OFFSET
1,2
COMMENTS
From Chai Wah Wu, Jul 20 2020: (Start)
Theorem: a(n) = 2*(10^(n-5)-1)/3 for n > 5.
Proof: clearly 2*(10^m-1)/3 are terms of this sequence. Next we show that all terms > 10 are of the form 2*(10^m-1)/3. Let k > 10 be a term of the sequence. Let x be the first digit (and thus also the last digit) of k. If x <> 6 then it is easy to show that the first and last digit of 2k-1 will not be the same. Thus x = 6. Let the digits of k be written as 6y****y6. Similarly if y <> 6 then again the second digit of 2k-1 will not be the same as the second to last digit of 2k-1. Continuing in this manner, this shows that k written in decimal is a sequence of 6's.
(End)
LINKS
FORMULA
From Chai Wah Wu, Jul 20 2020: (Start)
a(n) = 2*(10^(n-5)-1)/3 for n > 5.
a(n) = 11*a(n-1) - 10*a(n-2) for n > 7.
G.f.: x*(50*x^6 - 9*x^5 - 9*x^4 - 9*x^3 - 9*x^2 - 9*x + 1)/((x - 1)*(10*x - 1)).
(End)
EXAMPLE
66 is a member as 2*66 - 1 = 131 is also a palindrome.
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Amarnath Murthy, Apr 30 2002
EXTENSIONS
More terms from Hans Havermann, Jul 06 2002
STATUS
approved