[seqfan] Re: Digitsum and length
Lars Blomberg
larsl.blomberg at comhem.se
Wed Feb 19 13:21:45 CET 2020
Hello!
The sequence continues
1, 2, 10, 3, 100, 4, 1000, 5, 10000, 6, 100000, 7, 1000000, 8, 10000000, 9, 100000001, 11, 12, 101, 13, 1001, 14, 10001, 15, 100001, 16, 1000001, 17, 10000001, 18, 100000002, 102, 103, 1002, 104, 10002, 105, 100002, 106, 1000002, 107, 10000002, 108, 100000003, ...
Among the first 10000 terms the largest value is 10000000000000000000000003.
For 100000 terms it is 1000000000000000000000000000000004.
Computing 100000001 terms is unfortunately out of the reach of my computer.
/Lars B
-----Ursprungligt meddelande-----
Från: SeqFan <seqfan-bounces at list.seqfan.eu> För Éric Angelini
Skickat: den 18 februari 2020 12:08
Till: Sequence Discussion list <seqfan at list.seqfan.eu>
Ämne: [seqfan] Digitsum and length
Hello SeqFans,
say that "The digitsum of a(n) is the
length of a(n+1)" -- we then get (with S being the lexicographically earliest seq of such distinct integers):
S = 1,2,10,3,100,4,1000,5,10000,6,100000,
7,1000000,8,10000000,9,100000001,
11,12,101,12,1001,...
We see above after 9 that the next
integer cannot be 100000000 (length 9)
because this integer has digitsum 1 and
all integers with digitsum 1 have been
used already in S; thus 100000001
after 9 (with digitsum 2), followed by
11 (length 2), as 10 (same length) is
already in S.
S is infinite, of course, but in computing its terms one must be careful because of this "limited available lengths" story.
Any takers (if interested)? -- I'd love to see a 100000001-term graph (!) Best, É.
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