A307717 -id:A307717 - cp's OEIS frontend

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

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

A307752 Number of n-digit palindromic pentagonal numbers.

Original entry on oeis.org

3, 1, 0, 2, 1, 1, 2, 2, 0, 4, 0, 0, 3, 1, 1, 1, 3, 2, 4, 1, 3, 1, 1, 0, 3, 3, 2, 2, 2, 0, 2, 0, 0, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Robert Price, Apr 26 2019

Number of n-digit terms in A002069.

			There are only two 4-digit pentagonal number that are palindromic, 1001 and 2882. Thus, a(4)=2.

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

Cf. A000326, A002069, A028386, A059868, A263618, A307717.

A002069 = {0, 1, 5, 22, 1001, 2882, 15251, 720027, 7081807, 7451547, 26811862, 54177145, 1050660501, 1085885801, 1528888251, 2911771192, 2376574756732, 5792526252975, 5875432345785, 10810300301801, 264571020175462, 5292834004382925, 10808388588380801, 15017579397571051, 76318361016381367, 150621384483126051, 735960334433069537, 1003806742476083001, 1087959810189597801, 2716280733370826172};
Table[Length[Select[A002069, IntegerLength[#] == n  || (n == 1 && # == 0) &]], {n, 18}] (* Robert Price, Apr 26 2019 *)

def afind(terms):
  m, n, c = 0, 1, 0
  while n <= terms:
    p = m*(3*m-1)//2
    s = str(p)
    if len(s) == n:
       if s == s[::-1]: c += 1
    else:
      print(c, end=", ")
      n, c = n+1, int(s == s[::-1])
    m += 1
afind(14) # Michael S. Branicky, Mar 01 2021

a(19)-a(22) from Michael S. Branicky, Mar 01 2021

a(23)-a(40) from Bert Dobbelaere, Apr 15 2025

A307766 Number of palindromic hexagonal numbers of length n whose index is also palindromic.

Original entry on oeis.org

3, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Robert Price, Apr 27 2019

Is there a nonzero term beyond a(7)?

			There is only one palindromic hexagonal number of length 4 whose index is also palindromic, 55->5995. Thus, a(4)=1.

Patrick De Geest, Palindromic Squares in bases 2 to 17
Eric Weisstein's World of Mathematics, Palindromic Number

Cf. A000384, A054969, A054970, A082721, A307717, A307752, A307753.

A054969 = {0, 1, 6, 66, 3003, 5995, 15051, 66066, 617716, 828828, 1269621, 1680861, 5073705, 5676765, 1264114621, 5289009825, 6172882716, 13953435931, 1313207023131, 5250178710525, 6874200024786, 61728399382716, 602224464422206, 636188414881636, 1250444114440521, 16588189498188561, 58183932923938185, 66056806460865066, 67898244444289876, 514816979979618415, 3075488771778845703, 6364000440440004636, 15199896744769899151};
A054970 = {0, 1, 2, 6, 39, 55, 87, 182, 556, 644, 797, 917, 1593, 1685, 25141, 51425, 55556, 83527, 810311, 1620213, 1853942, 5555556, 17352586, 17835196, 25004441, 91071921, 170563673, 181737182, 184252876, 507354403, 1240058219, 1783816196, 2756800387};
Table[Length[ Select[A054970[[Table[ Select[Range[18], IntegerLength[A054969[[#]]] == n || (n == 1 && A054969[[#]] == 0) &], {n, 19}][[n]]]], PalindromeQ[#] &]], {n, 19}]

A307753 Number of palindromic pentagonal numbers of length n whose index is also palindromic.

Original entry on oeis.org

3, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Robert Price, Apr 26 2019

Is there a nonzero term beyond a(5)?

			There is only one palindromic pentagonal number of length 4 whose index is also palindromic, 44->2882. Thus, a(4)=1.

Patrick De Geest, Palindromic Squares in bases 2 to 17
Eric Weisstein's World of Mathematics, Palindromic Number

Cf. A000326, A002069, A028386, A059868, A263618, A307717.

A002069 = {0, 1, 5, 22, 1001, 2882, 15251, 720027, 7081807, 7451547, 26811862, 54177145, 1050660501, 1085885801, 1528888251, 2911771192, 2376574756732, 5792526252975, 5875432345785, 10810300301801, 264571020175462, 5292834004382925, 10808388588380801, 15017579397571051, 76318361016381367, 150621384483126051, 735960334433069537, 1003806742476083001, 1087959810189597801, 2716280733370826172};
A028386 = {0, 1, 2, 4, 26, 44, 101, 693, 2173, 2229, 4228, 6010, 26466, 26906, 31926, 44059, 1258723, 1965117, 1979130, 2684561, 13280839, 59401650, 84885761, 100058581, 225563533, 316882086, 700457153, 818049201, 851649306, 1345679688};
Table[Length[Select[A028386[[Table[Select[Range[18], IntegerLength[A002069[[#]]] == n  || (n == 1 && A002069[[#]] == 0) &], {n, 18}][[n]]]], PalindromeQ[#] &]], {n, 18}]

a(19)-a(35) from Chai Wah Wu, Sep 07 2019

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

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A307752 Number of n-digit palindromic pentagonal numbers.

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A307766 Number of palindromic hexagonal numbers of length n whose index is also palindromic.

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A307753 Number of palindromic pentagonal numbers of length n whose index is also palindromic.

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Mathematica

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