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Oh wow, so it is. Sorry, I didn't look at your code since you didn't mention that you had gone outside the parameters provided so I didn't think that you had done what I was doing.

In penance, here is a script that shows the repeat length for 1/$n only when that length isn't "boring". If $n is prime a repeat length of $n-1 is "boring". If $n is composite, the boring length is the max repeat length of its prime factors.

#!/usr/bin/perl -w use strict; Main(); exit(); sub repDig { my( $den, $num )= @_; $num ||= 1; my $rem= $num % $den; my %seen; my $rep= ''; while( 1 ) { $rem *= 10; last if exists $seen{$rem}; $seen{$rem}= length( $rep ); $rep .= int( $rem / $den ); $rem %= $den; } substr( $rep, 0, $seen{$rem} )= ''; return $rep; } sub factor { my( $r )= @_; my $f= ''; my @f; my $p= 2; while( 1 < $r ) { my $q= int( $r / $p ); last if $q < $p; my $e= 0; while( $r == $q*$p && $q ) { $e++; $r= $q; $q= int( $r / $p ); } if( $e ) { push @f, $p; $f .= "*$p"; $f .= "^$e" if 1 < $e; } $p += 2==$p ? 1 : 2; } if( 1 < $r || ! @f ) { push @f, $r; $f .= "*$r"; } substr( $f, 0, 1 )= ''; return $f, @f; } sub Main { my $dem= 1; my %r; while( 1 ) { my( $f, @f )= factor( ++$dem ); my $r= repDig( $dem ); $r= $r ? length($r) : 0; if( $f eq $dem ) { $r{$dem}= $r; next if $r == $dem-1; } else { my $max= 0; for( @f ) { $max= $r{$_} if $max < $r{$_}; } next if $r == $max; } printf "%8d: 1/%s\n", $r, $f; } }

and the first few lines of output:

0: 1/2 1: 1/3 0: 1/5 2: 1/11 6: 1/13 3: 1/3^3 15: 1/31 3: 1/37 5: 1/41 21: 1/43 42: 1/7^2 13: 1/53 3: 1/2*3^3 33: 1/67 35: 1/71 8: 1/73 13: 1/79 9: 1/3^4 41: 1/83 44: 1/89 42: 1/2*7^2 4: 1/101 34: 1/103 53: 1/107 3: 1/2^2*3^3 48: 1/7*17 22: 1/11^2 42: 1/127 3: 1/3^3*5 8: 1/137 46: 1/139 42: 1/3*7^2 75: 1/151 78: 1/157 66: 1/7*23 9: 1/2*3^4 81: 1/163 78: 1/13^2 43: 1/173 95: 1/191 42: 1/2^2*7^2 98: 1/197 99: 1/199 84: 1/7*29 30: 1/211 3: 1/2^3*3^3 30: 1/7*31 48: 1/13*17 113: 1/227 48: 1/2*7*17 7: 1/239 30: 1/241 22: 1/2*11^2 27: 1/3^5 42: 1/5*7^2 50: 1/251 3: 1/2*3^3*5 5: 1/271 69: 1/277 28: 1/281 141: 1/283 30: 1/7*41 272: 1/17^2 146: 1/293 42: 1/2*3*7^2 6: 1/3^3*11 66: 1/13*23 42: 1/7*43 153: 1/307 155: 1/311 79: 1/317 66: 1/2*7*23 144: 1/17*19 9: 1/2^2*3^4 138: 1/7*47 110: 1/331 78: 1/2*13^2 30: 1/11*31 294: 1/7^3 173: 1/347 116: 1/349 32: 1/353 48: 1/3*7*17 179: 1/359 342: 1/19^2 22: 1/3*11^2 78: 1/7*53 186: 1/373 84: 1/13*29 176: 1/17*23 42: 1/2^3*7^2 99: 1/397 200: 1/401 30: 1/13*31 9: 1/3^4*5 84: 1/2*7*29 6: 1/11*37 204: 1/409 174: 1/7*59 140: 1/421 215: 1/431 3: 1/2^4*3^3 30: 1/2*7*31 198: 1/19*23 219: 1/439 42: 1/3^2*7^2 48: 1/2*13*17 221: 1/443 32: 1/449 10: 1/11*41 152: 1/457 48: 1/3^3*17 154: 1/463 233: 1/467 66: 1/7*67 42: 1/11*43 48: 1/2^2*7*17 239: 1/479 66: 1/3*7*23 22: 1/2^2*11^2 27: 1/2*3^5 42: 1/2*5*7^2 112: 1/17*29 210: 1/7*71 78: 1/3*13^2 24: 1/7*73 52: 1/521 261: 1/523 240: 1/17*31

My favotite is: 42: 1/7^2

Now demonstrate your understanding by correctly predicting a number with a repeat length of 11 or 25.

- tye        


In reply to Re^3: Recurring Cycle of Fractions (same) by tye
in thread Recurring Cycle of Fractions by Limbic~Region

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