[seqfan] 2014 wishes -- with jumps

[Eric Angelini]

Hello SeqFans,

The idea behind the monotonically increasing sequence S is that the difference between a(n) and a(n+1) is given by the product z*a, "z" being the last digit of a(n) and "a" being the first digit of a(n+1).

An example of such a pair is (36,78); we see that |36-78|=42 and 42 is indeed 6*7.

My intuition is that S, whatever a(1), will always be finite. But, paradoxically, that the length of S can be arbitrarily long.

The finiteness of S comes from two facts:

- S stops immediately when "z", the last digit of a(n), is equal to 0;
- S cannot be extended, sometimes, because there is no product available to jump from a(n) to a(n+1). An example is given by 58: if you try to start a(n+1) with "a" = 1 or 2 or 3 or... 9, you will fail.

For a(1)=6, for instance, S stops after 6,12,14,18,58.

Please note that a(n) can be followed, sometimes, by more than one a(n+1). S will stop also after 6,12,14,18,66.

My question is: 
What is the smallest a(1) producing a 2014-term sequence?

Best wishes to Neil, his team and all Seqfans!
;-D
Best,
É.