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

Author

Patrick De Geest, Aug 15 1999

Keywords

• nonn
• easy
• base
• nice

Comments

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

Links

• 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).

Crossrefs

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.

Programs

GAP

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

Magma

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

Maple

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

Mathematica

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

PARI

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

PARI

\\ 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

PARI

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

PARI

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

Python

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

Formula

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