You are currently browsing the tag archive for the ‘sequences’ tag.

**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?

This is another interesting maths sequence puzzle:

When x = 1, y = 1, when x= 2, y = -1, when x = 3, y = 1,

a) if when x = 4, y = -1, what formula gives the nth term?

b) if when x = 4, y = 3, what formula gives the nth term?

Answer below in white text (highlight to see)

a) This is a nice puzzle when studying periodic graphs. Hopefully it should be clear that this is a periodic function – and so can be modelled with either sine or cosine graphs.

One possibility would be cos((n-1)pi)

b) This fits well when studying the absolute function – and transformations of graphs. Plotting the first 3 points, we can see they fit a transformed absolute value function – stretched by a factor of 2, and translated by (2,-1). So the function 2abs(x-2) -1 fits the points given.

This is a really nice puzzle we looked at at the IB HL workshop:

**When x = 1, y = 1, when x = 2, y = 2, when x = 3, y = 3 but when x = 4, y does not equal 4. Find a sequence which describes these points.**

There are an infinite number of answers, though not necessarily easy to find!

If you are interested in the solutions, the answer is written below in white text, highlight to reveal!

Answer:

Two possible ways of tackling the problem –

1) as a polynomial – if y = (x-1)(x-2)(x-3) + x this satisfies the original question – as the brackets all cancel to zero for 1,2,3 but will remain for x = 4 onwards.

2) modelling as a function with absolute value. Notice that -(abs x ) will satisfy the correct shape – ie a linear increase and then divergence from this. By transformations therefore we can get -(abs(x-3) ) +3. This fits the graph for y=x for 1,2,3 before altering for y=4.