A082275 Palindromes k such that k + 11 is also a palindrome.

11, 22, 33, 44, 55, 66, 77, 88, 191, 292, 393, 494, 595, 696, 797, 898, 1991, 2992, 3993, 4994, 5995, 6996, 7997, 8998, 19991, 29992, 39993, 49994, 59995, 69996, 79997, 89998, 199991, 299992, 399993, 499994, 599995, 699996, 799997, 899998, 1999991, 2999992, 3999993, 4999994, 5999995, 6999996, 7999997, 8999998

Offset

1,1

Comments

A pattern is visible.

From David A. Corneth, Jul 28 2025: (Start)

Terms can be constructed as follows: The first and last digit are 1 through 8 (and equal). Any digits in between are 9.

Proof: By adding 11 to a palindrome its last digit changes and its 10s digit changes. Therefore the first digit must change. This can only happen if there is carry to the through the first digit and only happens when the digits (if any) between the first and last digit are all 9. If the last digit of k is 9 then the last digit of k+11 is 0 but no palindrome exists ending in 0. There are no other palindromes with this property since we must have c*10^m - 10 <= k < c*10^m where 2 <= c <= 9. (End)

Links

Table of n, a(n) for n=1..48.

Example

595 + 11 = 606 is also a palindrome.

Mathematica

Select[Range[100000], AllTrue[#+{0, 11}, PalindromeQ]&]
Flatten[Table[FromDigits[Join[{n}, PadRight[{}, k, 9], {n}]], {n, 8}, {k, 0, 5}]]//Sort (* This program builds on the perceived pattern. It will generate an additional 8 terms for each increase in {k, 0, 5}. - Harvey P. Dale, Jan 29 2023 *) (* For a proof of this pattern, see David A. Corneth, Jul 28 2025 *)

Prog

(PARI) a(n) = {if(n <= 8, return(11*n), return(10*a(n-8) + 90 - 9*((n-1)%8+1)))} \\ David A. Corneth, Aug 04 2025

Crossrefs

Sequence in context: A044836 A338754 A033008 * A059544 A070022 A004940

Adjacent sequences: A082272 A082273 A082274 * A082276 A082277 A082278

Keyword

base,easy,nonn

Author

Amarnath Murthy, Apr 13 2003

Extensions

More terms from David A. Corneth, Aug 04 2025

Status

approved