[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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