[seqfan] Onwards a(n) = a(n-1) + first digit of a(n)

[Eric Angelini]

Hello SeqFans,
S=0,9,10,11,12,13,14,15,16,17,18,20,22,24,26,28,31,34,37,41,45,49,54,60,66,73,81,90,99,100,101,102,...

S was computed backwards, starting with a(1)=1000 and 
always subtracting to a(n) it's first digit.
Thus 1000,999,990,891,883,875,867,..., 12,11,10,9,0.

Is it possible to compute an infinite S onwards, starting with 0, without backtracking? No!
This is fascinating...

Best,
É.

Catapulté de mon aPhone