[seqfan] Extend S adding digits to a(n-1)

[Eric Angelini]

Hello SeqFans
We start S with a(1)=15.
We'll always extend S with a(n) such
that a(n)=a(n-1)+K.
K is the sum of the last "d" digits of S.
"d" is the first digit of a(n).

S = 15,21,24,31,39,59,96,102,104,108,116,122, etc.

Why is a(2)=21?
We have here a(n)=21, "d"=2, and K=1+5=6 --> thus 15+6 = 21

Why is a(3)=24?
We have here a(n)=24, "d"=2, and
K=2+1=3 --> thus  21+3 = 24

Why is a(4)=31?
We have here a(n)=31, "d"=3, and
K=1+2+4=7  --> thus 24+7 = 31

etc.

We always extend S with the smallest
available a(n) if this a(n) doesn't stop S.
Else we try the next smallest a(n).

Example of two possible extensions:

S = 15,21,24,31,39,...
S = 15,21,24,31,41,...

My efforts to find an infinite seq S failed...
S, in the best case, stops with 1000
(if I'm not wrong).

Is an infinite S possible? The challenge
is to jump from an x-digit integer to
an (x+1)-digit integer...

Best,
É.
Catapulté de mon aPhone