# Concatenate a(n) and [sum of digits of a(n)]

Franklin T. Adams-Watters franktaw at netscape.net
Wed Jun 15 21:01:31 CEST 2005

```The sum of the digits of an n-digit number is at most 9n (and at least 1).  So for numbers up to 11 digits, once you check all numbers with no more than 1 more digit, you know nothing else can be larger.

Regarding your second question - the answer is no, only one such decomposition is possible.  Consider dividing the number between some pair of digits; then compare the sum of the digits before the break to the number following the break.  As you move the break point to the right, the sum of the digits increases, while the remaining number decreases.  So they can only coincide at at most one point.  (Actually, that's a slight simplification - they can coincide at a sequence of points, but only when the digits in that range are all zeros.  Since we don't allow the sum of the digits to have leading zeros, only one of these breaks is valid.)

Note that you can use this for a quick test of whether a number has this form.  This is probably easier than generating all such numbers up to a point and then sorting them.

If you do add this to the OEIS, be sure to use the "base" keyword.

"Eric Angelini" <keynews.tv at skynet.be> wrote:
>Hello SeqFan,
>I've started from the beginning and got:
> 1-->11
>...
> 9-->99
>10-->101
>11-->112
>...
>18-->189
>19-->1910 (ouch!)
>20-->202
>...
>
>If I want to introduce in the OEIS the integers which
>are at the right end of the arrow [-->] I will always
>have the doubt that I've missed an integer somewhere
>("202" in the above enumeration appears _after_ "1910",
>though "202" will be in the sequence _before_ "1910")
>
>Could someone help?
>
>Second demand: are some numbers ambiguous? I mean, if
>I give a such large number to someone explaining it's
>hidden construction [concatenation of a(n) and sum of
>digits of a(n)], might he find _another_ "primitive"
>integer than me?
>
>Best,
>... and sorry if this is old hat...
>E.
>
>
>

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