A056810 -id:A056810 - cp's OEIS frontend

A186080 Fourth powers that are palindromic in base 10.

0, 1, 14641, 104060401, 1004006004001, 10004000600040001, 100004000060000400001, 1000004000006000004000001, 10000004000000600000040000001, 100000004000000060000000400000001, 1000000004000000006000000004000000001, 10000000004000000000600000000040000000001, 100000000004000000000060000000000400000000001

Offset: 1

Matevz Markovic, Feb 11 2011

See A056810 (the main entry for this problem) for further information, including the search limit. - N. J. A. Sloane, Mar 07 2011
Conjecture: If k^4 is a palindrome > 0, then k begins and ends with digit 1, all other digits of k being 0.
The number of zeros in 1x1, where the x are zeros, is the same as (the number of zeros)/4 in (1x1)^4 = 1x4x6x4x1.

P. De Geest, Palindromic cubes (The Simmons test is mentioned here) [broken link]
G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]

Cf. A002113, A168576, A056810, A002778, A002779.

[ p: n in [0..10000000] | s eq Reverse(s) where s is Intseq(p) where p is n^4 ];

Do[If[Module[{idn = IntegerDigits[n^4, 10]}, idn == Reverse[idn]], Print[n^4]], {n, 100000001}]

a(n) = A056810(n)^4.

a(11)-a(13) using extensions of A056810 from Hugo Pfoertner, Oct 22 2021

A263613 Palindromic numbers in base 4 that are cubes.

Original entry on oeis.org

0, 1, 1331, 1030301, 1003003001, 1000300030001, 1000030000300001, 1000003000003000001, 1000000300000030000001, 1000000030000000300000001, 1000000003000000003000000001, 1000000000300000000030000000001, 1000000000030000000000300000000001, 1000000000003000000000003000000000001

Offset: 1

N. J. A. Sloane, Oct 22 2015

There is an obvious pattern here, but it is not known how long it will continue.

The cube roots (written in base 4) are 0, 1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001 (coinciding with A056810 to this point).

Chai Wah Wu, Table of n, a(n) for n = 1..18
G. J. Simmons, On palindromic squares of non-palindromic numbers, J. Rec. Math., 5 (No. 1, 1972), 11-19. [Annotated scanned copy]

Cf. A056810, A168575.

Conjectures from Chai Wah Wu, Dec 25 2023: (Start)
a(n) = 1111*a(n-1) - 112110*a(n-2) + 1111000*a(n-3) - 1000000*a(n-4) for n > 6.
G.f.: x^2*(-7000000*x^4 + 6446000*x^3 - 336330*x^2 + 220*x + 1)/((x - 1)*(10*x - 1)*(100*x - 1)*(1000*x - 1)). (End)

a(11) from Chai Wah Wu, Oct 23 2015

a(12)-a(14) from Chai Wah Wu, Sep 26 2017

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

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A186080 Fourth powers that are palindromic in base 10.

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A263613 Palindromic numbers in base 4 that are cubes.

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

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