# [seqfan] More palindromes by concatenation

Eric Angelini Eric.Angelini at kntv.be
Sun Nov 24 18:56:28 CET 2013

```Hello SeqFans,
Start S with 1 and always extend S
with the smallest integer a(n) producing
a palindrome of concatenated digits
involving at least the first digit of a(n)
and the last digit of a(n-1); I guess S starts:

S=1,10,11,12,2,13,3,14,4,15,5,16,6,17,7,18,8,19,9,29,
20,21,22,23,24,25,26,27,28,38,30,31,32,33,34,35,36,37,39,
49,40,41,42,43,44,45,46,47,48,58,50,51,52,53,54,55,56,57,59,
69,60,61,62,63,64,65,66,67,68,78,70,71,72,73,74,75,76,77,78,79,
89,80,81,82,83,84,85,86,87,88,98,90,91,92,93,94,95,96,97,99,
119,100,101,102,120,103,130,104,140,105,106,160,107,170,...

Example:
S=1,10,... as 10 is the smallest integer not already in S producing
a palindrome: [11];
S=1,10,11,... as 11 produces [101];
S=1,10,11,12,... as 12 prod. [111];
S=1,10,11,12,2,... as 2 prod. [22];
S=1,10,11,12,2,13,... as 13 --> [1221], etc.

Remember: the palindrome must contain at least
the leftmost digit of the new term and the rightmost
digit of the previous one.

Best,
É.

Catapulté de mon aPhone
```