[seqfan] A fractal seq with palindrome pairs

[Eric Angelini]

Hello SeqFans,

Jps = 0,10,0,1,11,10,2,12,0,3,23,1,4,14,11,5,15,10,6,16,2,
7,17,12,8,18,0,9,19,3,10,1,23,11,211,1,12,21,4,13,1,14,
14,41,11,15,1,5,16,1,15,17,1,10,18,1,6,19,1,16,20,102,2,
21,12,7,22,2,17,23,32,12,24,2,8,25,2,18,26,2,0,27,2,9,
28,2,19,29,2,3,30,3,10,31,3,1,32,23,23,33,3,11,34,3,211,
35,3,1,36,3,12,37,3,21,38,3,4,39,93,13,40,4,1,42,24,14,
43,34,14,44,4,41,41,4,11,45,4,15,46,4,1,47,4,5,48,4,16,...

Erase all pairs of terms forming a palindrome of digits;
the remaining terms will reproduce the original seq.

Jps = (0,10),0,(1,11),10,(2,12),0,(3,23),1,(4,14),11,...

We see that the 4th palindrome (3,23) is not (3,13); this 
is because (3,13) would have produced an overlap with the
following "1" : (3,13),1, -- and we don't want overlaps. 
So we took the next available integer to complete the 
palindrome. Note that we don't accept palindromes like (3,3)
as we always want the two concatenated parts to be different.

How was Jps build? Jps is a succession of triplets p, q, r:

p  q  r
0,10,0,
1,11,10,
2,12,0,
3,23,1,
4,14,11,
5,15,10,
6,16,2,
7,17,12,
8,18,0,
9,19,3,
10,1,23,
11,211,1,
12,21,4,
13,1,14,...

p is always the smallest p not yet used and not producing
a palindrome with the previous term;
q is always the smallest term producing one palindrome
with p and zero palindrome with r; q is not equal to p;
r is imposed by Jps itself --no choice here-- as we want
the seq Jps to be "fractal".

Best,
É.


14,41,11,15,1,5,16,1,15,17,1,10,18,1,6,19,1,16,20,102,2,
21,12,7,22,2,17,23,32,12,24,2,8,25,2,18,26,2,0,27,2,9,
28,2,19,29,2,3,30,3,10,31,3,1,32,23,23,33,3,11,34,3,211,
35,3,1,36,3,12,37,3,21,38,3,4,39,93,13,40,4,1,42,24,14,
43,34,14,44,4,41,41,4,11,45,4,15,46,4,1,47,4,5,48,4,16,...