In general, regardless of base, the period of 1/n divides phi(n). So perhaps it would be interesting to provide phi(n)/period. On Sat, May 25, 2013 at 6:32 PM, Hans Havermann <gladhobo@teksavvy.com>wrote:
Bill Gosper:
In[209]:= RealDigits[1/499999] Out[209]= A very large output was generated. Here is a sample of it:
{{{2,0,0,0,0,0,4,0,0,0,0,0,8,0,0,0,0,1,6,0,0,0,0,3,2,0,0,0,0,6,4,0,0,0,1,2,8,0,0,0,2,5,
<<26587>>,
5,1,5,6,2,5,0,3,1,2,5,0,0,6,2,5,0,0,1,2,5,0,0,0,2,5,0,0,0,0,5,0,0,0,0,1,0,0,0,0,0}},<<1>>}
Mathematica 9 just gives:
A very large output was generated. Here is a sample of it: <<1>>
I would have expected shorter periods.
I wanted to chart the period lengths for numbers of the form 49..9:
Do[Print[Length[RealDigits[1/(10^n/2-1)][[1,1]]]],{n,2,16}] 42 498 357 641 26670 2499999 10248414 27737540 7001968 $Aborted
This is better:
Do[Print[MultiplicativeOrder[10,10^n/2-1]],{n,2,32}] 42 498 357 641 26670 2499999 10248414 27737540 7001968 24137931020 6247894877 2494908350060 7142857142856 249999999999999 1127586401304 7894736842105260 4671128568960 2448870123624540 7086158741278272 12095429958167388195 312056003600817627576 8668466022247153980 83333333331328827277944 81920903027779372603690 2521008403361344537815120 11523500839479696018300 1249999999999734482100644020 51349764590147678308496080 810869311801162457543134512 2288713183042871537476080 7142857142857142857142857142856 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun