**The Telephone Numbers – Graph Theory**

The telephone numbers are the following sequence:

1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…

(where we start from n=0).

This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.

To illustrate this idea, the graph below is for n=4. This is when we have 10 telephones:

Each red line represents a connection. So the first diagram is for when we have no connections (this is counted in our sequence). The next five diagrams all show a single connection between a pair of phones. The last three diagrams show how we could have 2 pairs of telephones connected at the same time. Therefore the 4th telephone number is 10. These numbers get very large, very quickly.

**Finding a recursive formula**

The formula is given by the recursive relationship:

**T(n) = T(n-1) + (n-1)T(n-2)**

This means that to find (say) the 5th telephone number we do the following:

**T(5) = T(5-1) + (5-1)T(5-2)**

**T(5) = T(4) + (4)T(3)**

**T(5) = 10 + (4)4**

**T(5) = 26**

This is a quick way to work out the next term, as long as we have already calculated the previous terms.

**Finding an nth term formula
**

The telephone numbers can be calculated using the nth term formula:

This is going to be pretty hard to derive! I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:

These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1. We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities. Therefore n choose 2 also generates the triangular numbers.

But then you would have to work out all the permutations which were allowed – not easy!

Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:

We have n = 5. The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).

Therefore T(5) = 5!/(2^{0}(5-0)!0!) + 5!/(2^{1}(5-2)!1!) + 5!/(2^{2}(5-4)!2!)

T(5) = 1 + 10 + 15 = 26.

**Finding telephone numbers through calculus**

Interestingly we can also find the telephone numbers by using the function:

y = e^{0.5x2+x}

and the nth telephone number (starting from n = 1) is given by the nth derivative when x = 0.

For example,

So when x = 0, the third derivative is 4. Therefore the 3rd telephone number is 4.

The fifth derivative of the function is:

So, when x =0 the fifth derivative is 26. Therefore the 5th telephone number is 26.

If you liked this post you might also like:

Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.

Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct. How?

Essential resources for IB students:

Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.

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2) Exploration Guides and Paper 3 Resources

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