[seqfan] A sum visible in the 1st integer

[Eric Angelini]

Hello SqFans,
Same idea as my prior post (that got not a single reply -- sigh!)

T = 10,100,101,90,91,80,81,70,71,60,61,50,51,40,41,30,31,20,21,102,
82,62,42,200,201,103,72,52,32,104,63,300,301,210,105,53,211,106,43,
107,302,108,220,221,109,110,112,92,73,43,113,83,54,114,74,310,115,
64,230,231,116,500,501,400,401,302,117,402,232,...

The rules for expanding T are:
# Say "a" and "b" are two consecutive terms of S;
# Add the last digit of "a" to the first digit of "b"
# The result must be a substring of "a"
# "b" cannot be a term already in S.

We see here that 
0+1 = 1 which is visible in 10
0+1 = 1 which is visible in 100
1+9 = 10 which is visible in 101
9+0 = 9 which is visible in 90
etc.
The term after 21 (end of first row) cannot be 10 (though 1+1 = 2 
which is visible in 21) because 10 was already used before. So 102,
being the next available integer, has been selected.

T is NOT a permutation of the integers > 0, of course.

This is tricky to compute by hand. Needing a lot of backtracking.
Hope no typos were left.

Best,
É.