A002778 -id:A002778 - cp's OEIS frontend

A059745 Palindromic squares of sporadic type.

676, 69696, 94249, 698896, 5221225, 6948496, 522808225, 617323716, 942060249, 637832238736, 1086078706801, 1615108015161, 4051154511504, 5265533355625, 9420645460249, 123862676268321, 144678292876441, 165551171155561, 900075181570009, 4099923883299904, 94206450305460249

Offset: 1

N. J. A. Sloane, Feb 21 2001

C. Ashbacher, More on palindromic squares, J. Rec. Math. 22, no. 2 (1990), 133-135. [A scan of the first page of this article is included with the last page of the Keith (1990) scan]
J. K. R. Barnett, "Tables of Square Palindromes in Bases 2 and 10," Journal of Recreational Mathematics, 23:1, pp. 13-18, 1991.
M. Keith, "Classification and Enumeration of Palindromic Squares," Journal of Recreational Mathematics, 22:2, pp. 124-132, 1990.
R. Ondrejka, "A Palindrome (151) of Palindromic Squares," Journal of Recreational Mathematics, 20:1, pp. 68-71, 1988.

Patrick De Geest, Full List of the Sporadic Square Palindromes (SSP), part of Palindromic Squares website on WorldOfNumbers.
M. Keith, Classification and enumeration of palindromic squares, J. Rec. Math., 22 (No. 2, 1990), 124-132. [Annotated scanned copy] (Has many more terms)
William Rex Marshall, Table up to 10^40

Cf. A059744, A016106, A028818, A060088.

a(n) = A059744(n)^2. - Hugo Pfoertner, Oct 03 2023

More terms from WorldOfNumbers website, communicated by Hugo Pfoertner, Oct 03 2023

A027829 Palindromic squares with an even number of digits.

Original entry on oeis.org

698896, 637832238736, 4099923883299904, 6916103777337773016196, 40460195511188111559106404, 4872133543202112023453312784, 9658137819052882509187318569, 46501623417708833880771432610564, 1635977102407987117897042017795361, 163296619873968186681869378916692361

Offset: 1

Keith Devlin, via Boon Leong (boon_leong(AT)hotmail.com)

			836^2 = 698896, which is palindromic, so 698896 is in the sequence.
1001^2 = 1002001, which is palindromic, but it has an odd number of digits, so it's not in the sequence.

Charles Ashbacher, More on palindromic squares, J. Rec. Math. 22, no. 2 (1990), 133-135. [A scan of the first page of this article is included with the last page of the Keith (1990) scan]

M. F. Hasler, Table of n, a(n) for n = 1..14
K. S. Brown, On General Palindromic Numbers
Patrick De Geest, Sporadic palindromic squares of even length
Michael Keith, Classification and enumeration of palindromic squares, J. Rec. Math., 22 (No. 2, 1990), 124-132. [Annotated scanned copy]

Cf. A002113, A002778, A002779, A016113.

Select[Range[1000000]^2, PalindromeQ[#] && OddQ[Floor[Log[10, #]]] &] (* Alonso del Arte, Oct 11 2019 *)

is_A027829(n)={issquare(n)&&Vecrev(n=digits(n))==n&&!bittest(#n, 0)} \\ This is faster than first checking for even length if applied to numbers known to have an even number of digits, as should be the case for a systematic search. For this, one should only consider squares, i.e., rather use is_A016113.  - M. F. Hasler, Jun 08 2014

from math import isqrt
from itertools import count, islice
def A027829_gen(): # generator of terms
    return filter(lambda n: (s:=str(n))[:(t:=(len(s)+1)//2)]==s[:-t-1:-1], map(lambda n: n**2, (d for l in count(2,2) for d in range(isqrt(10**(l-1))+1,isqrt(10**l)+1))))
A027829_list = list(islice(A027829_gen(),3)) # Chai Wah Wu, Jun 23 2022

def isPalindromic(n: BigInt): Boolean = n.toString == n.toString.reverse
val squares = ((1: BigInt) to (1000000: BigInt)).map(n => n * n)
squares.filter(n => isPalindromic(n) && n.toString.length % 2 == 0) // Alonso del Arte, Oct 07 2019

a(n) = A016113(n)^2. - M. F. Hasler, Jun 08 2014

Two new terms were recently found by Bennett from UK (communication from Patrick De Geest, Dec 1999 or before)

Edited by M. F. Hasler, Jun 08 2014

A186080 Fourth powers that are palindromic in base 10.

Original entry on oeis.org

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

A244411 Nonprimes n such that the product of its divisors is a palindrome.

Original entry on oeis.org

1, 4, 22, 26, 49, 111, 121, 202, 1001, 1111, 2285, 10001, 10201, 11111, 100001, 1000001, 1001001, 1012101, 1100011, 1101011, 1109111, 1111111, 3069307, 10000001, 12028229, 12866669, 100000001, 101000101, 110000011, 110091011, 200010002, 10000000001, 10011111001

Offset: 1

Derek Orr, Jun 27 2014

Primes trivially satisfy this property and are therefore not included in the sequence.
Numbers n such that A136522(A007955(n)) = 1.
A number is in the intersection of A002778 and A001358 iff it is in this sequence.
a(31) > 2*10^8.
a(32) > 4*10^8. - Chai Wah Wu, Aug 25 2015

			The divisors of 26 are 1,2,13,26. And 1*2*13*26 = 676 is a palindrome. Thus 26 is a member of this sequence.

Giovanni Resta, Table of n, a(n) for n = 1..47 (terms < 3.5*10^11)

Cf. A007955, A136522, A028980, A002778, A001358.

rev(n)={r="";for(i=1,#digits(n),r=concat(Str(digits(n)[i]),r));return(eval(r))}
for(n=1,2*10^8,if(!isprime(n),d=divisors(n);ss=prod(j=1,#d,d[j]);if(ss==rev(ss),print1(n,", "))))

import sympy
from sympy import isprime
from sympy import divisors
def rev(n):
  r = ""
  for i in str(n):
    r = i + r
  return int(r)
def a():
  for n in range(1,10**8):
    if not isprime(n):
      p = 1
      for i in divisors(n):
        p*=i
      if rev(p)==p:
        print(n,end=', ')
a()

from sympy import divisor_count, sqrt
A244411_list = [1]
for n in range(1,10**5):
    d = divisor_count(n)
    if d > 2:
        q, r = divmod(d,2)
        s = str(n**q*(sqrt(n) if r else 1))
        if s == s[::-1]:
            A244411_list.append(n) # Chai Wah Wu, Aug 25 2015

a(31) from Chai Wah Wu, Aug 25 2015

a(32)-a(33) from Giovanni Resta, Sep 20 2019

A263607 Base 3 numbers whose square is a palindrome in base 3.

Original entry on oeis.org

0, 1, 2, 11, 101, 102, 202, 211, 1001, 1021, 2002, 10001, 10022, 11012, 12201, 20002, 100001, 100201, 200002, 201102, 1000001, 1000222, 1002201, 1011221, 1101211, 1211201, 1212022, 2000002, 10000001, 10002001, 10200102, 10201121, 11011211, 12212101, 20000002, 20011002, 100000001, 100002222, 100022001

Offset: 1

N. J. A. Sloane, Oct 22 2015

Chai Wah Wu, Table of n, a(n) for n = 1..288
G. J. Simmons, On palindromic squares of non-palindromic numbers, J. Rec. Math., 5 (No. 1, 1972), 11-19. [Annotated scanned copy]

Cf. A002778, A029984, A263608, A029985.

A263608 Palindromes which are base-3 representations of squares.

Original entry on oeis.org

0, 1, 11, 121, 10201, 11111, 112211, 122221, 1002001, 1120211, 11022011, 100020001, 101212101, 122111221, 1012112101, 1100220011, 10000200001, 10111011101, 110002200011, 111221122111, 1000002000001, 1001221221001, 1012200022101, 1101202021011, 1221221221221, 10101111110101

Offset: 1

N. J. A. Sloane, Oct 22 2015

Robert Israel, Table of n, a(n) for n = 1..143
G. J. Simmons, On palindromic squares of non-palindromic numbers, J. Rec. Math., 5 (No. 1, 1972), 11-19. [Annotated scanned copy]

Cf. A002778, A003166, A029984, A262607, A029985.

Intersection of A001738 and A118594.

rev3:= proc(n) local L,i; L:= convert(n,base,3); add(L[-i]*3^(i-1),i=1..nops(L)) end proc:
c3:= proc(n) local L,i; L:= convert(n,base,3); add(L[i]*10^(i-1),i=1..nops(L)) end proc:
R:= 0,1: count:= 2:
for d from 2 while count < 100 do
    if d::odd then
      V:= select(issqr, [seq(seq(a*3^((d+1)/2) + b*3^((d-1)/2)+rev3(a),b=0..2),a=3^((d-3)/2) .. 3^((d-1)/2)-1)])
    else
      V:= select(issqr, [seq(a*3^(d/2) + rev3(a), a=3^(d/2-1) .. 3^(d/2)-1)]);
    fi;
    count:= count+nops(V);
    R:= R, op(map(c3,V));
od:
R; # Robert Israel, May 19 2024

Name edited by Robert Israel, May 19 2024

A263617 Number of numbers with at most n digits whose square is a palindrome.

Original entry on oeis.org

4, 7, 15, 20, 31, 37, 56, 70, 95, 113, 162, 193, 264, 310, 415, 486, 640, 741, 950, 1082, 1374, 1556, 1940, 2176, 2673, 2975, 3611, 3994, 4793, 5268, 6249, 6827, 8028, 8729

Offset: 1

N. J. A. Sloane, Oct 23 2015

G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]. See p. 95.

Partial sums of A263616.

Cf. A002778, A002779, A263618, A263619.

Table[Length[Select[Range[ 0, 10^n - 1], PalindromeQ[#^2] &]], {n, 6}] (* Robert Price, Apr 26 2019 *)

a(9)-a(10) from Chai Wah Wu, Oct 25 2015
a(11)-a(12) from Michael S. Branicky, May 23 2021
a(13)-a(22) (using A002778) from Chai Wah Wu, Sep 16 2021
a(23)-a(34) from Max Alekseyev, Apr 08 2025

A263619 Number of palindromic squares with at most n digits.

Original entry on oeis.org

4, 4, 7, 7, 14, 15, 20, 20, 31, 31, 36, 37, 56, 56, 69, 70, 95, 95, 113, 113, 161, 162, 193, 193, 263, 264, 308, 310, 415, 415, 485, 486, 639, 640, 738, 741, 950, 950, 1082, 1082, 1373, 1374, 1555, 1556, 1940, 1940, 2174, 2176, 2672, 2673, 2974, 2975, 3611, 3611, 3994, 3994, 4792, 4793, 5267, 5268, 6249, 6249, 6827, 6827, 8026, 8028, 8729

Offset: 1

N. J. A. Sloane, Oct 23 2015

G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy] See page 95.

Partial sums of A263618.

Cf. A002778, A002779, A263616, A263617, A263620.

Table[Length[Select[Range[0, Floor[Sqrt[10^n]]], PalindromeQ[#^2] &]], {n, 10}] (* Robert Price, Apr 26 2019 *)

a(13)-a(19) from Chai Wah Wu, Oct 24 2015
a(20) from Robert Price, Apr 26 2019
a(21)-a(44) using A263618 from Chai Wah Wu, Jun 14 2024
a(45)-a(67) added by Max Alekseyev, Apr 08 2025

A307717 Number of palindromic squares, k^2, of length n such that k is also palindromic.

Original entry on oeis.org

4, 0, 2, 0, 5, 0, 3, 0, 8, 0, 5, 0, 13, 0, 9, 0, 22, 0, 16, 0, 37, 0, 27, 0, 60, 0, 43, 0, 93, 0, 65, 0, 138, 0, 94, 0, 197, 0, 131, 0, 272, 0, 177, 0, 365, 0, 233, 0, 478, 0, 300, 0, 613, 0, 379, 0, 772, 0, 471, 0, 957, 0, 577, 0, 1170, 0, 698, 0, 1413, 0

Offset: 1

Robert Price, Apr 23 2019

			There are only two palindromic squares of length 3 whose root is also palindromic. 11^2=121 and 22^2=484. Thus, a(3)=2.

Patrick De Geest, Palindromic Squares in bases 2 to 17
M. Kauers and C. Koutschan, Guessing with Little Data, arXiv:2202.07966 [cs.SC], 2022.
Bertrand Teguia Tabuguia and Wolfram Koepf, FPS In Action: An Easy Way To Find Explicit Formulas For Interlaced Hypergeometric Sequences, arXiv:2207.01031 [cs.SC], 2022.
Bertrand Teguia Tabuguia, Hypergeometric-Type Sequences, arXiv:2401.00256 [cs.SC], 2023.
Eric Weisstein's World of Mathematics, Palindromic Number
Index entries for linear recurrences with constant coefficients, signature (0,0,0,4,0,0,0,-6,0,0,0,4,0,0,0,-1).

Cf. A002778, A002779, A034307, A034822, A218035.

Table[Length[Select[Range[If[n == 1, 0, Ceiling[Sqrt[10^(n - 1)]]], Floor[Sqrt[10^n]]], # == IntegerReverse[#] && #^2 == IntegerReverse[#^2] &]], {n, 15}]

From Christoph Koutschan, Feb 19 2022: (Start)
a(2n-1) = A218035(n).
a(n) is given by a quasi-polynomial (for a proof, see A218035):
a(1) = 4;
a(2n) = 0;
a(4n+1) = (n^3-3*n^2+11*n+6)/3 (n > 0);
a(4n+3) = (n^3+5*n+12)/6 (n >= 0). (End)

a(16)-a(20) from Robert Price, Apr 25 2019

a(21)-a(70) from Giovanni Resta, Apr 28 2019

A056810 Numbers whose fourth power is a palindrome.

Original entry on oeis.org

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

cp's OEIS Frontend

A059745 Palindromic squares of sporadic type.

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A027829 Palindromic squares with an even number of digits.

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A186080 Fourth powers that are palindromic in base 10.

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A244411 Nonprimes n such that the product of its divisors is a palindrome.

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PARI

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A263607 Base 3 numbers whose square is a palindrome in base 3.

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A263608 Palindromes which are base-3 representations of squares.

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Maple

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A263617 Number of numbers with at most n digits whose square is a palindrome.

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Mathematica

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A263619 Number of palindromic squares with at most n digits.

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