A069882 Numbers n such that n and 2n-1 are both palindromes.

1, 2, 3, 4, 5, 6, 66, 666, 6666, 66666, 666666, 6666666, 66666666, 666666666, 6666666666, 66666666666, 666666666666, 6666666666666, 66666666666666, 666666666666666, 6666666666666666, 66666666666666666, 666666666666666666

Offset: 1

Author

Amarnath Murthy, Apr 30 2002

Keywords

• nonn
• base
• easy

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)

Examples

			66 is a member as 2*66 - 1 = 131 is also a palindrome.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (11,-10).

Crossrefs

Cf. A002280, A069881, A185127, A259050.

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)

Extensions

More terms from Hans Havermann, Jul 06 2002