A001682

David Wilson davidwwilson at comcast.net
Thu Mar 10 15:57:49 CET 2005


(1) The first element of A001682 should be 0.

(2)  The name of the sequence:

%N A001682 3^n, 3^(n+1) and 3^(n+2) have same number of digits.

could be reduced to

%N A001682 3^n and 3^(n+2) have the same number of digits.

(3) The "magic number" you are looking for is

1/(1-log10(9)) = 21.854345326....


----- Original Message ----- 
From: "Richard Guy" <rkg at cpsc.ucalgary.ca>
To: <ham>; <seqfan at ext.jussieu.fr>; "Math Fun" 
<math-fun at mailman.xmission.com>
Cc: "Klamkin Memorial Volume -- Andy Liu" <aliu at math.ualberta.ca>; 
<bruce1 at math.mun.ca>; "David Langstroth" <dll at cs.dal.ca>; "Don Albers" 
<dalbers at maa.org>; "Graham Wright" <gpwright at cms.math.ca>; 
<jborwein at cs.dal.ca>
Sent: Wednesday, March 09, 2005 1:46 PM
Subject: A001682


> I'm collecting Murray Klamkin problems
> for a book, and have reached Amer
> Math Monthly 64(1957) 665 where Joe
> Lipman solves a Murray problem
> ``as long as tables of sufficient
> accuracy are available.''
>
> Of course, we don't use tables any more.
> I've corrected the arithmetic in the
> original, and in so doing am able to make
> a modest addition to A001682 which
> currently reads
>
> 21,42,65,86,109,130,151,174,195,218,239,262,283,
> 304,327,348,371,392,415,436,457,480,501,524,545,
> 568,589,610,633,654,677,698,721,742,763
>
> and to which may be added
>
> 786,807, 830, 851, 874, 895, 916, 939,
>     960, 983,1004,1027,1048,1069,1092,
>    1113,1136,1157,1180,1201,1222,1245,
>    1266,1289,1310,1333,1354,1375,1398,
>    1419,1442,1463,1486,1507,1528,1551,
>    1572,1595,1616.1639,1660,1681,1704,
>    1725,1748,1769,1792,1813,1834,1857,
>    1878,1901,1922,1945,1966,1987,2010,
>    2031,2054,2075,2098,2119,2140,2163,
>    2184,2207,2228,2249 (not 2251)
> where I've continued the calculation
> until the difference pattern
>   21   23   21   23   21   21   23
> is broken.  What's the magic number
> whose continued fraction expansion
> will tell me when to make a gear
> change?
>
> As always, someone should check
> my hand calculations!    R. 






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