Numbers with no finite representation on paper It occurred to me that there must be a lot of numbers without any form of finite representation on paper. Is there a name for these numbers?

For example...

Integers and rationals have a very simple representation e.g. 3/4

Irrational numbers obviously can also have a finite representation:
1.41421356... can be written as "the solution to the equation x^2 = 2"

Transcendental numbers can also have a finite representation:
e can be written as "the limit of (1 + 1/n)^n as n approaches infinity"

In other words, with a finite amount of effort one can give the reader enough information to calculate the value of the specified number exactly (to any degree of accuracy the reader chooses)

However, there must be a lot of numbers where this simply is not possible.

Consider the number 1.2736358762987349862379358... where this is just a string of (genuinely) random digits. There is no way to provide a finite definition that will specify this number to an arbitrary degree of accuracy. Similarly, there is no equation to which this number is a solution (I think, although I don't know how one would prove this).

Does this mean there are "gaps" in the real numbers. The number above is definitely somewhere between 1.2 and 1.3 but there is no way I can specify the value of this number (without writing an infinite number of digits). The number exists on the number line but I will never be able to do anything with it.

Is there a name for these numbers? Can anyone point me to some interesting resources on this topic?

I'm only asking as an interested hobbyist so apologies if this question isn't very scientific.
That depends on what you mean by "representation." One way to cash this out is to talk about the definable numbers. These are more general than the computable numbers, but they are still countable because there are still only countably many possible descriptions of a number in a language over a finite alphabet.
I think what you need is this Completeness Theorem:

Each contracting sequence of intervals has a common point; that is, the intersection of all the intervals of the family is not empty.

The basic idea is if the real number $c$ in the proof (below) is the number that you want a finite representation for (your randomly defined number for example), the construction given in the proof represents a finite way of defining the contracting sequence of intervals such that $c$ is in the intersection of all of them.

Here's the proof given in First Concepts of Topology, by Chinn and Steenrod.

Definition. an infinite sequence of closed intervals of real numbers $I_0, I_1, ..., I_n$ is called a contracting sequence if each interval contains the next and hence all intervals that follow: $$I_0 \supset I_1 \supset I_1 \supset ... \supset I_n \supset ...$$
A contracting sequence is called a regularly contracting sequence if $I_0 = [m, m+ 1]$ is the interval from an integer $m$ to the next integer $m + 1,$ and, for each $n &gt;= 1$ $I_n$ is one of the intervals obtained by partitioning $I_{n-1}$ into ten equal parts. Thus $I_1$ is a tenth part of $I_0,$ $I_2$ a tenth part of $I_1$ and so on. The length of $I_n$ is $10^{-n}.$

Consider first the case of a regularly contracting sequence. For each $n = 0, 1, 2, ...$ let $a_n$ denote the left-hand endpoint of $I_n.$ Then $a_0 = m$ is an integer. Since $I_n$ has length $10^{-n}$, $I_n = [a_n, a_{n+10^{-n}}].$ The points that divide $I_{n-1}$ into ten equal parts are

$$a_{n-1}, a_{n-1} + \frac{1}{10^n}, a_{n-1} + \frac{2}{10^n}, ..., a_{n-1}+\frac{10}{10^n}.$$

Since $I_n$ is one of these ten subintervals, its left endpoint must be $$a_n = a_{n-1} + \frac{k_n}{10^n},$$ where $k_n$ is one of the digits $0, 1, 2, ..., 9.$ In this way the sequence of intervals determines an integer $m$ and an infinite sequence of digits $k_1, k_2, ..., k_n, ....$ Let $c$ denote the real number 
$$c = m + \frac{k_1}{10} + \frac{k_2}{10^2} + ... + \frac{k_n}{10^n} + ....$$ 
(that is, the decimal expansion of $c$ is $m.k_1k_2,...$).

Since $$a_n = m + \frac{k_1}{10} + \frac{k_2}{10^2} + ... + \frac{k_n}{10^n},$$ it is clear that $a_n &lt;= c$. Now the decimal expansions of $a_n + \frac{1}{10^n}$ and of $c$ agree out to the n'th digits, but the n'th digit of $c$ is $k_n$ while that of $a_n + \frac{1}{10^n}$ is $k_n + 1$. It follows that $$a_n &lt;= c &lt;= a_n + \frac{1}{10^n}.$$ 

These inequalities assert that $c \in I_n$ and since they hold for every integer $n,$ it follows that $c$ lies in each interval of the sequence, hence in the intersection of all of them. This proves the theorem in case the sequence contracts regularly.

The case where the sequence does not contract regularly is similar, and the same book gives a proof. I am not going to enter it here right now, but would be happy to put the second part in also.