 1, k >= 2 such that m and m^k both are palindromes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python