A050683 -id:A050683 - cp's OEIS frontend

A050250 Number of nonzero palindromes less than 10^n.

9, 18, 108, 198, 1098, 1998, 10998, 19998, 109998, 199998, 1099998, 1999998, 10999998, 19999998, 109999998, 199999998, 1099999998, 1999999998, 10999999998, 19999999998, 109999999998, 199999999998, 1099999999998, 1999999999998, 10999999999998

Offset: 1

Eric W. Weisstein, Dec 11 1999

G. C. Greubel, Table of n, a(n) for n = 1..1000
Dr. Math, Palindromic Numbers.
Dr. Math, Palindromic Numbers.
G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Palindromic Number.
Index entries for linear recurrences with constant coefficients, signature (1,10,-10).

Cf. A002113, A002778, A050683.

A050250List := proc(len);  local s, egf, ser; s:= 11/(2*sqrt(10));
egf := -2*exp(x) + (1-s)*exp(-sqrt(10)*x) + (1+s)*exp(sqrt(10)*x);
ser := series(egf, x, len+2): seq(simplify(n!*coeff(ser,x,n)), n = 1..len) end:
A050250List(25); # Peter Luschny, Jun 11 2022 after Stefano Spezia

LinearRecurrence[{1,10,-10},{9,18,108},30] (* Harvey P. Dale, Jan 29 2012 *)
CoefficientList[Series[2Cosh[Sqrt[10]x]-2(Cosh[x]+Sinh[x])+11Sinh[Sqrt[10]x]/Sqrt[10],{x,0,25}],x]Table[n!,{n,0,25}] (* Stefano Spezia, Jun 11 2022 *)

a(n)=10^(n\2)*(13-9*(-1)^n)/2-2 \\ Charles R Greathouse IV, Jun 25 2017

def a(n):
  m = 10 ** (n >> 1)
  if n & 1 == 0:
    return (m - 1) << 1
  else:
    return (11 * m) - 2 # Darío Clavijo, Oct 16 2023

a(2*k) = 2*10^k - 2, a(2*k + 1) = 11*10^k - 2. - Sascha Kurz, Apr 14 2002
From Jonathan Vos Post, Jun 18 2008: (Start)
a(n) = Sum_{i=1..n} A050683(i).
a(n) = Sum_{i=1..n} 9*10^floor((i-1)/2).
a(n) = 9*Sum_{i=1..n} 10^floor((i-1)/2). (End)
From Bruno Berselli, Feb 15 2011: (Start)
G.f.: 9*x*(1+x)/((1-x)*(1-10*x^2)).
a(n) = (1/2)*10^((2*n + (-1)^n - 1)/4)*(13 - 9*(-1)^n) - 2. (End)
a(1)=9, a(2)=18, a(3)=108; for n>3, a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3). - Harvey P. Dale, Jan 29 2012
a(n) = 10*a(n-2) + 18. - R. J. Mathar, Nov 07 2015
E.g.f.: 2*cosh(sqrt(10)*x) - 2*(cosh(x) + sinh(x)) + 11*sinh(sqrt(10)*x)/sqrt(10). - Stefano Spezia, Jun 11 2022

More terms from Patrick De Geest, Dec 15 1999

a(24)-a(25) from Jonathan Vos Post, Jun 18 2008

A070252 Number of n-digit palindromes.

Original entry on oeis.org

10, 9, 90, 90, 900, 900, 9000, 9000, 90000, 90000, 900000, 900000, 9000000, 9000000, 90000000, 90000000, 900000000, 900000000, 9000000000, 9000000000, 90000000000, 90000000000, 900000000000, 900000000000, 9000000000000

Offset: 1

Amarnath Murthy, May 06 2002

Index entries for linear recurrences with constant coefficients, signature (0,10).

A variant of A050683, which is the principal entry for this sequence. Cf. A016115.

Partial sums give A070199.

LinearRecurrence[{0,10},{10,9,90},25] (* Stefano Spezia, Jun 11 2022 *)

def A070252(n): return 10 if n==1 else 9*10**(n-1>>1) # Chai Wah Wu, Jul 30 2025

From Colin Barker, Aug 19 2013: (Start)
a(n) = 10*a(n-2) for n>3.
G.f.: x*(10*x^2-9*x-10) / (10*x^2-1). (End)
E.g.f.: x + 9*(cosh(sqrt(10)*x) - 1 + sqrt(10)*sinh(sqrt(10)*x))/10. - Stefano Spezia, Jun 11 2022

A117855 Number of nonzero palindromes of length n (in base 3).

Original entry on oeis.org

2, 2, 6, 6, 18, 18, 54, 54, 162, 162, 486, 486, 1458, 1458, 4374, 4374, 13122, 13122, 39366, 39366, 118098, 118098, 354294, 354294, 1062882, 1062882, 3188646, 3188646, 9565938, 9565938, 28697814, 28697814, 86093442, 86093442, 258280326, 258280326, 774840978

Offset: 1

Martin Renner, May 02 2006

See A225367 for the sequence that counts all base 3 palindromes, including 0 (and thus also the number of n-digit terms in A006072). -- A nonzero palindrome of length L=2k-1 or of length L=2k is determined by the first k digits, which then determine the last k digits by symmetry. Since the first digit cannot be 0, there are 2*3^(k-1) possibilities. - M. F. Hasler, May 05 2013
From Gus Wiseman, Oct 18 2023: (Start)
Also the number of subsets of {1..n} with n not the sum of two subset elements (possibly the same). For example, the a(0) = 1 through a(4) = 6 subsets are:
  {}  {}   {}   {}     {}
      {1}  {2}  {1}    {1}
                {2}    {3}
                {3}    {4}
                {1,3}  {1,4}
                {2,3}  {3,4}
For subsets with no subset summing to n we have A365377.
Requiring pairs to be distinct gives A068911, complement A365544.
The complement is counted by A366131.
(End) [Edited by Peter Munn, Nov 22 2023]

			The a(3)=6 palindromes of length 3 are: 101, 111, 121, 202, 212, and 222. - _M. F. Hasler_, May 05 2013

Index entries for linear recurrences with constant coefficients, signature (0,3).

Cf. A050683 and A070252.
Bisections are both A025192.
A093971/A088809/A364534 count certain types of sum-full subsets.
A108411 lists powers of 3 repeated, complement A167936.
Cf. A004526, A004737, A008967, A038754, A046663, A167762, A365376, A366130.

With[{c=NestList[3#&,2,20]},Riffle[c,c]] (* Harvey P. Dale, Mar 25 2018 *)
Table[Length[Select[Subsets[Range[n]],!MemberQ[Total/@Tuples[#,2],n]&]],{n,0,10}] (* Gus Wiseman, Oct 18 2023 *)

A117855(n)=2*3^((n-1)\2) \\ - M. F. Hasler, May 05 2013

def A117855(n): return 3**(n-1>>1)<<1 # Chai Wah Wu, Oct 28 2024

a(n) = 2*3^floor((n-1)/2).
a(n) = 2*A108411(n-1).
From Colin Barker, Feb 15 2013: (Start)
a(n) = 3*a(n-2).
G.f.: -2*x*(x+1)/(3*x^2-1). (End)

More terms from Colin Barker, Feb 15 2013

A225367 Number of palindromes of length n in base 3 (A118594).

Original entry on oeis.org

3, 2, 6, 6, 18, 18, 54, 54, 162, 162, 486, 486, 1458, 1458, 4374, 4374, 13122, 13122, 39366, 39366, 118098, 118098, 354294, 354294, 1062882, 1062882, 3188646, 3188646, 9565938, 9565938, 28697814, 28697814, 86093442, 86093442, 258280326, 258280326, 774840978

Offset: 1

M. F. Hasler, May 05 2013

Also: The number of n-digit terms in A006072. See there for further comments.
A palindrome of length L=2k-1 or of length L=2k is determined by the first k digits, which then determine the last k digits by symmetry. Since the first digit cannot be 0 (unless L=1), there are 2*3^(k-1) possibilities for L>1.
Except for the initial term, this is identical to A117855, which counts only nonzero palindromes.

			The a(1)=3 palindromes of length 1 are: 0, 1 and 2.
The a(2)=2 palindromes of length 2 are: 11 and 22.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3).

Cf. A050683 and A070252 for base 10 analogs.

[n eq 1 select 3 else 2*3^Floor((n-1)/2): n in [1..40]]; // Bruno Berselli, May 06 2013

I:=[3,2,6]; [n le 3 select I[n] else 3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, May 31 2017

Join[{3}, LinearRecurrence[{0, 3}, {2, 6}, 40]] (* Vincenzo Librandi, May 31 2017 *)

PARI
```
A225367(n)=2*3^((n-1)\2)+!n
```

def A225367(n): return 3 if n==1 else 3**(n-1>>1)<<1 # Chai Wah Wu, Jul 30 2025

a(n) = 2*3^floor((n-1)/2) + [n=1].
a(n) = 3*a(n-2) for n>3.
G.f.: x*(3*x^2-2*x-3)/(3*x^2-1).
a(n) = (6-(1+(-1)^n)*(3-sqrt(3)))*sqrt(3)^(n-3) for n>1, a(1)=3. [Bruno Berselli, May 06 2013]

A050720 Number of nonzero palindromes of length n containing the digit '0'.

Original entry on oeis.org

0, 0, 9, 9, 171, 171, 2439, 2439, 30951, 30951, 368559, 368559, 4217031, 4217031, 46953279, 46953279, 512579511, 512579511, 5513215599, 5513215599, 58618940391, 58618940391, 617570463519, 617570463519, 6458134171671

Offset: 1

Patrick De Geest, Aug 15 1999

			For length 3 we find nine numbers: 101, 202, ... 909, so a(3) = 9.

Cf. A050683, A050686.

nzp[n_]:=Module[{k=Ceiling[n/2]},9*10^(k-1)-9^k]; Array[nzp,30] (* Harvey P. Dale, Jun 01 2019 *)

G.f.: 9*x^3*(x+1) / ((3*x-1)*(3*x+1)*(10*x^2-1)). - Colin Barker, Feb 15 2013

For n > 1, a(n) = 9*10^(k-1) - 9^k, where k = ceiling(n/2). - Jon E. Schoenfield, Sep 14 2013

More terms from Michael Lugo (mlugo(AT)thelabelguy.com), Dec 22 1999

Corrected by T. D. Noe, Nov 08 2006

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

A050686 Number of palindromes of length n and containing the digit 1 (or any other fixed nonzero digit).

Original entry on oeis.org

1, 1, 18, 18, 252, 252, 3168, 3168, 37512, 37512, 427608, 427608, 4748472, 4748472, 51736248, 51736248, 555626232, 555626232, 5900636088, 5900636088, 62105724792, 62105724792, 648951523128, 648951523128, 6740563708152

Offset: 1

Patrick De Geest, Aug 15 1999

			For length 3 we find 18 numbers: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 212, 313, 414, 515, 616, 717, 818, 919.

Cf. A050683, A050720.

Empirical g.f.: -x*(x-1)*(x+1)^2 / ((3*x-1)*(3*x+1)*(10*x^2-1)). - Colin Barker, Feb 15 2013
From Sela Fried, Dec 10 2024: (Start)
The conjectured g.f is correct.
a(n) = 9*10^(n/2 - 1) - 8*9^(n/2 - 1) if n is even
a(n) = 9*10^((n - 1)/2) - 8*9^((n - 1)/2) if n is odd. (End)

More terms from Michael Lugo (mlugo(AT)thelabelguy.com), Dec 22 1999

A117856 Number of palindromes of length n (in base 4).

Original entry on oeis.org

3, 3, 12, 12, 48, 48, 192, 192, 768, 768, 3072, 3072, 12288, 12288, 49152, 49152, 196608, 196608, 786432, 786432, 3145728, 3145728, 12582912, 12582912, 50331648, 50331648, 201326592, 201326592, 805306368, 805306368, 3221225472, 3221225472, 12884901888

Offset: 1

Martin Renner, May 02 2006

Index entries for linear recurrences with constant coefficients, signature (0,4).

Cf. A050683.

NestList[4#&,{3,3},20]//Flatten (* Harvey P. Dale, Dec 19 2016 *)

a(n) = 3*4^floor((n-1)/2).
From Colin Barker, Feb 15 2013: (Start)
a(n) = 4*a(n-2).
G.f.: -3*x*(x+1) / ((2*x-1)*(2*x+1)). (End)

More terms from Colin Barker, Feb 15 2013

A117857 Number of palindromes of length n (in base 5).

Original entry on oeis.org

4, 4, 20, 20, 100, 100, 500, 500, 2500, 2500, 12500, 12500, 62500, 62500, 312500, 312500, 1562500, 1562500, 7812500, 7812500, 39062500, 39062500, 195312500, 195312500, 976562500, 976562500, 4882812500, 4882812500, 24414062500, 24414062500, 122070312500

Offset: 1

Martin Renner, May 02 2006

Index entries for linear recurrences with constant coefficients, signature (0,5).

Cf. A050683.

A117857:=n->4*5^floor((n-1)/2): seq(A117857(n), n=1..40); # Wesley Ivan Hurt, Apr 18 2017

LinearRecurrence[{0,5},{4,4},40] (* or *) With[{c=NestList[ 5#&,4,20]},Riffle[ c,c]] (* Harvey P. Dale, Apr 18 2019 *)

a(n) = 4*5^floor((n-1)/2).

a(n) = 5*a(n-2). G.f.: -4*x*(x+1)/(5*x^2-1). [Colin Barker, Feb 15 2013]

More terms from Colin Barker, Feb 15 2013

A117861 Number of palindromes of length n (in base 9).

Original entry on oeis.org

8, 8, 72, 72, 648, 648, 5832, 5832, 52488, 52488, 472392, 472392, 4251528, 4251528, 38263752, 38263752, 344373768, 344373768, 3099363912, 3099363912, 27894275208, 27894275208, 251048476872, 251048476872, 2259436291848, 2259436291848, 20334926626632

Offset: 1

Martin Renner, May 02 2006

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,9).

Cf. A050683.

Cf. A055275.

Table[8 9^Floor[(n-1)/2],{n,50}] (* Harvey P. Dale, Oct 21 2011 *)

a(n) = 8*9^floor((n-1)/2).

G.f.: 8*x*(1+x)/(1-9*x^2). a(n) = 8*3^(n-2)*(2-(-1)^n). - Bruno Berselli, Oct 24 2011

More terms from Harvey P. Dale, Oct 21 2011

cp's OEIS Frontend

A050250 Number of nonzero palindromes less than 10^n.

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A070252 Number of n-digit palindromes.

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A117855 Number of nonzero palindromes of length n (in base 3).

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A225367 Number of palindromes of length n in base 3 (A118594).

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A050720 Number of nonzero palindromes of length n containing the digit '0'.

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

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A050686 Number of palindromes of length n and containing the digit 1 (or any other fixed nonzero digit).

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