Search: id:A036236
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%I A036236
%S A036236 1,0,3,4700063497,6,19147,10669,25,9,2228071,18,262279,3763,95,1010,481,
               20,
%T A036236 45,35,2873,2951,3175999,42,555,50,95921,27,174934013,36,777,49,140039,
%U A036236 56,2463240427,110,477,697,91,578,623,156,2453,540923,55,70,345119,287
%N A036236 a(n) = least positive number k such that the remainder when 2^k is divided 
               by k is n (the inverse of A015910), or 0 if no such k exists.
%C A036236 No n exists with 2^n mod n = 1.
%C A036236 a(3) was first computed by the Lehmers.
%C A036236 Labos Elemer (labos(AT)ana.sote.hu) asked on Sept 27, 2001 if all numbers 
               > 1 eventually appear in A015910.
%C A036236 a(n) > 10^11 for n = 69, 185, 231, 273, 309, 311, 405, 465, 581, 619, 
               649, 669, 675, 741, 771, 799, 849, 871, 881, 885, 939, 981, ... - 
               Hans Havermann (pxp(AT)rogers.com), Apr 19 2007
%C A036236 a(69) = 887817490061261 = 29 * 37 * 12967 * 63809371. [From Hagen von 
               Eitzen (math(AT)von-eitzen.de), Jul 26 2009]
%D A036236 P. Erdos and R. L. Graham, Old and new problems and results in combinatorial 
               number theory, Monographies de L'Enseignement Mathematique, 28, 1980.
%D A036236 R. K. Guy, Unsolved Problems in Number Theory, Section F10.
%H A036236 Joe K. Crump, <a href="http://www.immortaltheory.com/NumberTheory/2nmodn.htm">
               2^n mod n</a>
%H A036236 Hans Havermann, <a href="a036236.txt">Table of n, a(n) for n = 1..10000 
               with -1 for those entries where a(n) > 10^11</a>
%H A036236 Topology Q+A Board, <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2005;
               task=show_msg;msg=1441.0004">Prove that 2^n = 1 (mod n) is impossible 
               for an integer n > 1</a>
%H A036236 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               2.html">Link to a section of The World of Mathematics</a>
%F A036236 It's obvious that for each k, a(k)>k and we can easily prove that 2^(3^n)=3^n-1 
               (mod 3^n). So 3^n is the least k with 2^k mod k = 3^n-1. Hence for 
               each n, a(3^n-1)=3^n. - Farideh Firoozbakht (mymontain(AT)yahoo.com), 
               Nov 14 2006
%t A036236 a = Table[0, {75} ]; Do[ b = PowerMod[2, n, n]; If[b < 76 && a[[b]] == 
               0, a[[b]] = n], {n, 1, 5*10^9} ]; a
%t A036236 t = Table[0, {1000} ]; k = 1; While[ k < 6500000000, b = PowerMod[2, 
               k, k]; If[b < 1001 && t[[b]] == 0, t[[b]] = k]; k++ ]; t
%Y A036236 Cf. A015910, A015948.
%Y A036236 Cf. A078457, A119678, A119679, A127816, A119715, A119714, A127817, A127818, 
               A127819, A127820, A127821.
%Y A036236 Bisections: A122182, A124977.
%Y A036236 Sequence in context: A067481 A058433 A154998 this_sequence A058447 A058453 
               A058471
%Y A036236 Adjacent sequences: A036233 A036234 A036235 this_sequence A036237 A036238 
               A036239
%K A036236 nonn,nice
%O A036236 0,3
%A A036236 David W. Wilson (davidwwilson(AT)comcast.net)
%E A036236 More terms from Joe K. Crump (joecr(AT)carolina.rr.com), Sep 04 2000

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