[seqfan] Division with _all_ visible digits

[Eric Angelini]

re-Hello Seqfans,
the equivalent seq T where all digits of a(n) and a(n+1)
are visible in the result of a(n)/a(n+1) is nice too:

T = 1,6,13,17,7,8,14,19,23,...

I guess T is infinite and a permutation of the integers > 0.

Explanation:

1/6 = 0.166666666...   (where both "1" and "6" are visible)
6/13 = 0.46153...      (where "6", "1" and "3" are visible)
... here 13/10 doesn't fit as 13/10 = 1.3 NOT 1.30)
13/17 = 0.764705882352941... ("1", "3" and "7" are visible)
17/7 = 2.428571...     (where both "1" and "7" are visible)
7/8 = 0.875            (where both "7" and "8" are visible)
8/14 = 0.571428...     (where "8", "1" and "4" are visible)
14/19 = 0.73684210526315789...  ("1", "4", "9" are visible)
19/23 = 0.82608695652173...

Best,
É.