# [seqfan] Re: A158911

Charles Greathouse charles.greathouse at case.edu
Tue Aug 24 22:26:28 CEST 2010

```The sequence as written looks wrong.  I get

1, 3, 4, 7, 9, 15, 19, 24, 31, 39, 49, 63, 79, 99, 124, 127, 159, 199,
249, 255, 319, 399, 499, 511, 624, 639, 799, 999, 1023, 1249, 1279,
1599, 1999, 2047, 2499, 2559, 3124, 3199, 3999, 4095, 4999, 5119,
6249, 6399, 7999, 8191, 9999, 10239, 12499, 12799, 15624, 15999,
16383, 19999, 20479, 24999, 25599, 31249, 31999, 32767, 39999, 40959,
49999, 51199, 62499, 63999, 65535, 78124, 79999, 81919, 99999, 102399,
124999, 127999, 131071, 156249, 159999, 163839, 199999, 204799,
249999, 255999, 262143, 312499, 319999, 327679, 390624, 399999,
409599, 499999, 511999, 524287, 624999, 639999, 655359, 781249,
799999, 819199, 999999, 1023999, 1048575, 1249999, 1279999, 1310719,
1562499, 1599999, 1638399, 1953124, 1999999, 2047999, 2097151,
2499999, 2559999, 2621439, 3124999, 3199999, 3276799, 3906249,
3999999, 4095999, 4194303, 4999999, 5119999, 5242879, 6249999,
6399999, 6553599, 7812499, 7999999, 8191999, 8388607, 9765624,
9999999, ...

with the naive

is(n)=Mod(10,n+1)^n==0

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Tue, Aug 24, 2010 at 4:05 PM, Claudio Meller <claudiomeller at gmail.com> wrote:
> Hi,
>
> In http://www.research.att.com/~njas/sequences/A158911
> (Numbers n such that 10^n divided thru the number of digits of 10^n is an
> integer.)
> are 124, 249, 299, 624, 999,1249, 1549, 2499,3999, 4999 and 7999 terms of
> A158911?
> --
> Best
> Claudio Meller
>
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>

```