Of course I am extraordinarily fortunate in having so wonderful a quartet of children. Am I fortunate in another way:- that their ages allow such prime conjunctions? We shall soon see.

Let's consider a set of

*n*children (think of

*n*as being 2, 3 or 4 if you like). How likely is it that one day they will all have prime ages? I'm going to simplify matters in three ways. The first simplification is really unnecessary in practice: it is that we'll assume no two children were born at exactly the same time of day on the same day of the year. The second simplification is that we'll only worry about odd primes. Ignoring the prime 2 is not such a big deal. This special case is actually quite significant if you are just interested in whether the children have a prime conjunction at all but not if you are more concerned with whether they have lots of them. The third simplification is apparently quite bizarre but at the end of the post I will explain that it is not as bizarre as all that - so bear with me. This simplification is that rather than thinking about prime conjunctions we think about occasions (odd conjunctions) when all the children's ages are odd.

We want to think about their ages and we have to write down these ages in some order. Rather than that order being highest to lowest or lowest to highest I shall write them down in order of birthday throughout the year. For example three children born on 2 March 1991, 6 November 1994, 4 May 1997 will be ordered with the March birthday coming first, then the May birthday, then the November birthday. So, at the beginning of this year, their ages were 22, 15, 18. These remain their ages until 2 March when the 22 becomes 23. Another change (from 15 to 16 occurs on 4 May) and another change occurs on 6 November.

In terms of evenness and oddness it was Even, Odd, Even at the beginning of the year and then it went

Odd, Odd, Even

Odd, Even, Even

Odd, Even, Odd

and, continuing to track the changes in 2014,

Even, Even, Odd

Even, Even, Odd

Even, Odd, Odd

and then we come back to

Even, Odd, Even

Put more concisely we begin with a list of E's and O's. We change the first one, then we change the second, then we change the third, and then we go back to the start of the list and change the first (then the second, then the third).

EOE, OOE, OEE, OEO, EEO, EOO, EOE

and now the cycle repeats.

For this particular set of birthdays we never find OOO so these children are never all of an odd age (and therefore never all odd prime ages (as it happens, it doesn't matter in this case whether we allow the prime 2)).

OK. Let's jump to the general case. Now we have n children. As before we list their evenness or oddness of age by order of their birthday. Suppose at the beginning of the year that gives us a bunch of evens followed by a bunch of odds. At some stage in the year those initial evens will all have changed to odds and all of them will be an odd age. Instead suppose at the beginning of the year we have a bunch of odds followed by a bunch of evens. Part way through the year everyone will now be even and so, a year later, everyone will be odd.

But what if we don't have one of these types of initial odd/even lists. If that happens then either we begin with an even, later in the list we have some odd, and still later we have another even (or we might have a similar case with the roles of evens and odds exchanged):

E ... O ... E ...

We start our process of changing the evens and odds one by one from the beginning. It is easy to see that, no matter where we are in the process, the three symbols at these places cannot all be equal; so we can never get all odds.

It is not too difficult to work out that

*2n*even/odd lists have the property that they consist of a bunch of one type (even or odd) followed by a bunch of the other type (odd or even). But there are

*2*sequences in all. So what that means is that, with n children, the chance of their having an odd conjunction is

^{n}*n/2*. And notice that if they do have an odd conjunction then they will have others at two-yearly intervals - so many of them.

^{n-1}Therefore, given that I have 4 children, the chance of them having an odd conjunction is exactly 50%.

But, wait a minute, I began by asking about prime conjunctions! All I have done is analyse the chance of getting an odd conjunction. The passage from odd to prime leads us to some deep unsolved questions in number theory.

Suppose I have two children with the younger born between 1 and 3 years after the older. Then some of the time their ages will differ by 2 and every two years there will be a period when their ages are both odd. These children will have many prime conjunctions, the first few being (3, 5), (5, 7), (11, 13), (17, 19). The Twin Prime Conjecture states that there are infinitely prime conjunctions in this case (prime numbers differing by 2). It is a very long-standing unsolved problem and most mathematicians believe it is true.

What if I have two children born more than a year apart? Then every two years their ages will differ by an even number

*k*. Now we are in the realm of Polignac's Conjecture: whether there are infinitely many primes that differ by some specific even integer

*k*. With more than two children we come to an even more general conjecture by Dickson.

What all this means is that if your children do not have an odd conjunction (and we have seen above how this may be tested, and how likely it is) then you will never see a prime conjunction. But, more usefully, if you are looking for a prime conjunction, you should first find an odd conjunction (if any) and then starting adding two to it repeatedly hoping to find a set of odd numbers that are indeed all prime. This is not guaranteed (e.g. 7, 9, 11 - all increments include a multiple of 3). However Dickson's theorem (which I will write about in a future blog) provides some refined help if needed. Finally, unless a large number of mathematicians are going to be very surprised indeed, if your children have "several" prime conjunctions, they will have infinitely many prime conjunctions.

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