[seqfan] Re: Sum of a(n) first digits of S is a(n+1)

Maximilian Hasler maximilian.hasler at gmail.com
Wed Jan 6 04:13:25 CET 2010


The numbers with the property that
the constant sequence  (k,k,k,....) is a solution to the original problem
(without the requirement of being strictly increasing, of course),
are:
1, 11, 20, 102, 111, 120, 201, 210, 300, ...

This matches
A061384 : Numbers n such that sum of digits = number of digits.

but the next term in A061384, 1003, does not have any more the above property.

However, it is easy to see that the terms of A061384 which have the
additional property of being a multiple of their sum-of-digits,
(satisfied by all of the above terms except for 11),
do have the initially mentioned property ((k,k,k,...) is a solution).
That subsequence continues
1012, 1120, 1300, 2020, 2110, 2200, 3100, 4000, 10040, 10130, ...

The number 11 as well as 1111 lead to a (constant) solution in the
same way as does any repunit 11...11 = (10^m-1)/9.

Maximilian

On Tue, Jan 5, 2010 at 9:59 PM,  <hv at crypt.org> wrote:
> Assuming there is an additional rule that a(n+1) > a(n), I knocked up some
> (very hacky) code to search for longer sequences, attached below.
>
> Running it for half an hour, it gets as far as:
>  4 10 50 68 89 91 100 110 120 210 1000 2000 10000 20000 100000 100010 100080 100313 102000
> (and the initial seven terms were fixed after the first 10 seconds).
[snip]




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