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Code and ciphers: Julius Caesar, the Enigma and the internet

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<strong>and</strong> so<br />

2�4�6�12<br />

�(4095)�4095� �1728<br />

3�5�7�13<br />

<strong>and</strong> <strong>the</strong> number of binary linear recurrences of order 12 which generate<br />

binary sequences of maximum length 4095 is <strong>the</strong>refore<br />

�144<br />

as stated in Chapter 8. Note that although 4095 is divisible by 3 2 <strong>the</strong> fraction<br />

in <strong>the</strong> Euler function is two-thirds, not eight-ninths.<br />

Similarly, when k�23 <strong>the</strong> number of binary linear recurrences of order<br />

23 which generate binary sequences of maximal length (2 23 �1) is<br />

which is<br />

1728<br />

12<br />

�(2 23 � 1)<br />

23<br />

�(47�178 481)<br />

23<br />

<strong>and</strong> 47 <strong>and</strong> 178 481 are primes (that 178 481 is a prime follows from <strong>the</strong><br />

fact that it is not divisible by any prime less than its square root, i.e. by<br />

any prime less than 422). We <strong>the</strong>refore find that <strong>the</strong> number of such<br />

sequences is<br />

46�178 480<br />

23<br />

�356 960<br />

which was also mentioned in Chapter 8.<br />

(For a proof of <strong>the</strong> <strong>the</strong>orem quoted above see [8.2].)<br />

M12 Recovery of a binary linear recurrence from a<br />

stretch of key<br />

Example<br />

The following 15 binary digits of key have been recovered from a cipher<br />

message:<br />

10101 00110 00100<br />

Ma<strong>the</strong>matical aspects 199<br />

<strong>the</strong> most recent digits being to <strong>the</strong> right. Verify that <strong>the</strong>se can be generated<br />

by a linear recurrence of order 5.

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