[seqfan] Re: A158911
Charles Greathouse
charles.greathouse at case.edu
Tue Aug 24 23:47:02 CEST 2010
I certainly agree with keeping 0 in the sequence. I omitted it inadvertently.
The description of the sequence, if it is kept, should be rewritten
entirely. Perhaps
Numbers n such that 10^n is divisible by n+1.
or
Numbers n such that 10^n is divisible by the number of digits in 10^n.
depending on whether we decide that it needs the base keyword. I
think it needs the keyword, so I'd go for the second wording.
My Pari code is significantly more efficient than Rick's. I feel like
there should be a better way to generate terms than testing sequential
numbers, though. Any ideas?
Charles Greathouse
Analyst/Programmer
Case Western Reserve University
On Tue, Aug 24, 2010 at 5:35 PM, Rick Shepherd <rlshepherd2 at gmail.com> wrote:
> 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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>>
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>>
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