A186080 -id:A186080 - cp's OEIS frontend

A056810 Numbers whose fourth power is a palindrome.

0, 1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001

Offset: 1

Robert G. Wilson v, Aug 21 2000

Suppose a number is of the form a=10...01 (see A000533) then a^2=10..020..01, so a^2 is always a palindrome. a^3=10..030..030..01, so a^3 is always a palindrome. Similarly we also have a^4=10..040..060..040..01, so a^4 is always a palindrome. However, a^5 is in general not a palindrome, for example 101^5=10510100501. - Dmitry Kamenetsky, Apr 17 2009
The sequence contains no term with digit sum 3. - Vladimir Shevelev, May 23 2011.  Proof: There are four possibilities for n:
1) 1+10^k+10^m, 00, 3) 2+10^s, s>0, 4) 3*10^t, t>=0.
In the last two cases n^4 is trivially not a palindrome.
For r>=2, in the second case we have n^4 = (1 + 2*10^r)^4 = 1 + 8*10^r + 4*10^(2*r) + 2*10^(2*r + 1) + 2*10^(3*r) + 3*10^(3*r + 1) + 6*10^(4*r) + 10^(4*r + 1)
which cannot be a palindrome.
If r=1, we have 1+8*10+...9*10^4+10^5 which also is not a palindrome.
The proof for the first case is similar. QED - Vladimir Shevelev, Oct 24 2015
Does every term have the structure 100...0001? Referring to the Simmons (1972) paper, we can also ask, if n is a number whose cube is a palindrome in base 4, must the base-4 expansion of n have the form 100...0001? - N. J. A. Sloane, Oct 22 2015

G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]
G. J. Simmons, On palindromic squares of non-palindromic numbers, J. Rec. Math., 5 (No. 1, 1972), 11-19. [Annotated scanned copy]

Cf. A186080.

palQ[n_] := Block[{}, Reverse[idn = IntegerDigits@ n] == idn]; k = 0; lst = {}; While[k < 1000000002, If[ palQ[k^4], AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Oct 23 2015 *)

def ispal(n): s = str(n); return s == s[::-1]
def afind(limit):
    for k in range(limit+1):
        if ispal(k**4): print(k, end=", ")
afind(10000001) # Michael S. Branicky, Sep 05 2021

a(11) from Robert G. Wilson v, Oct 23 2015

a(12)-a(13) from Michael S. Branicky, Sep 05 2021

A002108 4th powers written backwards.

Original entry on oeis.org

1, 61, 18, 652, 526, 6921, 1042, 6904, 1656, 1, 14641, 63702, 16582, 61483, 52605, 63556, 12538, 679401, 123031, 61, 184491, 652432, 148972, 677133, 526093, 679654, 144135, 656416, 182707, 18, 125329, 6758401, 1295811, 6336331, 5260051, 6169761

Offset: 1

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A004086, A000583, A002942, A004165.

{This sequence} Intersection A000583 = A186080 (palindromes).

FromDigits[Reverse[IntegerDigits[#]]]&/@(Range[40]^4) (* Harvey P. Dale, May 03 2012 *)

a(n) = fromdigits(Vecrev(digits(n^4))); \\ Michel Marcus, Jun 04 2019

From Michel Marcus, Jun 04 2019: (Start)
a(n) = A004086(A000583(n)).
a(n) = A002942(n^2). (End)
a(n * 10^k) = a(n) for k >= 1. - Bernard Schott, Jun 04 2019

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

A265449 Palindromes that are the sums of consecutive fourth powers.

Original entry on oeis.org

0, 1, 353, 979, 14641, 16561, 998899, 2138312, 104060401, 1004006004001, 10004000600040001, 85045192129154058, 100004000060000400001, 1000004000006000004000001, 10000004000000600000040000001

Offset: 1

Chai Wah Wu, Jan 29 2016

Subsequence of A217844 and supersequence of A186080.

			353 = 2^4 + 3^4 + 4^4
979 = 1^4 + 2^4 + 3^4 + 4^4 + 5^4
16561 = 9^4 + 10^4
998899 = 19^4 +...+ 23^4
2138312 = 10^4 +...+ 25^4
85045192129154058 = 5582^4 +...+ 5666^4

Cf. A186080, A217844.

import heapq
def ispal(n): s = str(n); return s == s[::-1]
def afind():
  print("0, ") # special case
  N, T = 4, 1  # power, min number of terms
  sigma = sum(i**N for i in range(1, T+1))
  h = [(sigma, 1, T)]
  nextcount = T + 1
  while True:
    (v, s, l) = heapq.heappop(h)
    if ispal(v): print(f"{v}, [= Sum_{{i = {s}..{l}}} i^{N}]")
    if v >= sigma:
      sigma += nextcount**N
      heapq.heappush(h, (sigma, 1, nextcount))
      nextcount += 1
    v -= s**N; s += 1; l += 1; v += l**N
    heapq.heappush(h, (v, s, l))
afind() # Michael S. Branicky, May 16 2021 after Bert Dobbelaere in A344338

a(13)-a(15) from Giovanni Resta, Aug 27 2019

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A056810 Numbers whose fourth power is a palindrome.

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A002108 4th powers written backwards.

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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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A265449 Palindromes that are the sums of consecutive fourth powers.

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