A225367 -id:A225367 - cp's OEIS frontend

A050683 Number of nonzero palindromes of length n.

9, 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

Patrick De Geest, Aug 15 1999

In general the number of base k palindromes with n digits is (k-1)*k^floor((n-1)/2). (See A117855 or A225367 for an explanation.) - Henry Bottomley, Aug 14 2000

This sequence does not count 0 as palindrome with 1 digit, see A070252 = (10,9,90,90,...) for the variant which does. - M. F. Hasler, Nov 16 2008

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dr. Math, More info 1.
Dr. Math, More info 2.
Index entries for linear recurrences with constant coefficients, signature (0,10).

Cf. A002113, A050250, A050251, A070252, A070199.
Cf. A016116 for numbers of binary palindromes, A016115 for prime palindromes.
Cf. A117855 for the base 3 version, and A225367 for a variant.

a:=[9,9];; for n in [3..30] do a[n]:=10*a[n-2]; od; a; # Muniru A Asiru, Oct 07 2018

[9*10^Floor((n-1)/2): n in [1..30]]; // Vincenzo Librandi, Aug 16 2011

seq(9*10^floor((n-1)/2),n=1..30); # Muniru A Asiru, Oct 07 2018

With[{c=9*10^Range[0,20]},Riffle[c,c]] (* or *) LinearRecurrence[{0,10},{9,9},40] (* Harvey P. Dale, Dec 15 2013 *)

A050683(n)=9*10^((n-1)\2) \\ M. F. Hasler, Nov 16 2008

\\ using M. F. Hasler's is_A002113(n) from A002113
is_A002113(n)={Vecrev(n=digits(n))==n}
for(n=1,8,j=0;for(k=10^(n-1),10^n-1,if(is_A002113(k),j++));print1(j,", ")) \\ Hugo Pfoertner, Oct 03 2018

is_palindrome(x)={my(d=digits(x));for(k=1,#d\2,if(d[k]!=d[#d+1-k],return(0)));return(1)}
for(n=1,8,j=0;for(k=10^(n-1),10^n-1,if(is_palindrome(k),j++));print1(j,", ")) \\ Hugo Pfoertner, Oct 02 2018

a(n) = if(n<3, 9, 10*a(n-2)); \\ Altug Alkan, Oct 03 2018

def A050683(n): return 9*10**(n-1>>1) # Chai Wah Wu, Jul 30 2025

a(n) = 9*10^floor((n-1)/2).
From Colin Barker, Apr 06 2012: (Start)
a(n) = 10*a(n-2).
G.f.: 9*x*(1+x)/(1-10*x^2). (End)
E.g.f.: 9*(cosh(sqrt(10)*x) + sqrt(10)*sinh(sqrt(10)*x) - 1)/10. - Stefano Spezia, Jun 11 2022

A118594 Palindromes in base 3 (written in base 3).

Original entry on oeis.org

0, 1, 2, 11, 22, 101, 111, 121, 202, 212, 222, 1001, 1111, 1221, 2002, 2112, 2222, 10001, 10101, 10201, 11011, 11111, 11211, 12021, 12121, 12221, 20002, 20102, 20202, 21012, 21112, 21212, 22022, 22122, 22222, 100001, 101101, 102201, 110011, 111111, 112211, 120021

Offset: 1

Martin Renner, May 08 2006

The number of n-digit terms is given by A225367. - M. F. Hasler, May 05 2013 [Moved here on May 08 2013]
Digit-wise application of A000578 (and also superposition of a(n) with its horizontal OR vertical reflection) yields A006072. - M. F. Hasler, May 08 2013
Equivalently, palindromes k (written in base 10) such that 4*k is a palindrome. - Bruno Berselli, Sep 12 2018

Michael S. Branicky, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Palindromic Number
Eric Weisstein's World of Mathematics, Ternary

Cf. A007089, A014190, A057148, A118595, A118596, A118597, A118598, A118599, A118600, A002113.

(* get NextPalindrome from A029965 *) Select[NestList[NextPalindrome, 0, 1110], Max@IntegerDigits@# < 3 &] (* Robert G. Wilson v, May 09 2006 *)
Select[FromDigits/@Tuples[{0,1,2},8],IntegerDigits[#]==Reverse[ IntegerDigits[ #]]&] (* Harvey P. Dale, Apr 20 2015 *)

{for(l=1,5,u=vector((l+1)\2,i,10^(i-1)+(2*i-11&&i==1,2]), print1(v*u",")))} \\ The n-th term could be produced by using (partial sums of) A225367 to skip all shorter terms, and then skipping the adequate number of vectors v until n is reached.  - M. F. Hasler, May 08 2013

from itertools import count, islice, product
def agen(): # generator of terms
    yield from [0, 1, 2]
    for d in count(2):
        for start in "12":
            for rest in product("012", repeat=d//2-1):
                left = start + "".join(rest)
                for mid in [[""], ["0", "1", "2"]][d%2]:
                    yield int(left + mid + left[::-1])
print(list(islice(agen(), 42))) # Michael S. Branicky, Mar 29 2022

from sympy import integer_log
from gmpy2 import digits
def A118594(n):
    if n == 1: return 0
    y = 3*(x:=3**integer_log(n>>1,3)[0])
    return int((s:=digits(n-x,3))+s[-2::-1] if nChai Wah Wu, Jun 14 2024

[int(n.str(base=3)) for n in (0..757) if Word(n.digits(3)).is_palindrome()] # Peter Luschny, Sep 13 2018

More terms from Robert G. Wilson v, May 09 2006

a(40) and beyond from Michael S. Branicky, Mar 29 2022

A006072 Numbers with mirror symmetry about middle.

Original entry on oeis.org

0, 1, 8, 11, 88, 101, 111, 181, 808, 818, 888, 1001, 1111, 1881, 8008, 8118, 8888, 10001, 10101, 10801, 11011, 11111, 11811, 18081, 18181, 18881, 80008, 80108, 80808, 81018, 81118, 81818, 88088, 88188, 88888, 100001, 101101, 108801, 110011, 111111, 118811, 180081

Offset: 1

N. J. A. Sloane

Apparently this sequence and A111065 have the same parity. - Jeremy Gardiner, Oct 15 2005
Obviously, terms of this sequence also have the same parity (and also the same digital sum mod 6) as those of A118594, see below. - M. F. Hasler, May 08 2013
The number of n-digit terms is given by A225367 -- which counts palindromes in base 3, A118594. The terms here are the base 3 palindromes considered there, with 2 replaced by 8 (which means this sequence A006072 arises from A118594 not only by taking the 3rd power of each digit, but also by superposing the number with its horizontal or vertical reflection, somehow remarkably given the symmetry of numbers considered here). - M. F. Hasler, May 05 2013 [Part of the comment moved from A225367 to here on May 08 2013]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1450 from Vincenzo Librandi)
Eric Weisstein's World of Mathematics, Tetradic Number

Subsequence of A000787. Cf. A045574.

NextPalindrome[n_] := Block[{l = Floor[Log[10, n] + 1], idn = IntegerDigits[n]}, If[ Union[idn] == {9}, Return[n + 2], If[l < 2, Return[n + 1], If[ FromDigits[ Reverse[ Take[idn, Ceiling[l/2]]]] > FromDigits[ Take[idn, -Ceiling[l/2]]], FromDigits[ Join[ Take[idn, Ceiling[l/2]], Reverse[ Take[idn, Floor[l/2]]]]], idfhn = FromDigits[ Take[idn, Ceiling[l/2]]] + 1; idp = FromDigits[ Join[ IntegerDigits[ idfhn], Drop[ Reverse[ IntegerDigits[ idfhn]], Mod[l, 2]]]]]]]]; np = 0; t = {0}; Do[np = NextPalindrome[np]; If[Union[Join[{0, 1, 8}, IntegerDigits[np]]] == {0, 1, 8}, AppendTo[t, np]], {n, 1150}]; t (* Robert G. Wilson v *)
TetrNumsUpTo10powerK[k_]:= Select[FromDigits/@ Tuples[{0, 1, 8}, k],IntegerDigits[#] == Reverse[IntegerDigits[#]] &]; TetrNumsUpTo10powerK[7] (* Mikk Heidemaa, May 21 2017 *)

{for(l=1,5,u=vector((l+1)\2,i,10^(i-1)+(2*i-11&&i==1,2]), print1((v+v\2*6)*u", ")))} \\ The n-th term could be produced by using (partial sums of) A225367 to skip all shorter terms, and then skipping the adequate number of vectors v until n is reached. - M. F. Hasler, May 05 2013

from itertools import count, islice, product
def agen():
    yield from [0, 1, 8]
    for d in count(2):
        for start in "18":
            for rest in product("018", repeat=d//2-1):
                left = start + "".join(rest)
                for mid in [[""], ["0", "1", "8"]][d%2]:
                    yield int(left + mid + left[::-1])
print(list(islice(agen(), 42))) # Michael S. Branicky, Mar 29 2022

a(n) = digit-wise application of A000578 to A118594(n). - M. F. Hasler, May 08 2013

More terms from Robert G. Wilson v, Nov 16 2005

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

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A050683 Number of nonzero palindromes of length n.

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A118594 Palindromes in base 3 (written in base 3).

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A006072 Numbers with mirror symmetry about middle.

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

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