The ever popular Manfred Mann song,
5-4-3-2-1 (a sixties hit and more recently part of a chocolate advert) surprisingly features references to the Charge of the Light Brigade (“onward rode the 600”) and Helen of Troy. Sung by Paul Jones, who joined the band after failing to complete his Oxford English degree, Paul was replaced by Mike d’Abo. Paul went on to have a varied solo career, currently still active as a music broadcaster and respected harmonica player, still playing the clubs. And looks hardly a day older even today!

The Mathemateer’s interest is not confined to the music, as the sequence “5-4-3-2-1” could one day feature – if not done already – as a GCSE question, set at around Grade 4 Level.

Namely, prove that the nth term of the sequence 5-4-3-2-1 is –n + 6.  (See below for the methods to prove this.)

Paul still plays in various bands, including the Manfreds and the Blues Band, and was interviewed on Breakfast TV with fellow original member Tom McGuiness (yes, that one). When a clip from “Do Wah Diddy Diddy” was played, Paul was asked whether he still did the famous knee-knocking dance style (think Mick Jagger, perhaps a precursor to Dad Dancing) and he replied “No, but I still play the maracas!”

There is a thriving club scene for bands – original and tributes – from the 60’s, 70’s and 80’s. For instance one of Paul’s other bands, the Blues Band, soon plays Blackheath Halls, which typically eatures comedians as well, such as Arthur Smith.

South West London features strongly in the history of rock’n’roll, and for instance the Half-Moon in Putney is still going strong, with Eddie and the Hotrods playing soon.  The Boom Boom Club in Sutton is soon host to Curved Air, and tribute bands like Alter Eagles (love that name) and Absolute Bowie. The Clapham Grand  will soon be featuring Like the Jam, with original member, that great bass player Bruce Foxton (As an aside, The Mathemateer went up to Somerset House recently to see the wonderful About the Young Idea retrospective of the Jam’s career. It reminded him that although the above websites are great, he misses the paper bill posters, showing for instance early Jam on the same bill as the Clash).

So, back to Maths questions: how on earth is 5,4,3,2,1 represented by “ – n + 6 ”?

This is intriguing because no less than three methods of solution are available in GCSE textbooks, summarised below. The first is the most often recommended, and the others are also of interest.

Method 1 The sequence reduces by 1 each time so there must be a “–n” in the answer. Then for n =1, what must be done to get to the first term of the actual sequence? To go from -1 to +5 you have to add 6.

 So answer is that an expression for the nth term is –n + 6. Check for n= 2, the answer should be -2 + 6, which as expected equals 4.

 Method 2 The formula for the nth term of a liner sequence like this is the nth term = dn + (a-d) where d is the difference in successive terms (-1) and a is the first term ( 5). So -1n + (5 –(-1))  implies the nth term is –n + 6, as before.

 Method 3. Form two simultaneous equations for the first two terms using a for the n part and b for the number. Using the first and second terms, 

For n = 1  a + b = 5
For n = 2  2a + b = 4

 Solving these, a = -1 and b = 6.,  So the answer as before is that the nth term (known as Un) = -n + 6

 Always double check the answer. For example, feed n= 5 (the 5^th term) in. U (5) = -5 + 6 = 1

 It’s a strange but true answer. “5-4-3-2-1” expressed as –n + 6 ! But correct!

 And finally, Paul Jones sings “onward rode the 600”.  

 A supplementary foundation question might be: what is the 600^th term? Answer of course is -600 + 6, namely -594.   

Recent news from the Government that a new (10^th wave) of Free Schools is to be approved in England prompted me to look into the questions, what exactly is a Free School and do we have any in the TW area?

Free Schools are similar to Academies (like the excellent Waldegrave School) in that they are in the State system but not directly controlled by the Local Education Authority, so they are “free” in that sense, as well as making no charge to parents and having no academic selection criteria for admission.  But they differ in that normally they are new schools, sponsored or run by an education or learning charitable trust.

In the TW area, two Free Schools are opening as we speak, namely Turing House, run by the Russel Education Trust, and Twickenham Primary Academy at Heath Gate House Twickenham Green, run by the GEMS education trust, who are also next year opening a Primary Academy in Kingston. GEMS is a successful worldwide provider of education services, started in 1959.

The Turing House school is to open in a temporary site in Teddington, but is possibly later putting down roots in Whitton. This is strange for two reasons, first the idea was sold as filling a gap in secondary education in the Teddington area, and indeed there is some opposition in Whitton because of the effect upon traffic. You can argue that it is still within a couple of miles, but a trip across the A316 (either way) seems like a different place altogether. Second, the name Turing comes of course from Professor Alan Turing, associated with the NPL in Teddington, who many believe was the father of modern computers and artificial intelligence.

Turing, who graduated from Kings College Cambridge, was a talented marathon runner who regularly ran the 40 miles from Bletchley Park to London for meetings.  He lived at one point in Hampton, where you can see his Blue Plaque.

His life was recently portrayed memorably in the film The Imitation Game  (named after his “Turin Test” for artificial intelligence) by Benedict Cumberbatch, plotting his triumph in cracking the Enigma code through to his tragic death after undergoing treatment for his homosexuality, then illegal.

At one point Cumberbatch’s character says “There are 159 million, million, million possible Enigma settings…it is 20 million years to check each of the settings (manually)”

This links to a potentially typical GCSE foundation question which might be:

Express 159 million, million, million in Standard Form. Ans. Each million has 6 zeroes i.e. 10⁶, and when you multiply such numbers you add the powers.
So it is 159 x 10¹⁸  and finally in standard form 1.59 x 10²⁰