Working backwards from a term to its position, or to a or d

0:00The rule for the entry in position n is a plus, in brackets, n minus one, times d. Count the letters. The first entry a. The step d. The position n. And the entry itself. Four quantities held by one equation, which means that if you know three of them it will generally hand you the fourth. That is not one technique. It is four questions wearing the same coat.

0:29Give me the first entry, the step and a position, and I give you the entry. That is the way we came. Give me an entry instead and I give you its position; give me two entries and I give you the first entry and the step together. Three of those are ordinary algebra, and the answer is whatever comes out. Solving for the position is not ordinary, and the whole of this video is why.

0:58Start with an easy one. Twenty-one, eighteen, fifteen, downwards: first entry twenty-one, step minus three. Somewhere along it sits minus eighty-one. Which position? Write the rule with the entry known and the position unknown. Twenty-one minus three times n minus one equals minus eighty-one. Take the twenty-one across. Three times n minus one is a hundred and two. So n minus one is thirty-four, and n is thirty-five. Minus eighty-one is the thirty-fifth entry, and we did not write out the thirty-four before it.

1:39You can check by walking, and I did, one addition at a time. Same list, and a question that looks different and is not. Does it ever hit zero exactly? It is falling by three, so it will certainly go negative. The question is whether it lands on nothing on the way past. Same equation. Twenty-one minus three times n minus one equals nought. Three times n minus one is twenty-one, so n minus one is seven, and n is eight.

2:14The eighth entry is zero, and there is nothing special about it. A falling list passes through zero the way it passes anything else, and keeps going. What matters is the shape of the answer. Eight is a counting number, and that is what made it a position. Hold on to that, because the next question hands us something that is not. Here is five, eleven, seventeen, twenty-three, and onwards. First entry five, step six.

2:47Is three hundred and one on it? Same move. Five plus six times n minus one equals three hundred and one. Six times n minus one is two hundred and ninety-six. So n minus one is two hundred and ninety-six over six. And two hundred and ninety-six does not divide by six. It leaves a remainder of two. The position comes out as a hundred and fifty-one thirds. That is not a mistake in your working. The fraction is the answer, and it says three hundred and one is not on the list.

3:27Because a position is not an ordinary number. There is no such place as the hundred and fifty-first and a third. The equation answered the question, by producing something a position is not allowed to be. Now the wrong way to that verdict, which is right here and will not be next time: look at three hundred and one and think, too big. But the list does not stop. Two hundred and ninety-nine is on it, at position fifty. Three hundred and five is on it, at position fifty-one.

4:04It is not too big. It is in the gap between two entries the list walks straight past. Size is not the test. But dividing exactly is not the whole of the test either, and this is the part that is easy to miss. Take the same list and ask about minus one. Minus one is five take away six: exactly one step before the list begins. Put it in the equation and it divides perfectly. n comes out as nought.

4:36Nought is a whole number. And there is no entry at position nought, because the list starts at position one. So the position carries two demands, not one. It must be a whole number, and it must be at least one. Drop the first and every number between two entries slips through. Drop the second and the entire backwards extension slips through, dividing perfectly all the way. Turn the equation round once more and it counts things.

5:10How many two-digit numbers does three divide? The first is twelve, the last is ninety-nine, and in between they go up in threes. Now the unknown is not a value. It is how many entries there are, which is the position of the last one. Twelve plus three times n minus one equals ninety-nine. Three times n minus one is eighty-seven, so n minus one is twenty-nine, and n is thirty.

5:40Thirty two-digit numbers. And be careful of the thing we were careful of last time. Eighty-seven over three is twenty-nine, and twenty-nine counts the steps, not the entries. Gaps between fenceposts, not fenceposts. Add the one back. Now a question with two unknowns in it, which sounds worse and is not. Suppose all you are told is that the third entry is five and the seventh entry is nine. Neither the first entry nor the step. But two facts about two unknowns is a pair of equations.

6:19The third entry is a plus two d, and that is five. The seventh is a plus six d, and that is nine. Subtract one from the other and the first entry cancels: four d is four, so d is one. Put that back and a is three. The list is three, four, five, six, seven, rebuilt from two entries in the middle of it, and the count of steps between them is what did the work: from the third to the seventh is four steps.

6:51And if you are handed the same position twice, there is nothing to solve. One fact is one equation, and one equation will not settle two unknowns. Here is the question that ambushes almost everybody. Ten, seven, four, down to minus sixty-two, where it stops. First, how long is it? Ten minus three times n minus one equals minus sixty-two gives n minus one equals twenty-four, so twenty-five entries. Now: which entry is the eleventh from the end?

7:30The tempting answer is twenty-five take away eleven, which is fourteen. It is not fourteen. It is fifteen. Here is why. Counting backwards, the last entry is not the zeroth from the end. It is the first from the end. So the eleventh from the end has ten entries after it, and twenty-five minus ten is fifteen. As a rule: the kth from the end is the n minus k plus one from the start.

8:02The fifteenth entry is ten minus three times fourteen, which is minus thirty-two. Had you taken fourteen you would have got minus twenty-nine, which is a real entry of a real list, just the wrong one, one step short. There is a second way, and it makes the off-by-one disappear rather than defeating it. Read the list backwards. Backwards it starts at minus sixty-two and climbs by three, because reversing a list flips the sign of its step.

8:37And now the eleventh from the end is simply the eleventh entry of this one. Minus sixty-two plus ten threes. Minus sixty-two plus thirty is minus thirty-two. The same answer, reached without ever counting how many entries there were. That is not a trick. A finite list read the other way round is another finite list, with the first and last entries swapped and the step negated. Sometimes nobody gives you a list at all.

9:11A thousand is put aside at eight per cent simple interest a year. How much interest has it earned by the end of each year? Simple interest means it is worked out on the original thousand every year, never on the interest. The first year earns eighty. By the end of the second the total is a hundred and sixty; by the end of the third, two hundred and forty.

9:38Eighty, a hundred and sixty, two hundred and forty. A progression, first entry eighty, step eighty. So the total by the end of the thirtieth year is eighty plus twenty-nine eighties, which is two thousand four hundred. Here the step and the first entry happen to be equal, which is a feature of simple interest, not a general rule. And it is worth seeing what breaks it. If the interest itself earned interest, the yearly gains would be eighty, then eighty-six point four, then ninety-three point three one two.

10:15Those are not equal and never will be. Interest on the original sum gives a progression; interest on the growing sum does not. The whole topic fits on one small table. Five rows, each the same equation, three quantities given, the fourth blank, and the blank moves. First entry seven, step three, position eight. Seven plus seven threes is twenty-eight. First entry minus eighteen, position ten, entry nought. Nine steps have to cover eighteen, so the step is two.

10:54Step minus three, position eighteen, entry minus five. Going back seventeen steps of minus three from minus five gives forty-six. First entry minus eighteen point nine, step two point five, entry three point six. The gap is twenty-two point five, which is nine steps, so the position is ten. And the fifth row hands the blank back where the first row had it, after the step, the first entry and the position have each had a turn.

11:28But look at its step. It is nought. The list is three point five, three point five, three point five, all the way to position a hundred and five. That row is safe, because it is the entry that is blank. Had it been the position, there would have been nothing to solve: on a list that never moves every position holds the same number, and the equation cannot single one out.

11:58One last thing about that equation, and it is why it is worth having at all. It does not need the list. One, four, seven, up to three thousand and one. How many entries? Three thousand over three is a thousand, so a thousand and one. Or five, eleven, seventeen, up to two thousand nine hundred and ninety-nine: five hundred. Or a hundred, ninety-seven, ninety-four, falling to minus eight hundred: three hundred and one.

12:31I asked a routine to write those lists out and look, with a hundred and twenty entries to work with. It found none of the three. The equation answered all three, in the same three lines each time. That is the trade. Writing the list out is always right and bounded by your patience. The equation has no horizon. Which is why the two demands matter so much. When the list is not in front of you, the whole number and the at-least-one are all you have.

13:06None of that was taken on trust, and the obvious way to check it would have proved nothing. Work an entry out with the rule, solve for its position with the rule, and be pleased when the position comes back: that is one equation used twice. So every list here was walked instead, one addition per step, and every position found by looking along it, one entry at a time. No division in the looking, no formula in the walking.

13:38Seventy-two progressions were built, and three and a half thousand numbers put to them by four routes through one line of code: walking and looking; solving and demanding a whole number at least one; solving and accepting any whole number; and judging by size. The first two agreed on all three thousand five hundred and twenty-eight. The third let in two hundred and sixteen numbers that are on no list, and every one of them divides perfectly and sits before the first entry.

14:14The size test threw away two thousand five hundred and ninety-two genuine entries. The guards were handed what they must refuse. A step of nought went to the position solver, on a list where the number really is at all forty positions, and it refused. Two entries naming the same position went to the recovery five hundred and four times, and it refused every one. And the routes wrong on purpose were not merely recorded as wrong: where they land was recorded. The one that forgets to reduce the position lands on the entry one step before the first, all one thousand five hundred and twelve times.

15:01One equation, four quantities, and the blank can go anywhere in it. Three of the four are ordinary algebra. Solving for the position is not. The answer must be a whole number, and it must be at least one, and those two demands catch two different mistakes. A fraction is not an error in your working. It is the answer, and the answer is no. Size tells you nothing, because the list does not stop where you stopped reading.

15:32Counting from the far end costs you one, because the last entry is the first one you count. And two entries anywhere in a progression are two equations, which is enough to rebuild the whole thing. Every entry reachable and every entry testable. What is missing is the total, and adding a thousand entries one at a time is exactly the work we keep refusing to do.