A057135 -id:A057135 - cp's OEIS frontend

A057136 Palindromes whose square root is a palindrome.

0, 1, 4, 9, 121, 484, 10201, 12321, 14641, 40804, 44944, 1002001, 1234321, 4008004, 100020001, 102030201, 104060401, 121242121, 123454321, 125686521, 400080004, 404090404, 10000200001, 10221412201, 12102420121, 12345654321, 40000800004

Offset: 1

Henry Bottomley, Aug 12 2000

Always contain an odd number of digits.

			a(8) = 14641 since 14641 = 121^2 and 121 is also a palindrome

Robert Israel, Table of n, a(n) for n = 1..10000 (n=1..412 from N. J. A. Sloane based on R. J. Mathar's b-file for A057135)

Cf. A000290, A002113, A002779, A057135 (the square roots).

dmax:= 7: # to get all terms with up to dmax digits
Res:= 0,1,2^2,3^2,11^2,22^2:
Po:= [[0],[1],[2],[3]]: Pe:= [[0,0],[1,1],[2,2]]:
for d from 1 to dmax do
  if d::odd then
    Po:= select(t -> add(s^2,s=t) < 10, [seq(seq([i,op(t),i], t=Po),i=0..2)]);
    Res:= Res, op(map(proc(p) if p[1] <> 0 then add(p[i]*10^(i-1),i=1..nops(p))^2 fi end proc, Po))
  else
    Pe:= select(t -> add(s^2,s=t) < 10, [seq(seq([i,op(t),i], t=Pe),i=0..2)]);
    Res:= Res, op(map(proc(p) if p[1] <> 0 then add(p[i]*10^(i-1),i=1..nops(p))^2 fi end proc, Pe))
  fi;
od:
Res; # Robert Israel, Jun 21 2017

Select[Range[0, 10^6], PalindromeQ[#] && PalindromeQ[#^2] &]^2 (* Robert Price, Apr 26 2019 *)

a(n) = A057135(n)^2

A128921 Palindromes m such that reverse of m^2 is also a square.

Original entry on oeis.org

0, 1, 2, 3, 11, 22, 33, 99, 101, 111, 121, 202, 212, 1001, 1111, 2002, 10001, 10101, 10201, 11011, 11111, 11211, 20002, 20102, 100001, 101101, 110011, 111111, 200002, 1000001, 1001001, 1002001, 1010101, 1011101, 1012101, 1100011, 1101011

Offset: 1

Zak Seidov, Mar 02 2005, definition corrected Sep 16 2007

Most terms have a palindromic square; for the rare exceptions see A133901. - Klaus Brockhaus and Zak Seidov, Sep 29 2007

			33 and 99 are terms because 33^2=1089 => 9801=99^2 and 99^2=9801 => 1089=33^2.

Klaus Brockhaus, Table of n, a(n) for n = 1..360

Cf. A002113, A057135, A057136, A133901.

A128921=Select[Range[0, 100000], IntegerQ[Sqrt[FromDigits[Reverse[IntegerDigits[ #^2 ]]]]]&&FromDigits[Reverse[IntegerDigits[ # ]]]==#&]

from sympy.ntheory.primetest import is_square
from itertools import chain, count, islice
def A128921_gen(): # generator of terms
    return filter(lambda n:is_square(int(str(n**2)[::-1])),chain((0,),chain.from_iterable(chain((int((s:=str(d))+s[-2::-1]) for d in range(10**l,10**(l+1))), (int((s:=str(d))+s[::-1]) for d in range(10**l,10**(l+1)))) for l in count(0))))
A128921_list = list(islice(A128921_gen(),20)) # Chai Wah Wu, Jun 23 2022

More terms from Klaus Brockhaus, Sep 23 2007

A218035 Number of n-digit palindromes with squares that are also palindromes.

Original entry on oeis.org

4, 2, 5, 3, 8, 5, 13, 9, 22, 16, 37, 27, 60, 43, 93, 65, 138, 94, 197, 131, 272, 177, 365, 233, 478, 300, 613, 379, 772, 471, 957, 577, 1170, 698, 1413, 835, 1688, 989, 1997, 1161, 2342, 1352, 2725, 1563, 3148, 1795, 3613, 2049, 4122, 2326, 4677, 2627, 5280, 2953

Offset: 1

David Zvirbulis, Oct 19 2012

Number of n-digit terms in A057135.
From Chai Wah Wu, Apr 03 2021: (Start)
The conjectures in the formula section are true.
Theorem: a(n) = (n^3-6*n^2+32*n+48)/48 if n is even.
a(n) = (n^3-9*n^2+59*n-3)/24 if n > 1 is odd.
Proof: For n < 9, this is true by inspection.
The set of palindromes whose square are palindromic are the numbers whose squares of the digits sums to less than 10 (see A057135). For n >= 9, the nonzero digits are from one of the following 12 sets:
(1, 1, 1, 1, 1, 1, 1, 1), (1, 2, 2), (1, 1, 1, 1), (1, 1, 2), (1, 1, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1, 1, 1, 1), (2, 2), (1, 1, 1), (1, 1, 1, 1, 1), (1, 1), (1, 1, 1, 1, 2), (1, 1, 1, 1, 1, 1)
For odd n >= 9:
(1, 1, 1, 1, 1, 1, 1, 1): number of palindromes with 8 1's and n-8 0's is C((n-3)/2,3) = (n-3)(n-5)(n-7)/48. This is because all palindromes must start and end with a nonzero digit and the middle digit is necessarily 0, so only the (n-3) remaining digits are permuted with 6 1's and (n-9) 0's. By symmetry of the palindromes the number of combinations is C((n-3)/2,3).
(1, 2, 2): 1 palindrome of the form 20..010..2.
(1, 1, 1, 1): number of palindromes with 4 1's and n-4 0's is C((n-3)/2,1) = (n-3)/2.
(1, 1, 2): 1 palindrome of the form 10..020..1.
(1, 1, 1): 1 palindrome of the form 10..010..1.
(1, 1, 1, 1, 1): number of palindromes with 5 1's and n-5 0's is C((n-3)/2,1) = (n-3)/2.
(1, 1, 1, 1, 2): C((n-3)/2,1) = (n-3)/2.
(1, 1, 1, 1, 1, 1, 1): C((n-3)/2,2).
(1, 1, 1, 1, 1, 1, 1, 1, 1): C((n-3)/2,3).
(2,2): 1 palindrome of the form 200...002.
(1,1): 1 palindrome of the form 100...001.
(1,1,1,1,1,1): number of palindromes with 6 1's and n-6 0's is C((n-3)/2,2) = (n-3)(n-5)/8.
Thus A218035(n) = 2*(n-3)(n-5)(n-7)/48 +2*(n-3)(n-5)/8 +3*(n-3)/2 + 5 = (n^3-9n^2+59n-3)/24.
For even n >= 9:
(1, 2, 2), (1, 1, 2), (1, 1, 1), (1, 1, 1, 1, 1), (1, 1, 1, 1, 2), (1, 1, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1, 1, 1, 1): no palindromes are possible.
(1, 1, 1, 1, 1, 1, 1, 1): number of palindromes 8 1's and n-8 0's = C((n-2)/2,3) = (n-2)(n-4)(n-6)/48.
(1, 1, 1, 1): number of palindromes with 4 1's and n-4 0's is C((n-2)/2,1) = (n-2)/2.
(2,2): 1 palindrome of the form 200...002.
(1,1): 1 palindrome of the form 100...001.
(1,1,1,1,1,1): number of palindromes with 6 1's and n-6 0's is C((n-2)/2,2) = (n-2)(n-4)/8.
Thus A218035(n) = (n-2)(n-4)(n-6)/48 + (n-2)(n-4)/8 + (n-2)/2 +2 = (n^3-6n^2+32n+48)/48. QED
(End)

			For n=4, the solutions are:
1001, 1001^2 = 1002001,
1111, 1111^2 = 1234321,
2002, 2002^2 = 4008004.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Manuel Kauers and Christoph Koutschan, Guessing with Little Data, arXiv:2202.07966 [cs.SC], 2022.
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A050683, A070252, A307717.

from itertools import product
def ispal(n): s = str(n); return s == s[::-1]
def pals(n):
  midrange = [[""], [str(i) for i in range(10)]]
  for p in product("0123456789", repeat=n//2):
    left = "".join(p)
    if len(left) and left[0] == '0': continue
    for middle in midrange[n%2]: yield left+middle+left[::-1]
def a(n): return sum(ispal(int(strpal)**2) for strpal in pals(n))
print([a(n) for n in range(1, 13)]) # Michael S. Branicky, Apr 02 2021

def A218035(n): return 4 if n == 1 else (n**3-9*n**2+59*n-3)//24 if n % 2 else (n**3-6*n**2+32*n+48)//48 # Chai Wah Wu, Apr 03 2021

Conjecture: a(n) = n^3/48 - n^2/8 + 2n/3 + 1 if n even, see A011826.
Conjecture: a(n) = n^3/24 - 3n^2/8 + 59n/24 - 1/8 if n odd, n > 1.
From Chai Wah Wu, Apr 03 2021: (Start)
a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n > 9.
G.f.: x*(2*x^8 - x^7 - 5*x^6 + 5*x^5 + 12*x^4 - 5*x^3 - 11*x^2 + 2*x + 4)/((x - 1)^4*(x + 1)^4). (End)

a(19)-a(20) from Michael S. Branicky, Apr 02 2021

More terms from Chai Wah Wu, Apr 03 2021

A133901 Numbers in A128921 whose square is not a palindrome.

Original entry on oeis.org

33, 99, 1119111, 10110901101, 11110901111, 101109901101, 1011109011101, 1110109010111, 10111099011101, 100110090011001, 100111090111001, 101011090110101, 111001090100111

Offset: 1

Klaus Brockhaus and Zak Seidov, Sep 29 2007

Also terms of A128921 that do not consist completely of digits 0, 1, 2, however excluding A128921(4) = 3.

Conjecture: Sequence is infinite.

			A128921(41) = 1119111, 1119111^2 = 1252409430321 is not a palindrome, but its reverse is also a square: 1230349042521 = 1109211^2.

Cf. A128921 (palindromes p such that reverse of p^2 is also a square), A002113 (palindromes in base 10), A057135 (palindromes whose square is a palindrome), A057136 (palindromes whose square root is a palindrome).

A225603 Palindromic primes whose square is also a palindrome.

Original entry on oeis.org

2, 3, 11, 101, 100111001, 110111011, 111010111, 1100011100011, 1100101010011, 1101010101011, 100110101011001, 101000010000101, 101011000110101, 101110000011101, 10000010101000001, 10011010001011001, 10100110001100101, 10110010001001101, 10111000000011101

Offset: 1

Jayanta Basu, May 11 2013

Subsets of A002385, A057135 and A065378.

Palindromes in A161721. Conjecture: a(n) for n >=3 consists only of the digits 0,1. - Chai Wah Wu, Jan 06 2015

			101 is a member since it is a palindromic prime such that 101^2=10201 is a palindrome.

Chai Wah Wu, Table of n, a(n) for n = 1..27

Cf. A002385, A057135, A065378.

palQ[n_]:=FromDigits[Reverse[IntegerDigits[n]]]==n; t={}; Do[If[palQ[p=Prime[n]] && palQ[p^2],AppendTo[t,p]],{n,10^7}]; t

from _future_ import division
from sympy import isprime
def paloddgenrange(t,l,b=10): # generator of odd-length palindromes in base b of 2*t <=length <= 2*l
    if t == 0:
        yield 0
    else:
        for x in range(t+1,l+1):
            n = b**(x-1)
            n2 = n*b
            for y in range(n,n2):
                k, m = y//b, 0
                while k >= b:
                    k, r = divmod(k,b)
                    m = b*m + r
                yield y*n + b*m + k
A225603_list = [2,3,11]
for i in paloddgenrange(1,10):
    s = str(i*i)
    if s == s[::-1] and isprime(i):
        A225603_list.append(i) # Chai Wah Wu, Jan 06 2015

a(15)-a(19) from Giovanni Resta, May 11 2013

A241096 Palindromes in base 16 whose squares are also palindromes.

Original entry on oeis.org

1, 2, 3, 11, 22, 101, 111, 121, 131, 202, 212, 222, 1001, 1111, 1221, 2002, 2112, 10001, 10101, 10201, 10301, 11011, 11111, 11211, 11311, 12021, 12121, 12221, 20002, 20102, 20202, 21012, 21112, 21212, 100001, 101101, 102201, 110011, 111111, 112211, 120021

Offset: 1

J. Lowell, Apr 26 2014

			131 is a term of this sequence because (unlike in base 10, where squaring 131 carries a 1 into the thousands place so that 131^2 is the non-palindromic number 17161) in base 16, 131^2 is 16B61.
141 is not a term because, even in base 16, a 1 is carried into the next place, so the result (19281) is not palindromic.

Lars Blomberg, Table of n, a(n) for n = 1..10000

Cf. A057135.

L=[]
for x in [1..100000]:
    M=x.digits(base=16)
    N=M[::-1]
    if N == M:
        d=x^2
        D=d.digits(base=16)
        E=D[::-1]
        if D == E:
            MM=(str(x) for x in M)
            L.append(Integer(''.join(MM)))
L # Tom Edgar, Apr 29 2014

a(35)-a(41) from Lars Blomberg, Oct 23 2014

A343098 Number of palindromes < 10^n whose squares are also palindromes.

Original entry on oeis.org

1, 4, 6, 11, 14, 22, 27, 40, 49, 71, 87, 124, 151, 211, 254, 347, 412, 550, 644, 841, 972, 1244, 1421, 1786, 2019, 2497, 2797, 3410, 3789, 4561, 5032, 5989, 6566, 7736, 8434, 9847, 10682, 12370, 13359, 15356, 16517, 18859, 20211, 22936, 24499, 27647, 29442, 33055

Offset: 0

Chai Wah Wu, Apr 04 2021

Partial sum of A218035. Number of terms in A057135 < 10^n.

			a(2) = 6 since the only palindromes < 100 whose square are palindromes are 0,1,2,3,11,22.

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).

Cf. A000290, A002113, A002779, A057136, A057135, A218035.

A343098_list = [1, 4, 6, 11, 14, 22, 27, 40, 49, 71]
for i in range(10**3): A343098_list.append(A343098_list[-1] + 4*A343098_list[-2] - 4*A343098_list[-3] - 6*A343098_list[-4] + 6*A343098_list[-5] + 4*A343098_list[-6] - 4*A343098_list[-7] - A343098_list[-8] + A343098_list[-9])

a(n) = #{i:A057135(i)<10^n}.
For n > 0, a(n) = Sum_{i=1..n} A218035(i).
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) - 6*a(n-4) + 6*a(n-5) + 4*a(n-6) - 4*a(n-7) - a(n-8) + a(n-9) for n > 9.
G.f.: (-x^9 + x^7 - x^6 - 6*x^5 - x^4 + 7*x^3 + 2*x^2 - 3*x - 1)/((x - 1)^5*(x + 1)^4).
a(n) = 1491 + 904*n + 510*n^2 - 52*n^3 + 6*n^4 + (-1)^n * (45 - 296*n + 42*n^2 - 4*n^3) for n>0. - Greg Dresden, Jun 20 2021

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

cp's OEIS Frontend

A057136 Palindromes whose square root is a palindrome.

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A128921 Palindromes m such that reverse of m^2 is also a square.

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A218035 Number of n-digit palindromes with squares that are also palindromes.

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A133901 Numbers in A128921 whose square is not a palindrome.

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A225603 Palindromic primes whose square is also a palindrome.

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A241096 Palindromes in base 16 whose squares are also palindromes.

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Sage

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A343098 Number of palindromes < 10^n whose squares are also palindromes.

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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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