A002780 -id:A002780 - cp's OEIS frontend

A002781 Palindromic cubes.

0, 1, 8, 343, 1331, 1030301, 1367631, 1003003001, 10662526601, 1000300030001, 1030607060301, 1334996994331, 1000030000300001, 1033394994933301, 1331399339931331, 1000003000003000001, 1003006007006003001, 1331039930399301331

Offset: 1

N. J. A. Sloane

a(9) = 1066252601 = 2201^3 is the unique known palindromic cube that has a non-palindromic rootnumber (see comments in A002780 and Penguin reference). - Bernard Schott, Oct 21 2021

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Revised Edition), Penguin Books, 1997, entry 10662526601, page 188.

T. D. Noe, Table of n, a(n) for n = 1..89 (from De Geest)
Patrick De Geest, Palindromic Cubes
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]
G. J. Simmons, Palindrome cubes: Problem B-183, Fibonacci Quart. 8 (1970), no. 5, p. 551.

Cf. A002780.

Intersection of A000578 and A002113.

Select[Range[0,12*10^5]^3,PalindromeQ[#]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 02 2017 *)

ispal(x) = my(d=digits(x)); d == Vecrev(d); \\ A002113
lista(nn) = my(list = List(), c); for (n=0, sqrtnint(nn, 3), if (ispal(c=n^3), listput(list, c));); Vec(list); \\ Michel Marcus, Oct 21 2021

a(n) = A002780(n)^3.

Thanks to Pierre Genix (Pierre.Genix(AT)wanadoo.fr) and Harvey P. Dale who pointed out that there were errors in earlier versions of this sequence.

A069748 Numbers k such that k and k^3 are both palindromes.

Original entry on oeis.org

0, 1, 2, 7, 11, 101, 111, 1001, 10001, 10101, 11011, 100001, 101101, 110011, 1000001, 1001001, 1100011, 10000001, 10011001, 10100101, 11000011, 100000001, 100010001, 100101001, 101000101, 110000011, 1000000001, 1000110001, 1010000101, 1100000011, 10000000001

Offset: 1

Joseph L. Pe, Apr 22 2002

For an arithmetical function f, call the pairs (x,y) such that y = f(x) and x, y are palindromes the "palinpairs" of f. {a(n)} is then the sequence of abscissae of palinpairs of f(n) = n^3.
Perhaps this sequence is the same as A002780, except for 2201. - Dmitry Kamenetsky, Apr 16 2009
For n >= 5, there are no terms with digit sum 5. Conjecture: all terms belong to one of 3 disjoint classes of the following forms: 10^k+1, 10^(2*t)+10^t+1, t > 0, and (10^u+1)*(10^v+1), u,v > 0, with digit sums 2, 3 and 4 correspondingly. - Vladimir Shevelev, May 31 2011

Michael S. Branicky, Table of n, a(n) for n = 1..117
Vladimir Shevelev, Re: numbers whose cube is a palindrome, seqfan list, May 25 2011

Intersection of A002113 and A002780.

isPalin[n_] := (n == FromDigits[Reverse[IntegerDigits[n]]]); Do[m = n^3; If[isPalin[n] && isPalin[m], Print[{n, m}]], {n, 1, 10^6}]

ispal(n) = my(d=digits(n)); d == Vecrev(d);
isok(n) = ispal(n) && ispal(n^3); \\ Michel Marcus, Dec 16 2018

a(29) and beyond from Michael S. Branicky, Aug 06 2022

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

Original entry on oeis.org

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

A135066 Primes p such that p^3 is a palindrome.

Original entry on oeis.org

2, 7, 11, 101

Offset: 1

Alexander Adamchuk, Nov 16 2007

Note that all first 4 listed terms are the palindromes. Corresponding palindromic cubes a(n)^3 are listed in A135067 = {8, 343, 1331, 1030301, ...}. PrimePi[ a(n) ] = {1, 4, 5, 26, ...}.

No further terms less than 1.29 * 10^10. - Michael S. Branicky, Feb 07 2021

			a(3) = 11 because 11^3 = 1331 is a palindrome.

Patrick De Geest, Palindromic Cubes

Cf. A002780 (cube is a palindrome), A069748 (n and n^3 are both palindromes), A002781 (palindromic cubes), A135067 (palindromic cubes of primes).

Do[ p = Prime[n]; f = p^3; If[ f == FromDigits[ Reverse[ IntegerDigits[ f ] ] ], Print[ {n, p, f} ]], {n, 1, 200000} ]

from sympy import nextprime
def ispal(n): s = str(n); return s == s[::-1]
p = 2
while True:
  if ispal(p**3): print(p)
  p = nextprime(p) # Michael S. Branicky, Feb 07 2021

a(n) = A135067(n)^(1/3).

A135067 Palindromic cubes p^3, where p is a prime.

Original entry on oeis.org

8, 343, 1331, 1030301

Offset: 1

Alexander Adamchuk, Nov 16 2007

Corresponding primes p such that a(n) = p^3 are listed in A135066 = {2, 7, 11, 101, ...} = Primes p such that p^3 is a palindrome. PrimePi[ a(n)^(1/3) ] = {1, 4, 5, 26, ...}.

No further terms up to the 100,000th prime. - Harvey P. Dale, Jan 26 2021

			a(3) = 1331 because 11^3 = 1331 is a palindrome and 11 is a prime.

P. De Geest, Palindromic Cubes

Cf. A002780 = Cube is a palindrome. Cf. A069748 = Numbers n such that n and n^3 are both palindromes. Cf. A002781 = Palindromic cubes. Cf. A135066 = Primes p such that p^3 is a palindrome.

Do[ p = Prime[n]; f = p^3; If[ f == FromDigits[ Reverse[ IntegerDigits[ f ] ] ], Print[ {n, p, f} ]], {n, 1, 200000} ]
Select[Prime[Range[200]]^3,PalindromeQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 26 2021 *)

a(n) = A135066(n)^3.

A190839 a(n) is the greatest prime divisor of 10^(2*n)+10^n+1.

Original entry on oeis.org

37, 37, 333667, 9901, 2906161, 333667, 10838689, 99990001, 440334654777631, 2906161, 1344628210313298373, 999999000001, 900900900900990990990991, 10838689, 4185502830133110721, 9999999900000001, 13168164561429877, 440334654777631, 3931123022305129377976519, 39526741

Offset: 1

Vladimir Shevelev, May 21 2011

(10^(2*n)+10^n+1)^3 is palindrome.

Amiram Eldar, Table of n, a(n) for n = 1..158
factordb, 10^(2*n)+10^n+1.

Cf. A002780, A006530, A066138.

f[n_]:=Max[Transpose[FactorInteger[10^(2n)+10^n+1]][[1]]]; Array[f,20] (* Harvey P. Dale, Jun 10 2011 *)

u(n)=10^(2*n)+10^n+1; for(n=1,30,f=factor(u(n));print1(f[matsize(f)[1],1],", ")) \\ Joerg Arndt, May 27 2011

a(n) = A006530(A066138(n)). - R. J. Mathar, Oct 30 2015

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

A087988 Palindromic numbers whose squares and cubes are equally palindromic.

Original entry on oeis.org

0, 1, 2, 11, 101, 111, 1001, 10001, 10101, 11011, 100001, 101101, 110011, 1000001, 1001001, 1100011, 10000001, 10011001, 10100101, 11000011, 100000001, 100010001, 100101001, 101000101, 110000011, 1000000001, 1000110001

Offset: 1

Labos Elemer, Oct 01 2003

Numbers n such that n, n^2 and n^3 are all palindromes.

Essentially A002780 with two terms removed, 7 and 2201.

			11^2=121, 11^3=1331.

Cf. A002779, A002781.

Intersection of A002113, A002778 and A002780.

rev:=proc(a) local aa,ct: aa:=convert(a,base,10): ct:=nops(aa): add(10^(ct-j)*aa[j],j=1..ct) end: p:=proc(n) if rev(n)=n and rev(n^2)=n^2 and rev(n^3)=n^3 then n else fi end: seq(p(n),n=0..12*10^5); # Emeric Deutsch, May 01 2005

ispal(n) = my(d = digits(n)); Vecrev(d) == d;
isok(n) = ispal(n) && ispal(n^2) && ispal(n^3); \\ Michel Marcus, Oct 25 2015

More terms from Ray Chandler, Oct 05 2003

Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar

A191355 Indices of terms in A069748 with two decimal digits 1 and all others 0.

Original entry on oeis.org

5, 6, 8, 9, 12, 15, 18, 22, 27, 31, 37, 43, 49, 56, 64, 71, 80, 89, 98, 108

Offset: 1

Vladimir Shevelev, May 31 2011

Cf. A002780, A069748, A105637.

Conjectural formula: a(n) = 5 + Sum_{i=1..n-1} A105637(i).

Equivalent conjecture: a(n) = -(-1)^n/8 + A049347(n)/3 + 115/24 + n/6 + n^2/4. - R. J. Mathar, Jul 11 2011

a(18)-a(20), using A002780 b-file, added by Michel Marcus, Apr 17 2023

A191356 Interpret the terms of A069748 having only decimal digits 0 and 1 as binary numbers and then convert those numbers to decimal.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 17, 21, 27, 33, 45, 51, 65, 73, 99, 129, 153, 165, 195, 257, 273, 297, 325, 367, 513, 561

Offset: 1

Vladimir Shevelev, May 31 2011

May be the same sequence obtained from reading the terms of A002780 having only digits 0,1 in base 2 and converting to decimal.

			A069748(10)=10101 is the eighth number of A069748 having only digits 0 and 1. The binary number 10101 equals 21 in decimal. Thus a(8)=21.

Cf. A002780.

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A002781 Palindromic cubes.

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A069748 Numbers k such that k and k^3 are both palindromes.

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

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A135066 Primes p such that p^3 is a palindrome.

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A135067 Palindromic cubes p^3, where p is a prime.

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A190839 a(n) is the greatest prime divisor of 10^(2*n)+10^n+1.

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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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