Prime conjunctions, and ruling families

This is positively my last post on prime conjunctions and I expect it to be read by nerds only. My purpose is to give two concrete examples of the more technical discussions in my two previous posts on prime conjunctions.
Let's begin with the children of Elizabeth II. They are (listed not by age but by the order in the year that their birthdays occur):

• Andrew b. 19/2/1960
• Edward b. 10/3/1964
• Anne b. 15/8/1950
• Charles b. 14/11/1948

Their ages at the beginning of 2018 were 57, 53, 67, 69. There's a stroke of luck - they are all odd. Indeed they are odd during the time interval 14/11/1999 to 18/2/2000. Such an odd conjunction period occurs every two years so maybe we shall find many prime conjunctions. Sadly, that set of ages has remainders 0, 2, 1, 0 on division by 3. In other words the 3-condition of my previous post fails - whenever the ages are odd, therefore, at least one age will be divisible by 3. So the only chance of a prime conjunction is when one of the is 3 itself and that would have to Edward's age. When he was 3 during the odd conjunction period the set of ages was 7, 3, 17, 19 and they are all prime. So the Royal siblings have had a prime conjunction (which they no doubt celebrated uproariously) but there will never be another.

What about 5 individuals who I shall call (in the order of their birthdays throughout the year)

• Barron b. 20/3/2006
• Melania b. 26/4/1970
• Donald b. 14/6/1946
• Tiffany b. 13/10/93
• Ivanka b. 30/10/1981

At the beginning of 2018 their ages are 11, 47, 71, 36, 24. Oh dear - not all prime and neither are they all odd. However they do have the property that they are bunch of odds followed by a bunch of evens and so we know that eventually they will have an odd conjunction. The nearest odd conjunctive period is 14/6/2017 to 29/10/2017 when their ages are 11, 47, 71, 35, 23. Nearly all prime but not quite - maybe this will also prove impossible.

Since there are 5 people in this case we have to check both the 3-condition and the 5-condition. The 3-residues are 2, 2, 2, 2, 2 while the 5-residues are 1, 2, 1, 0, 3. Both conditions hold! So Dickson's conjecture would predict that, not only will these 5 individuals have a prime conjunction, they will have infinitely many. How do we find them?

Their earliest prime conjunction occurred when the ages were 7, 43, 67, 31, 19. To find later prime conjunctions we must add a number d to all of them that has the properties

• d is even
• d has remainder 0 or 1 when divided by 3
• d is exactly divisible by 5

By a procedure called the Chinese Remainder Theorem these 3 conditions are equivalent to the condition d has remainder 0 or 10 when divided by 30. So the choices for d are 0, 10, 30, 40, 60, 70 etc. In fact d = 30 works (37, 73, 97, 61, 79). The next successful value of d is 40 and I'll leave you to find others.