A002781 -id:A002781 - cp's OEIS frontend

A035090 Non-palindromic squares which when written backwards remain square (and still have the same number of digits).

144, 169, 441, 961, 1089, 9801, 10404, 10609, 12544, 12769, 14884, 40401, 44521, 48841, 90601, 96721, 1004004, 1006009, 1022121, 1024144, 1026169, 1042441, 1044484, 1062961, 1212201, 1214404, 1216609, 1236544, 1238769, 1256641

Offset: 1

Patrick De Geest, Nov 15 1998

Squares with trailing zeros not included.

Sequence is infinite, since it includes, e.g., 10^(2k) + 4*10^k + 4 for all k. - Robert Israel, Sep 20 2015

Robert Israel, Table of n, a(n) for n = 1..798
Patrick De Geest, Palindromic Squares in bases 2 to 17
Index entry for sequences related to reversing digits of squares

Reversing a polytopal number gives a polytopal number:
cube to cube: A035123, A035124, A035125, A002781;
square to square: A161902, A035090, A033294, A106323, A106324, A002779;
square to triangular: A181412, A066702;
tetrahedral to tetrahedral: A006030;
triangular to square: A066703, A179889;
triangular to triangular: A066528, A069673, A003098, A066569.
Cf. A319388.

rev:= proc(n) local L,i;
L:= convert(n,base,10);
add(L[-i]*10^(i-1),i=1..nops(L))
end proc:
filter:= proc(n) local t;
  if n mod 10 = 0 then return false fi;
  t:= rev(n);
t <> n and issqr(t)
end proc:
select(filter, [seq(n^2, n=1..10^5)]); # Robert Israel, Sep 20 2015

Select[Range[1200]^2,!PalindromeQ[#]&&IntegerLength[#]==IntegerLength[ IntegerReverse[ #]] && IntegerQ[Sqrt[IntegerReverse[#]]]&] (* Harvey P. Dale, Jul 19 2023 *)

a(n) = A035123(n)^2. - R. J. Mathar, Jan 25 2017

A002780 Numbers whose cube is a palindrome.

Original entry on oeis.org

0, 1, 2, 7, 11, 101, 111, 1001, 2201, 10001, 10101, 11011, 100001, 101101, 110011, 1000001, 1001001, 1100011, 10000001, 10011001, 10100101, 11000011, 100000001, 100010001, 100101001, 101000101, 110000011, 1000000001, 1000110001

Offset: 1

N. J. A. Sloane

a(8)=2201 is the only known non-palindromic rootnumber.
There are no further non-palindromic terms (other than 2201) up to 10^11. - Matevz Markovic, Apr 04 2011. There are none up to 10^15, by direct search. - Charles R Greathouse IV, May 16 2011
There are no non-palindromic terms in the range 10^15 to 10^20 with digits from the set {0,1,2}. - Hans Havermann, May 18 2011.
From Vladimir Shevelev, May 23 2011: (Start)
Using the table by Noe-De Geest, I noticed that all numbers {a(n)=A002780(n);  11<=a(n)<=10^17+10^16+11}, except 2201, allow a partition into 3 disjoint classes of terms of the following forms: 10^k+1, 10^(2*k)+10^k+1, and (10^u+1)*(10^v+1).
Does there exist a term a(n)>10^17+10^16+11 which is in none of these classes?
If there is no such term, then we conclude that the sum of digits of a(n) does not exceed 4 (more exactly, it is i+1 where i is the number of class).
One can prove that the sequence contains no term (other than 2201) with sum of digits = 5. (End)

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).

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]

Cf. A002781 (cubes of these numbers).

isok(k) = my(d=digits(k^3)); Vecrev(d) == d; \\ Michel Marcus, Aug 02 2022

def ispal(s): return s == s[::-1]
def ok(n): return ispal(str(n**3))
print([k for k in range(10**7) if ok(k)]) # Michael S. Branicky, Aug 02 2022

More terms from Patrick De Geest

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.

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

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A035090 Non-palindromic squares which when written backwards remain square (and still have the same number of digits).

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A002780 Numbers whose cube is 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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A348429 Perfect powers m^k, m >= 1, k >= 2 such that m and m^k both are palindromes.

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Examples

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Mathematica

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A087988 Palindromic numbers whose squares and cubes are equally palindromic.

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Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

PARI

Extensions