[seqfan] Re: A158911

[Rick Shepherd]

Hi,

Some comments:
1) A158911(n) = A003592(n) - 1 (i.e., should be true -- by the first
comment in A158911 and the name of A003592).
2) A003592 already has a b-file with 1000 terms.
3) I also confirmed Charles' computations (through 102399) , before seeing
1) above.
4) I vote for keeping 0 in the sequence because of 1) above and because it
fits the definition.
5) PARI: is(n)=(10^n)%(n+1)==0.
6) I think Charles' one-liner of code had a typo (with the ^n misplaced).
7) I don't understand the comments in A158911, if taken literally,
such as "thus
the sequence which gives the number of digits of K^n (written in base 10) is
periodic."  What was intended and would someone like to clarify those in
A158911 itself?
8) Nit-picking(?):  While we're at it, how about "divided by" instead of
"divided thru"?
9) Nit-picking:  The name of the sequence should begin with a capitalized
word: "Numbers"

Regards,
Rick

On Tue, Aug 24, 2010 at 4:31 PM, Claudio Meller <claudiomeller at gmail.com>wrote:

> Ok, thanks Charles!
> Claudio
>
> 2010/8/24 Charles Greathouse <charles.greathouse at case.edu>
>
> > 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<
> http://www.research.att.com/%7Enjas/sequences/A158911<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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> >
> >
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> >
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> >
>
>
>
> --
> Claudio
> http://grageasdefarmacia.blogspot.com
> http://todoanagramas.blogspot.com/
> http://simplementenumeros.blogspot.com/
>
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