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

1,2

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

Table of n, a(n) for n=1..23.

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

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

Cf. A002280, A069881, A185127, A259050.

Sequence in context: A024643 A166099 A004857 * A329580 A037331 A037438

Adjacent sequences: A069879 A069880 A069881 * A069883 A069884 A069885

Keyword

nonn,base,easy

Author

Amarnath Murthy, Apr 30 2002

Extensions

More terms from Hans Havermann, Jul 06 2002

Status

approved