(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@cpsc.ucalgary.ca> To: <ham>; <seqfan@ext.jussieu.fr>; "Math Fun" <math-fun@mailman.xmission.com> Cc: "Klamkin Memorial Volume -- Andy Liu" <aliu@math.ualberta.ca>; <bruce1@math.mun.ca>; "David Langstroth" <dll@cs.dal.ca>; "Don Albers" <dalbers@maa.org>; "Graham Wright" <gpwright@cms.math.ca>; <jborwein@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.