An explicit rule computes any term straight from n

0:00Suppose someone hands you the odd numbers and asks for the term in position one thousand one hundred and seventy. You could walk. Start at one, add two, add two again, and keep going. It works, and it takes one thousand one hundred and sixty-nine additions before you arrive. Reaching position three the same way takes two. So the cost of walking is not a fixed price. It grows with how far you are going.

0:30And every step depends on the step before it, which means you cannot start in the middle. That is what we are about to get rid of - not to save arithmetic, but to change which questions you can ask. Here is the alternative. A machine with one slot. Put a position into the slot, and a term comes out. For the odd numbers, that machine is two n minus one.

0:57The letter n is the slot. It is not a number yet; it is the place where the position goes. So the machine says: take the position, double it, subtract one. That is an explicit rule, and the word explicit is doing real work here. It computes the term from the position alone. Nothing else goes in. In particular, no earlier term goes in. The machine need not be this simple, either. n times n is a rule of exactly the same kind, and its gaps are not all the same.

1:36Let us check that it is the machine we wanted. Put in one: two times one is two, minus one is one. Put in two: four minus one is three. Put in three: six minus one is five. One, three, five, and it carries on seven, nine, eleven. That is the list we were handed. Notice what each of those cost: one multiplication and one subtraction. And notice what none of them needed. The term before.

2:07Position three never consulted position two. It could have been asked first. Which means we can ask for any of them, in any order, at the same price. The fifty-third term is two times fifty-three minus one, a hundred and five. The hundred and eighth term is two hundred and fifteen. And the one we started with, position one thousand one hundred and seventy, is two thousand three hundred and thirty-nine.

2:38One multiplication and one subtraction, three times over. The walk to that last one was one thousand one hundred and sixty-nine additions. The rule got there in two operations, and it would reach the millionth term in two as well. Now for the thing the rule really buys you. So far the arrow points one way. A position goes in, a term comes out. Turn it round. Put a term in, and ask which position it came from.

3:11The machine says: take the position, double it, subtract one, and hand me the term. So the reverse says: take the term, add one, halve it, and you have the position. That is not a new idea. It is the same rule read as an equation, with the position as the unknown, and solving it is something you already know how to do. Try it. Is a hundred and thirty-seven an odd number, and if it is, where does it sit?

3:43Write two n minus one equals a hundred and thirty-seven. Add one to both sides. Two n is a hundred and thirty-eight. Halve it. n is sixty-nine. So yes, and it sits in position sixty-nine. Now say carefully what you have found. Not the value - the value was handed to you. You have found the position. The answer to which term is a hundred and thirty-seven is sixty-nine, and mixing up those two numbers is the commonest way to get this wrong.

4:19A second machine, so that the shape becomes visible. Five n minus two. Put in one, two and three, and you get three, eight, thirteen, then eighteen, twenty-three, twenty-eight. The gaps are five, every time. And five is the multiplier. The first term is three, which is five minus two: the multiplier plus the constant. So in a rule of this shape, the multiplier is the size of the step and the constant decides where the list begins.

4:51Its hundredth term is four hundred and ninety-eight. Its thousandth is four thousand nine hundred and ninety-eight. One more of the same shape: three n minus two gives one, four, seven, ten. Step of three, starting at one. Still two operations, either time. Now the question all of this was for. Is three hundred and eight in that list? Write five n minus two equals three hundred and eight. Add two: five n is three hundred and ten.

5:26Divide by five. n is sixty-two. Yes, and it is the sixty-second term. Now try four hundred and seventy-one. Five n is four hundred and seventy-three. Divide by five, and n is ninety-four point six. Something has gone wrong, and it has gone wrong in a very particular way. The equation solved perfectly well. The arithmetic is not in doubt. What came out was simply not a position. Here is why that settles the matter.

6:02Draw the positions. One, two, three, and onwards - ticks, with nothing at all between them. There is a ninety-fourth term and there is a ninety-fifth term. There is no ninety-four-point-sixth place in a list. Look at what sits on those two ticks. The ninety-fourth term is four hundred and sixty-eight. The ninety-fifth is four hundred and seventy-three. So four hundred and seventy-one falls between them, and between is not a place.

6:33Which is why you must not round. Rounding gives ninety-five, and the ninety-fifth term is four hundred and seventy-three, a different number from the one you asked about. There is a second way for this test to fail, and it is easy to miss. Take a third rule: three n minus seven. Its early terms are minus four, minus one, two. This one starts below zero. Ask whether minus seven is in it. Three n equals zero, so n equals zero.

7:08That is a whole number. There is no fraction anywhere. And it is still not a term, because there is no position zero. The list starts at one. Ask about minus ten and you get n equals minus one, which is further outside still. So the test is not whether a whole number came out. It is whether a counting number came out. Two warnings before we finish. The first: a membership question is not a trick with a fixed answer.

7:42Is five hundred and fifty-seven a term of three n minus seven? Three n is five hundred and sixty-four, so n is a hundred and eighty-eight. Yes, and there is its position. The method decides. The phrasing does not. The second warning: being large enough is not the same as being reachable. Between three and four hundred and seventy-three there are four hundred and sixty-nine whole numbers. Five n minus two lands on ninety-three of them, and steps clean over the other three hundred and seventy-six.

8:21And sometimes every answer is yes. In five n minus three, ninety-seven is the twentieth term and a hundred and seventy-two is the thirty-fifth. Last, an honest limit. Here are the first ten primes: two, three, five, seven, eleven, thirteen, seventeen, nineteen, twenty-three, twenty-nine. Their gaps run one, two, two, four, two, four, two, four, six. Nothing settles. Is there a rule? Of the shape we have been using, exactly one matches the first two primes.

8:59It says four comes next, where the primes say five. No rule of that shape matches even the first three. And if you allow a curve instead of a straight line, you can fit the first four exactly. That fit then says eight for the fifth, where the primes say eleven. Matching the front of a list is evidence. It is never proof.