Mathematical calculation to regenerate numeric series I am searching for a mathematical technique by which I can perform the following,
Let me explain first what i want to do.
consider a numeric series:
 23, 45, 42, 56, 12, 56 ...

Now, I want to perform any calculation so that at final I will get two numbers at the end.
lets say 33,22 .
These two numbers a such, so that I can regenerate the above series again by applying some calculations.
23, 45, 42, 56, 12, 56 ...

Is it possible in mathematics.
 A: Whether or not this is possible depends entirely on the details of the problem. Here are a few cases.
The sequence contains finitely many integers, and the two numbers must be 
integers.
This is possible. Let $M$ be the largest integer in the sequence; $M$ will also be your first number. Then let
$$N = \sum_i a_i M^i$$ 
where the $a_i$ are the entries of your sequence. $N$ is your second number. To decode the sequence, repeatedly take $N$ mod $M$ followed by division by $M$.
The sequence contains infinitely many integers, and the two numbers must be integers.
Not possible. The cardinality of the set of pairs of integers is $\aleph_0$, while the cardinality of the set of infinite sequences of integers is at least $2^{\aleph_0} > \aleph_0$.
The sequence contains infinitely many integers, and the two numbers can be reals.
This is possible. First, encode the sequence of integers as a sequence of positive integers (using, for instance, the usual map that takes positive integers $n$ to $2n$, and non-positive integers to $-2n+1$) and then encode the sequence of positive integers as the continued fraction
$$N = \alpha_1 + \frac{1}{\alpha_2 + \frac{1}{\alpha_3 + \frac{1}{\ddots}}},$$
which always converges.
(No second real number is needed.)
The sequence contains infinitely many real numbers, and the two numbers must be reals.
Again, not possible, for the same reason as the second case above; $2^{2^{\aleph_0}} > 2^{\aleph_0}.$
