OFFSET
1,2
COMMENTS
Suggested by A214927.
Conjecture: a(n) = A222811(n) - 9*10^floor((n-1)/2). - Lars Blomberg, Jul 03 2014
Proof of this conjecture: A number N with n digits which equals its own reversal has the first and identical last digit from {1, 2, ..., 9}. If n is even the other n-2 digits come in n/2 - 1 pairs of equal numbers from {0, 1, ..., 9}. If n is odd with n >= 3 then the other n-2 numbers come in (n-3)/2 pairs of equal numbers from {0, 1, ..., 9} and an additional middle digit also from {0, 1, ..., 9}. Therefore there are 9*10^(n/2-1) such numbers N for even n, and 9*10^((n-1)/2) for odd n, fitting 9*10^floor((n-1)/2). - Wolfdieter Lang, Jul 13 2014
EXAMPLE
Some of the smallest solutions are:
[10, 20, 30, 40, 50, 60, 70, 80, 90] (so a(2) = 9),
[100, 110, 200, 220, 300, 330, 400, 440, 500, 510, 540, 550, 600, 660, 700, 770, 800, 810, 880, 900, 990] (so a(3) = 21),
[1000, 1010, 1100, 1110, 1210, 1310, 1410, ...].
PROG
(PARI) a(n) = sum(i=10^(n-1), 10^n-1, (irev=eval(concat(Vecrev(Str(i))))) && irev!=i && !(i % irev)); \\ Michel Marcus, Jul 03 2014
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
N. J. A. Sloane, Mar 10 2013
EXTENSIONS
a(7)-a(12) from Lars Blomberg, Jul 03 2014
a(13) from Giovanni Resta, Aug 15 2019
STATUS
approved