[seqfan] Extend S adding digits to a(n-1)
Eric Angelini
Eric.Angelini at kntv.be
Sun Oct 26 00:39:08 CEST 2014
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
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