My wife and I have recently had our third child - hence I've not been blogging much lately. However, I've been thinking about blog ideas, or more specifically, I've been thinking of mathematical investigations that I could explore and then, if they were interesting, share on my blog.

Our house is full of family photographs - almost every room has family photographs in it - and in particular on one wall, we have three photographs: one of our daughter; one of our older son, and one of the two of them together.

In mathematical notation, let's call my daughter A and my older son B; so we have A, B and AB.

Now we have three children, and I have started considering how many photos we would need to show the same range of variations... and it's more than I thought. Let's introduce the baby as C.

We would have:

A, B, C - each child individually

AB, AC, BC - each child with one other sibling

ABC - all three children together.

So the total number of pictures has gone from 3 to 7.

Let's suppose we have four children, A, B, C and D, and we want the same range of photos, with all variations. The list grows dramatically:

A, B, C and D - individual photos - 4

AB, AC, AD, BC, BD, CD - pairs - 6

ABC, ABD, ACD, BCD - trios - 4

ABCD - group - 1

Total = 15

Children Photos

1 1

2 3

3 7

4 15

In order to work out the nature of the series, I looked at the differences between terms, and then the second differences (i.e. the differences between the differences).

3-1 = 2

7-3 = 4

15-7 = 8

4-2=2

8-4 = 4

What became clear to me at this point is that the sequence is expanding exponentially or logarithmically, and not quadratically. And then it very quickly followed that each n

I had not expected a logarithmic series from this starting point; in fact I had not expected expected the number of photographs to increase so quickly - the volume more than doubles each time, as we have to account for every combination incorporating the new child. I was expecting something similar to a Fibonacci series - but that's more about multiplying rabbits, not children!

Our house is full of family photographs - almost every room has family photographs in it - and in particular on one wall, we have three photographs: one of our daughter; one of our older son, and one of the two of them together.

In mathematical notation, let's call my daughter A and my older son B; so we have A, B and AB.

Now we have three children, and I have started considering how many photos we would need to show the same range of variations... and it's more than I thought. Let's introduce the baby as C.

We would have:

A, B, C - each child individually

AB, AC, BC - each child with one other sibling

ABC - all three children together.

So the total number of pictures has gone from 3 to 7.

Let's suppose we have four children, A, B, C and D, and we want the same range of photos, with all variations. The list grows dramatically:

A, B, C and D - individual photos - 4

AB, AC, AD, BC, BD, CD - pairs - 6

ABC, ABD, ACD, BCD - trios - 4

ABCD - group - 1

Total = 15

Children Photos

1 1

2 3

3 7

4 15

In order to work out the nature of the series, I looked at the differences between terms, and then the second differences (i.e. the differences between the differences).

3-1 = 2

7-3 = 4

15-7 = 8

4-2=2

8-4 = 4

What became clear to me at this point is that the sequence is expanding exponentially or logarithmically, and not quadratically. And then it very quickly followed that each n

*th*term is 2^{n}-1 (the series 2^{n}is immediately recognisable - 1, 2, 4, 8, 16, 32 etc). The need to introduce the -1 suggests to me that we're excluding the photo which has no children in it.I had not expected a logarithmic series from this starting point; in fact I had not expected expected the number of photographs to increase so quickly - the volume more than doubles each time, as we have to account for every combination incorporating the new child. I was expecting something similar to a Fibonacci series - but that's more about multiplying rabbits, not children!