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

Linear recurrences
The sequences looked at above are examples of sequences generated by
means of what are known as linear recurrences. Since each new term
involved adding together multiples of the two preceding terms they are
more specifically known as linear recurrences of order 2. More generally, a
linear recurrence of order k is one in which each new term is the sum of multiples
of the k preceding terms. So, for example, if we let U n denote the nth
term of a sequence then
U n �U (n�1) �2U (n�2) �U (n�3)
is a linear recurrence of order 3 and
U n �U (n�3) �U (n�5)
is a linear recurrence of order 5. The fact that in the second case some of the
preceding terms are not involved doesn’t matter; five preceding terms are
required in order to find the next term but three of the terms, U (n�1) , U (n�2)
and U (n�4) , have multipliers of 0. Had the term U (n�5) not been present
however the recurrence would not have been of order 5. The multipliers,
for our purposes, are always integers but may be positive, negative or zero.
It is assumed that in a linear recurrence of order k the term U (n�k) is
present, with either a positive or a negative multiplier, but not zero.
The terms of linear recurrences usually grow very rapidly and
although they often have interesting arithmetical properties they are
only suitable for cryptographic purposes when the terms themselves are
replaced by their values (mod 2), that is the terms are replaced by 0 if they
are even and by 1 if they are odd, thus producing a binary sequence.
Calculation of the terms of a linear recurrence (mod 2) is particularly easy,
there is no need to compute the actual value of the terms and then replace
them by 0 or 1. Each term is simply replaced by 0 or 1 as soon as it is calculated;
we then only have to add up a number of 0s and 1s which is a lot
easier than adding increasingly large integers. The resulting binary
sequence is identical to the one which would be obtained by computing
each term exactly and then replacing it by 0 or 1. So, for example, the
linear recurrence of order 2
U n �3U (n�1) �2U (n�2)
with the initial values U 0 �0, U 1 �1 continues
0, 1, 3, 7, 15, 31, 63, 127, 255, 511,.....
Producing random numbers and letters 99