OFFSET
1,1
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).
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
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 *)
PROG
(PARI) A050683(n)=9*10^((n-1)\2) \\ M. F. Hasler, Nov 16 2008
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
(Magma) [9*10^Floor((n-1)/2): n in [1..30]]; // Vincenzo Librandi, Aug 16 2011
(GAP) a:=[9, 9];; for n in [3..30] do a[n]:=10*a[n-2]; od; a; # Muniru A Asiru, Oct 07 2018
CROSSREFS
KEYWORD
nonn,easy,base,nice
AUTHOR
Patrick De Geest, Aug 15 1999
STATUS
approved