Times tables chess (where Bragg meets Eratosthenes)

This page is really just a general undirected, bottom-up dabbling, just for the fun of seeing where it might lead, and irrespective of how much it involves re-inventing of the wheel, and treading of ground that others have already trod. It is drawn from the pages of my log-book, dating back to May 1996 (for HCF and the Babylonian function) and March 1995 for the proposal for the Bottle experiment. Being bottom-up, this page does tend to go off at a tangent, and to do a bit of a Montecarlo walk.

Times tables revisited

Remembering back to primary-school days, a times table, in the singular, is a simple column of the selected integer multiplied into each of the other integers in turn. It says litterally what happens if you add together the given integer that number of times, using repeated addition. So, "4 times 7" is a shorthand for "7+7+7+7".

In the past, in the UK, children had to learn the times tables up to 12, since the 12 times table was so useful when having to calculate the number of inches in a foot, pennies in a shilling, months in a year, eggs in a dozen.

The times tables, in the plurial, is the result of laying each of the twelve times tables, side by side, to form a grid. Thus the four-times table, above, is found in the fourth column of this grid. Later, this grid is likened to a chess board, but it is convenient, straight away, to refer to the elements of the grid as "cells" rather than as "squares", since there will be a need to talk about those cells that contain integers that are a perfect squares.

Lessons learned

One of the first lessons learned, at primary school, is the commutative property of multiplication. The rows are symmetrical with the colums, "x times y" is the same as "y times x", and there is a consequent mirror symmetry about the leading diagonal.

Secondly, learning how to fill in this grid, by heart, was a major exercise, filled in a bit like a crossword puzzle, doing the easy rows and columns first (the first, second, fifth, tenth and eleventh rows, and the first, second, fifth, tenth and eleventh columns) and then filling in the gaps in increasing order of apparent difficulty. Ultimately, the cells for "8 times 7" and for "7 times 8" remained as the last ones to be filled in. But, this made those two cells more and more memorable, until they were no longer the most difficult cells to fill in, in the same way that there is a proof that there are no uninteresting natural numbers.

Thirdly, it became apparent that there were many, many intricate patterns to be found. The "10 times table" and "11 times table" are the most blatant (after the "1 times table"). The patterns in the "9 times table" were particularly noted, not least as a way of remembering the table (for "9 times x", the tens digit is (x-1) and the units digit is 9 minus the tens digit).

Fourthly, the leading diagonal therefore became singled out as an interesting part of these patterns as a consequence of the commutivity property of multiplication. And so, the special nature of the perfect squares was highlighted (in the grid below, in violet). But this, then, showed up a pattern of near misses of the perfect squares, such as "8 times 6 is 48" being one less than the nearby "7 times 7 is 49". These have been highlighted in cyan, in the grid below, and are discussed further for the bishop's moves.

Jacquard looms for Wilton carpets

One way of highlighting the patterns in the times tables is to find ways of colouring in the cells. One way is to look at the units digit in each cell (the result of "a times b, modulo 10"). The cell would then be coloured in with one of ten specific colours. This could then be used as the instructions for creating the cards of a Jacquard loom, such as for making a large Axminster or Wilton carpet. In the following table, an extra initial column and row has been added for the "0 times table" partly to make the overall pattern more apparent.

The grid can be extended indefinitely, of course, to cover as many columns and rows as there are integers (of which there is a countably infinite number). So, the patten within the black grid (the "0 times table", the "10 times table", the "20 times table", and so on) is repeated indefinitely in the horizontal and vertical directions.

Since the pattern repeats indefinitely, in both directions, we can determine the frequency of occurrence of each colour just by analysing one 10-by-10 section of it. We find that black(0) occurs with a frequency of 27%, brown(1)=4%, red(2)=12%, orange(3)=4%, yellow(4)=12%, green(5)=9%, blue(6)=12%, violet(7)=4%, grey(8)=12% and white(9)=4%. If the carpet were hung like a tapestry on the wall, and used as a dart-board, these would be the frequencies of the scores. We also note the symmetry from 1,2,3,4 against 9,8,7,6. 75% of the numbers in the time tables, therefore, are even numbers (27% of them end in 0, and 48% in one of the other even digits), and 25% are odd numbers (9% of them end in 5, and 16% in one of the other odd digits). This is consistent with the observation that only ODDxODD gives an ODD number, while ODDxEVEN, EVENxODD and EVENxEVEN all give EVEN numbers.

Triangle numbers

Since the perfect square numbers seem important, it is then natural to wonder about the perfect triangle numbers. Since n² amounts to n+n+n+..+n (where there are n terms added together), triangle(n) would need to be n+(n-1)+..+3+2+1 (again with n terms added together). Since the area of a n-by-n square is n², it would at first seem reasonable to assume that the area of the triangle described by the triangle numbers would be ½n². However, when stacking up wooden cubes in a triangle, for example, there is a quantisation error; each row has an extra half a block pertrudes from the hypothenuse of the triangle. Thus the triangle function equates to ½n²+½n, namely n.(n+1)/2.

The more traditional way at arriving at the same conclusion is to notice that two such block-based triangles slot together to make a rectangle that is n units wide and (n+1) units long.

Recurring decimal fractions

All rational fractions (one positive integer divided by another) can be represented as decimal fractions, provided that we have a notation to indicate "recurring". Normally, this is found simply by going ahead with the procedure for long division, and stopping when the same state as earlier occurs again (not just the same digit contributed to the quotient, but the same remainder carried down to the next step). When the divisor only has 2s and 5s as its prime factors, the recurring part is merely trailing 0s.

Another test is to write out the times table for the denominator, all the way up to the first product expressed only with 9s. For example, for 368/471, this would involve writing: 1x471=471; 2x471=942; 3x471=1413; 4x471=1884; … Nx471=999..999. In reality, this can be done the other way round: first check that 9 is not divisible by 471, then try 99, then 999, and so on (where "not divisible by" means "leaves a non-zero remainder'). Then, count how many 9s there are in the last entry. That is the number of decimal places before the recurring pattern starts again. In the case of this example, it appears that the last entry would be an integer with seventy-eight 9s (that is, (10⁷⁸-1). So, 368/471 is represented as a decimal fraction that is recurring in its last seventy-eight digits (remembering to pad the numerator out with leading zeros if it does not have the right number of digits, as in 1/11 being 09/99 not 9/99).

Babbage differences

Although he was not the inventor of the differences method, Babbage is the one who really put them on the history-of-computing map.

The first row of differences, (D₀ D₁ D₂ D₃ D₄ D₅ D₆ ...) for y=x³ is (0 1 6 6 0 0 0 ...), for y=x² is (0 1 2 0 0 0 0 ...), for y=x¹ is (0 1 0 0 0 0 0 ...), for y=x⁰ is (1 0 0 0 0 0 0 ...). (Incidentally, it is not a coincidence that, for an n^th-order polynomial, the last (constant) value before the sea of zeros, in the D_n column, has the same value as dⁿ(xⁿ)/dxⁿ, namely fac(n).) The differences settings can be combined by simple, linear superposition, so the first row of differences for y=3x³-4x²+5x-1 is (-1 4 10 18 0 0 0 ...).

In the context of the leading diagonal of the times tables, any given perfect square integer, x², can be visualised as a x-by-x array of cubies (to borrow a term from the Rubrik's cube). The next perfect square, (x+1)², involves first adding x cubies to one side, to make a (x+1)-by-x rectangle, then adding (x+1) cubies to the adjacent side, to make the (x+1)-by-(x+1) square. Thus, to go from (x+1)² to (x+2)² involves adding x+1+x+2 more cubies, which is an odd number of cubies, and is two more than for going from x² to (x+1)². This explains the "2" in the Babbage differences for x², and also the relationship between the perfect squares and the sum of the odd numbers.

Babbage's later, more adventurous project would indeed have been a veritable weaver of numbers, capable of a representation of the Mandelbrot set, or an abstract representation of any other computation; but even the Difference Engine could generate the carpet examples that are illustrated on this page. It all highlights some of the underlying patterns thrown up by mathematics.

The n-star tables

Defining multiplication to be a one-star operator, x*y, the one-star tables are simply those shown above, namely the multiplication tables. The two-star tables are then those for the exponentiation operator, x**y (and hence the origins of the notation can be seen to be rooted in those of Fortran-IV). The zero-star tables are those of the addition operator, x+y, and gives the following carpet pattern:

The carpet (modulus-10 or last digit) view for the exponentiation operator finally highlights that y is traditionaly platted along the horizontal axis, and x along the vertical axis, consistent with the way that the times tables are normally listed vertically, and then placed next to each other, side-by-side. This is not noticeable for the multiplication and addition tables, due to the property of commutivity.

The carpet view for the exponentiation operator shows that there are indeed repeating patterns. The to-the-power-of-three table cycles through the sequence (0 1 8 7 4 5 6 3 2 9), the to-the-power-of-two table cycles through the sequence (0 1 4 9 6 5 6 9 4 1). and the to-the-power-of-five table cycles through the sequence (0 1 2 3 4 5 6 7 8 9). This latter can be proved using the method of Babbage differences, and is really just a special case of proof by induction. It leads to the further observation that all values of n⁵-n are divisible by 30. The rows are even simpler, being either a string of 0s, 1s, 5s, 6s, or alternating (4 6) or (9 1), or (2 4 8 6), (3 9 7 1), (7 9 3 1) or (8 4 2 6); and, of course, there is no symmetry between rows/columns.

Another point brought home by this is that only the zero-star and one-star operators are commutative (x+y=y+x and x*y=y*x) and further that all the higher star-count operators must be treated as right-associative: x***3=x**x**x=x^(xx). Thus, for the zero-star operator, 3+4=7; for the one-star operator, 3*4=12; for the two-star operator 3**4=81; and for the three-star operator, 3***4=3**(3**(3**3)) = 3**(3**27) = 3**7625597484987 = 1.258x10^{3638334640024}. The definition of (op) would be x(op)y=x(op-1)x(op)(y-1), noting that op-zero, the zero-star operator, is addition, and x(op)0=1 for all op except for the zero-star operator. Similarly, 1(op)y=1 for all op except for the zero-star operator, while 0(op)y=0 and x(op)1=x for all op.

The usual test for commutivity is whether x(op)y-y(op)x=0. In the physical world, this fails even for multiplication of quantum operators (as opposed to mulltiplication of natural numbers). For situations in which geometric superposition is required, instead of linear (arithmetic) superposition, the test would be whether x(op)y/y(op)x=1.

Both of these rely on the availability of an inverse operation; and as soon as these are contemplated, forbidden regions are brought to light. For example, z=x+y works for all choices of natural number for x and y, to find a value for z, but fails for some choices of x and z to find a value for y. In a sense, the minus sign stands for an error message for the algorithm, and indeed at primary school was initially phrased as "can't be done" (for example for "4 take-away 7"). As soon as that error message has a magnitude attached to it, though (such as the 3 of -3) it is possible to start opening up a new branch of mathematics. Similarly for multiplication, z=x*y, where the error message encountered initially at primary school was "won't go" (for example for "7s into 12") and represented by the remainder sign, r, or vulgar fraction (rational fraction) sign, or decimal point. Again, attaching a magnitude to that error message then allows a new mathematics to be performed.

Because of the property of commutivity, there is only one inverse operation for the zero-star and one-star operators, but two for all the others: finding x or y from z=x+y is the process of subtraction; finding x or y from z=x*y is the process of division; finding x or y from z=x**y is the process of taking the y^th root or taking logs (to base-x). The root of x*2=z is z*0.5, and that of x**2=z is z**0.5, but this does not work for x+2=z; however, let us assume that it continues to work for all other op, and that z***0.5 is the correct representation for the inverse operator to x***2=z; if not, then the following table merely needs to have the label of the final column changed to the correct representation. The values in the final column grow very slowly, only reaching the value of 3 in the 27^th row, 4 in the 256^th row, 5 in the 3125^th row, and so on.

If the labelling of the above table is indeed correct, it seems that with positive values of x, and rational values of y (and, by interpolation, for irrational values, too) x***y evaluates to positive values. If the inverse operator is applied to a negative value, an error message will initially be invoked. For exponentiation, that message is symbolised by the i sign (the square-root of minus-one) both for x^2=-6 and 2^y=-6, for example.

ln(x) is a member of the xⁿ family

It is possible to get a long way towards deriving Stirling's formula for an approximation of the factorial function simply by noting that the unit-step secants immediately after and before fac(n) have gradients of fac(n).((n+1)-1) and fac(n).(1-1/n), respectively. It is to be assumed that the tangent at factorian(n) therefore has a gradient that is some sort of a mean of these. Instintively, the geometric mean would appear to be the appropriate one, but this merely leads the formula back in a tautological circle. Instead, by taking the aritmetic mean, we obtain:

d(approxfac(n))/dn = fac(n).√(n-1)

This yields:

approxfac(n) = (2/3).exp( (n-1)(3/2) )

As often noted, fac(n) has a value that is between exp(n) and exp(n²). The above approximation is along the right lines, but the correct derivation of Stirling"s approximation gives the exponential term as exp(n.ln(n)).

That x.ln(x) lies between x¹ and x² is further supported by the generalisation of ∫xⁿ.dn. This resolves to x^n-1/(n-1) for all n, except n=-1, where it resolves to ln(x). Once again, ln(x) is the case between x^-1 and x¹. This is also confirmed further when noting that a mean (in general) can be defined as:

f(mean) = Σf(xi) / n
So:
mean = inv.f( Σf(xi) / n )

For most of the commonly used means, f=[x]{x^m}. The arithmetic mean has m=1, the harmonic mean has m=-1, and the root-mean-square has m=2. Between the first two, it turns out that this definition is not helpful in generating the geometric mean for m=0; instead, it requires f to be ln.

Chess-board view of the times tables

It is useful to invert the times tables, so that they build from the bottom upwards instead of from the top downwards. This places the origin in the bottom left corner, consistent with Cartessian coordinate notation, and the white player's view of a chess-board. Also, it is convenient just to handle up to the "10 times table", on the understanding that it actually continues out to infinity in both directions.

To obtain a chess-board, or chequer-board, pattern, the Jacquard loom can be given the instruction "if( IsOdd(RowNumber()+ColumnNumber()), White, Black )", where the IsOdd(n) function can be replaced by n%2, and the plus sign is effectively performing an exclusive-or operation. (In general, the IsNotDivisableByN(n) function can be replaced by n%N.) The carpet pattern in an earlier section would be generated from something along the lines of "switch(n%10){0:Black; 1:Brown; 2:Red; 3:Orange; 4:Yellow; 5:Green; 6:Blue; 7:Purple; 8:Grey; 9:White}".

It is worth noting, too, at this point that the period of the pattern in the carpets shown on this page are an artifact of our arbitrarily chosen number base. Each patten repeats every tenth row or column because of the 10 in the %10, modulus division operator, that we have chosen at the last moment before rendering the pattern. Incidentally, x%y is just another operation that could have its carpet view depictd like the other before, and that "lastdigit=[y] {x%y} 10" is an example of a higher-order function that requires two arguments, but receives only one, so remains as a one-argument function in this sugared lambda calculus notation; and designing a programming language that can assign a multiple-byte stretch of machine code, representing a function definition, to a variable (in this case one called 'lastdigit') is no more surprising that to have one to assign a multiple-byte object, such as "Hello, World." to a variable called 'mystring'.

Rook's moves

A rook can move any number of cells horizontally (±k,0) or vertically (0,±k). Starting from the origin, a rook's move of (x,0) followed by one of (0,y) arrives at a cell that contains the value "x times y". Alternatively, if the (0,y) is made first, followed by (x,0), the rook arrives at the same cell.

Starting from the cell (x,y), with a value of x*y, the rook will end up in a cell (x±k,y) or (x,y±k), with a corresponding value y*(x±k) or x*(y±k).

So, the commutative property of travelling round two sides of a rectangle two commutative properties of multiplication, since we could land in a cell with the same value by tracing out the path (y,0) followed by (0,x), and hence alternatively yet again by following the path (0,x) followed by (y,0).

This seems to be related, in an abstract way, to the closed form of mathematics: that there are often multiple ways to perform a derivation, and hence to check the results of one again the others. This is seen in Maxwell's equations, which can be applied in their integral form or differential form, to arrive at the same result. Coumatative properties are also used in the error-checking techniques like double-entry bookkeeping. Parallax is the manifestation of non-commutivity: the subject's view of the object is different when passing through differing routes. Alice performing an experiment, while Bob performs the control experiment (travelling close to a blackhole, or near to the speed of light, or subjecting plants to different conditions in a biology experiment). Watching world news by comparing broadcasts from different countries.

Bishop's moves

Similarly, a bishop can move (±n,±n) or (±n,-±n). In its first move, starting off in the (0,0) cell, it can reach any of the cells on the leading diagonal (the perfect squares). From there, in its second move, it can either continue to travel on the SW-NE leading diagonal, or can branch off perpendicularly on one of the SE-NW or NW-SE diagonals. In two moves, the chess bishop can reach any of the black cells on the board. The bishop can only reach half of the cells on the board. The other half (the white cells) are not accessible to it. If the bishop moves by i cells to the NE on its first move, it ends up on the (i,i) cell, whose value is i². If it then moves by j cells to the SE on its second move, it ends up on the (i+j, i-j) cell, whose value is (i+j)(i-j), or i²-j². (If, instead, it had moved j cells to the NW on its second move, it would have ended up on the (i-j)(i+j) cell, whose value is also i²-j². This, of course, is because of the symmetry of the times tables about the leading diagonal, and the commutative property of multiplication, in which x*y=y*x).

Of course, this merely observes that (i+1).(i-1)=i²-1. The cyan cells in the table, eearlier, represent (i+1).(i-1)=i²-1², and the violet cells represent (i+0).(i-0)=i²-0². Generalising further, we can colour in all of the cells, the same colour, that represent (i+j).(i-j)=i²-j². This is what led to the drawing out of the multiplication tables on a chess board, with 0x0 at the (0,0) cell, in the bottom-left, black cell. The perfect squares lie on the leading diagonal, a single bishop's move away, and all black cells. Moving from one to the next involves the value of the cell incrementing by 2i-1 (1, 3, 5, 7,...). Note that Sigma(i=0,n; 2n-1) equals (2.m²-2m)/2-m. If the bishop's next move is at 90 degrees to the first, it travels towards one of the axes at a 45-degree angle. Now, the value, in subsequent cells, decreases in steps of 2j-1 (-1, -3, -5, -7,...). Half of the cells in the times tables are of the form i²-j², and accessible in two moves by a bishop starting at (0,0), and are arranged as a chequered pattern, like the black cells. These cells are all EVENxEVEN or ODDxODD. For these, p.q = ((p+q)/2)²-((p-q)/2)². The white cells are EVENxODD and ODDxEVEN, and need a 1-cell rook's move (+ or - n or m of the times table that the piece is currently in). We notice, looking along the 1-times table column (the second one from the left) that all the odd numbers (and hence all of the prime numbers, except the value 2) lie on black cells, and hence are of the form n²-(n-1)², which factorises to 2n-1. All numbers that can be expressed as EVENxEVEN, are divisible by 4, and are of the form n²-(n-2)², which can be factorised to 4n-4.

Starting from the cell (x,y), with a value of x*y, the rook bishop end up in a cell (x±k,y±k), with a corresponding value (x±k).(y±k), where the two plus/minus signs are independent.

Supercharged-knight's moves

We can now imagine a chess piece that can move along lines that are less angled than the diagonals, such as (x±k,y±2k) or (x±2k,y±k), or even (x±k,y±3k) or (x±3k,y±k). In effect, the knight's move is a restricted version of the former, where k can only be 0 or 1.

Let us define a type-i chess-piece that simulates its overall move by decomposing it into two Cartessian, rook moves, with either the first i times bigger than the second, or the second i times bigger than the first.

new position = (x±i.d, y±d) or (x±d, y±i.d)

From this, a bishop is a type-1 chess piece, a rook is a type-0 chess piece, and a super-charged knight (able to travel indefinitely far in a single move) is a type-2 piece. The embroidery example, later, traces out the moves of pieces of type-0, type-11, type-5, type-3, type-2, and type-1, not forgetting the type-1.4 piece for the move from (0,7) to (5,0). However, it is probably better to leave this last one expressed as a rational fraction, a chess piece of type-7/5 (one that moves 7n cells parallel to one axis for every 5n cells parallel to the other axis).

new position = (x±i.d, y±j.d) or (x±j.d, y±i.d)

Emulated moves

Rooks take two moves to emulate a bishop's move, (x±k,y) plus (x',y'±k),

Similarly, bishops take two moves to emulate a rook's move, (x±k/2,y±k/2) plus (x'±k/2,y'-±k/2). When k is even, and so k/2 is integer, the piece ends up on the same-coloured cell; while if it is odd, so that k/2 is half-integer, the piece ends up on the opposite-coloured cell.

However, though the destinations are the same in each case, the journeys are not, when logging the values of each of the cells that are traversed. The double rook move, for example, passes through the cells on the adjacent and opposite sides of a right-angled triangle, whereas the single bishop move passes through the cells on the hypotenuse.

The one-times table

The one-times table has turned out to serve as the indexing column and row for the matrix.

All of the odd cells in the one-times table column and row are coloured. So, starting from the (0,0) cell, the chess bishop can reach any cell whose value is an odd number in exactly two moves. Thus, every odd number (and every prime number, except 2) can be expressed in the form n=i²-j². We can further note that j=(i-1) since the two moves are almost symmetrically equal, except that the first move starts in the row for the 0-times table, and the second move ends in the row for the 1-times table. This resolves to ((n+1)/2)2-((n-1)/2)², which evaluates, as expected to n.

We can see that even numbers, too, can be made to fit ((n+1)/2)²-((n-1)/2)², simply by letting the bishop take half steps as well as whole steps, provided that the second move balances out the half steps again, to lead to a whole number of steps over the two moves.

Testing for prime numbers

It is inevitable that any exploration around the times tables will eventually looking at the prime numbers, at some point.

The prime numbers can be found in only two places: in the second column from the left (the column for the one-times table) and in the second row from the bottom (the row for the something-times-1 commutative equivalent). (From now on, because of this symmetry, caused by the commutative property of multiplication, the rows will be treated as times tables, just as the columns are).

Every natural number appears in the one-times table. Prime numbers are those that only appear in the one-times table, and no-where else in the body of the matrix. This supports the view of Hofstadter 1980) that the prime numbers are merely the ground that remains after the figure (the values in the later times-tables) has been removed. There is no function that will return the value of the i-th prime number, p_i

With one exception (the value 2), all of the prime numbers are odd, and all of the odd cells in this column and row are coloured. So, starting from the (0,0) cell, the chess bishop can reach any prime number cell, except that for 2, in exactly two moves.

The traditional search for the first factor of n involves "FOR i:=2 TO sqrt(n) DO IF (n%i)=0 THEN abort;" (where the % sign represents the modulus-division, or remainder, operator). This can be further sped up, to "IF (n%2)=0 THEN abort ELSE FOR i:=3 TO sqrt(n) STEP 2 DO IF (n%i)=0 THEN abort". This is a search for n=a.b where a can be from n to √n, and b can be from 1 to √n.

Powers of triangle numbers

The triangle function is to addition as the factorial function is to multiplication: triangle=[n] {if(n<1;0;n+triangle(n-1)} and evaluates to n.(n+1)/2.

Taking the modulus-2 of these values yields the repeating sequence (0 1 1 0). Taking the modulus-4 yields the repeating sequence (0 1 3 2 2 3 1 0). Taking the modulus-8 yields the repeating sequence (0 1 3 6 2 7 5 4 4 5 7 2 6 3 1 0). Taking the modulus-16 yields a repeating sequence that starts with (0 1 3 6 A F 5 C 4 D 7 2 E B 9 8 and then back down again).

Further, we can define the pwrtriang function: pwrtriang=[n,m] {if(n<1;0;n**m+pwrtriang(n-1,m)}. For this, define a function to return the i^th coeffient from the n^th row of Pascal's triangle, pt=[n,i] {fac(n)/(fac(i)*fac(n-i))}.

pwrtriang(n,0)=pt(n,1)
pwrtriang(n,1)=pt(n+1,2)
pwrtriang(n,2)=pt(n+2,3)+pt(n+1,3)
pwrtriang(n,3)=pt(n+3,4)+4.pt(n+2,4)+pt(n+1,4)
pwrtriang(n,4)=pt(n+4,5)+11.pt(n+3,5)+11.pt(n+2,5)+pt(n+1,5)
pwrtriang(n,5)=pt(n+5,6)+26.pt(n+4,6)+66.pt(n+3,6)+26.pt(n+2,6)+pt(n+1,6)
pwrtriang(n,6)=pt(n+6,7)+57.pt(n+5,7)+302.pt(n+4,7)+302.pt(n+3,7)+57.pt(n+2,7)+pt(n+1,7)

Tools for calculating mean path lengths across a grid

The following functions were originally defined for helping to find the mean path length for data packets in a distributed grid with ranxomy-placed faulty nodes, but can be used more generally for paths across a chess board of cells of different values.

Sum of an arithmetic progression, sap(a,b,n):
= a + (a+b) + (a+2b) + (a+3b) + ... + (a+(n-1).b)
= a.n + n.(n-1).b/2

Sum of a geometric progression, sgp(r,n):
= 1 + r + r2 + r3 + ... + rn-1
= rn-1/(r-1)

Sum of an offset geometric progression, sogp(s,f,a,b,r):
= Sum{ a.rb+i ; i= s to f }
= a.rs+b.sgp(r,f-s+1)

Sum of a triangle series, triangle(n):
= sap(1,1,n)
= n.(n+1)/2

Sum of a triangle with each term raised to a power, pwrtriang(n,d):
= Sum{ id; i= 1 to n }
pwrtriang(n,0) = n
pwrtriang(n,1) = triangle(n)
pwrtriang(n,2) = (n-1).n.(n+1)/3 + n.(n+1)/2
pwrtriang(n,3) = triangle(n)2
pwrtriang(n,4) = (n5-n)/5 + 2n2(n+1)2/4 - 2n.(n-1)(n+1)/3 - n.(n+1)/2

Sum of a triangle-weighted geometric progression, stwgp(r,n):
= 1.r0 + 2.r1 + 3.r2 + 4.r3 + ... + n.rn-1
= Sum{ ri-1.sgp(r,m-i+1); i= 1 to n }
= ( n.rn-1 - (n+1).rn + 1 ) / (r-1)2

Sum of an offset triangle-weighted geometric progression, sotwgp(s,f,a,b,r):
= Sum{ (a+i).rb+i ; i= s to f }
= rb+s.( (a+s-1).sgp(r,f-s+1) + stwgp(r,f-s+1) )

Sum of a quadratic-triangle-weighted geometric progression, sqtwgp(r,n):
= 1.r0 + 4.r1 + 9.r2 + 16.r3 + ... + n2.rn-1
= Sum{ (2n-1).ri-1.sgp(r,m-i+1); i= 1 to n }
= n.(n+1).rn/(r-1) - (r+1).stwgp(r,n)/(r-1)

Sum of an offset quadratic-triangle-weighted geometric progression, soqtwgp(s,f,a,b,r):
= Sum{ (a+i)2.rb+i ; i= s to f }
= rb+s.( (a+s-1)2.sgp(r,f-s+1) + 2.(a+s-1).stwgp(r,f-s+1) + sqtwgp(r,f-s+1) )

Testing for perfect squares

Now we see that the test for primes could be turned into a search for n=c²-d², where c can be from (n+1)/2 to √n and d can be from √n to 0. This involves testing if c²-n is a perfect square each time (perhaps allowing half steps, as indicated for the "emulated moves"). By representing the integers in binary, we know that those ending in 10 or 11 (1..10 and 1..11) are definitely not perfect squares. Those ending in 00 need to be divided by four (shifted two places to the right) and then tested again. Integers that end in 01 (1..01) might or might not be perfect squares, and need to be analysed further. (This property comes as a result of f²-f-m=0 having solutions at f=(-1±√(1+4m²))/2).

Setting the D₀ column to the list of the perfect squares (0 1 4 9 16 25...), the D₁ column of the Babbage differences starts (1 3 5 7 9...) so the D₂ column consists of the sequence (2 2 2 2...) so all subsequent columns beyond that just consist of 0s. If we consider only the sequence of odd squares (1 9 25 49 81...) the D₂ column becomes (3+5 7+9 11+13...): that is, 1+2n+1+2(n+1), which is 4((n+1). The values in this column are all divisible by eight. Since the first odd perfect square is 1, it follows that for all the others, n²-1 is divisible by 8. It follows that n²-1 cannot be a perfect square for sufficiently large odd values of n (n>1, in fact). (The binary integer that represents n² would end in b..xyz001, so subtracting 1 from it, and dividing by 4 would give an integer ending in b..xyz0, unless z is 0, in which case we could divide by 4 again). It also follows that n²-4 cannot be a perfect square for sufficiently large odd values of n (n>1, in fact). (The binary integer that represents n² would end in xxx001, so subtracting 4 from it would give an integer ending in yyy101). It also follows that n²-9 cannot be a perfect square for sufficiently large odd values of n (n>1, in fact). (The binary integer that represents n² would end in b...xyz001, so subtracting 9 from it, and dividing by 4 would give an integer ending in b...xyz0, unless z is 0, in which case we could divide by 4 again).

All binary integers ending in 010, 011, 100, 101, 110 or 111 are definitely not perfect squares. Those ending in 00 need to be divided by four (shifted two places to the right) and then tested again (which is the same as saying that squares of even integers are divisible by four, so to test whether a given integer is a perfect square, it is divided by 4 as many times as it will go, until an even integer that is only divisible by two (and hence not a perfect square) is obtained, or else an odd integer that might be a perfect square of an odd number, which can only be the case if the binary integer ends with 001). The "might be" arises because the condition, though necessary, is not sufficient.

The next step would involve testing for membership of 1+8.triangle(i), which still involves taking the square-root in the usual equation for solving quadratics. At least the problem has been made smaller, though; the above tests filter out the cases that are obviously not perfect (integer) squares.

More on the double bishop’s move

Continuing from the bishop's move and its further compounding to a double bishop's move, x = m²–(m-n)², where m = (1/2).( x/n + n ). Starting at (0,0), the bishop moves m places along the leading diagonal, and then n places perpendicular to it, arriving in a cell in the n-times table (black cells only). If n is odd, we get integer values of m for all odd multiples of n. If n is even, we get integer values of m for all even multiples of n. Therefore, we can get all odd values of x, setting n=1, and all multiples of 4, setting n=2. No other values of n can get us to more values of x than this. This is reminiscent of the test for perfect squares, in the previous section, except this time we can get any value of x from the form 1...00 (binary), or 1...01, or 1...11, but not those of the form 1...10.

n = ((n+1)/2)²–((n-1)/2)², says that all of the integers in the one-times table can be reached from (0,0) by a bishop’s move up the leading diagonal, followed by a second move perpendicular to this (though this might entail a half-integer move if the target is an even number, hence on a white cell). If n is also divisible by m, where m>2, then (n/m)+m-1=(((n/m)+m)/2)²-(((n/m)+m-2)/2)² over-shoots, and lands in the one-times table. Instead, we stop at n=(n/m).m =(((n/m)+m)/2)²-(((n/m)+m-2m)/2)² =(((n/m)+m)/2)²-(((n/m)-m)/2)². n is odd, m is odd (since it is a prime that is greater than 2) and n/m must also be odd (and also an integer). (n/m + m -1) is the point to which we over-shot, and is also odd.

Starting from the origin, the bishop's first move incurs a repeated incementation by (1, 3, 5, 7, 9, 11...), namely the D1 Babbage differences column for x². If hte second move is orhogonal to the first, it incurs a repeated decementing by (1, 3, 5, 7, 9, 11...); else if it continues on from the first move, it incurs a continuation of the incrementing (13, 15, 17, ...); else if it involves going back the way the bishop had original come, it incurs a repeated decementing by (11, 9, 7, 5...).

Equi-perimeter rectangles

The point (8,7) defines a rectangle, of area 56, from the origin (0,0). The point (9,6) defines a rectangle, of area 54, but with the same perimeter as the previous rectangle (8+7+8+7=9+6+9+6). The remote apex has been moved one cell to the SE.

SE: ΔA=n.(n+i)-(n-1).(n+i+1)=i+1
NW: ΔA=n.(n+i)-(n+1).(n+i-1)=-i+1
NE: ΔA=n.(n+i)-(n+1).(n+i+1)=2n-i-1
SW: ΔA=n.(n+i)-(n-1).(n+i-1)=2n+i-1

Prime numbers only exist on the table at (1,p) and (p,1). They can be got to by a SE move from (p-1,2) or a SW move from (p+1,2). The ΔA is p-2 or p+2 smaller, or by ratio 2-2/p or 2+2/p smaller. But, this does not really help, since in this, p does not need to be prime, and is true for all integers. The near-misses to the perfect squares also use these equi-perimeter rectangles, and also all of the destination cells of the bishop’s second move along the NW-SE diagonal.

Sieve of Eratosthenes

We could implement the sieve of Eratosthenes by defining a function that generates the infinite list of positive integers greater than 1, then passing it through successive applications of a function outputlist=eliminateevery(p,inputlist). The correct way of doing this would be for the eliminateevery function to set the eliminated members of the list to null; the preferred solution would be for it simply to remove those elements from the list, thereby leaving just a list of prime numbers, without null values dispersed between them.

Unfortunately, this does not work. It works for outputlist=eliminateevery(2,initiallist); it also works for outputlist=eliminateevery(3,eliminateevery(2,initiallist)) because the two-times table takes the form of a symmetrical square-wave, and can have the eliniation of what remains of the three-times table from that. This, though, generates an asymmetric result that does not allow for a simple process for the elimination of what remains of the five-times table.

Let us trace the state of the list at successive applications of the eliminateevery function, starting with the initial list:

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ... 

First, we take off the head of the list, and add it to the list of prime numbers, and then eliminate every 2nd element after that:

3 5 7 9 11 13 15 17 19 21 23 25 27 28 31 33 35 37 ... 

Again, we take off the head of the list, and add it to the list of prime numbers, and then eliminate every 3rd element after that:

5 7 11 13 17 19 23 25 28 31 35 37 41 43 47 49 53 ... 

We still take off the head of the list, and add it to the list of prime numbers, but this time we alternately eliminate every 7th or 3rd element after that:

7 11 13 17 19 23 28 31 37 41 43 47 49 53 59 ... 

So, the first question is how the algorithm is to know to remove the 7th element first: the answer is because it is visibly the distance to the 5² term. The next question is how the algoritm is to know that it is the 3rd element after that that is to be eliminated; the answer is because it is visibly the distance to the 5x7 element, and the algorithm already knows that 7 is going to be the next prime number since it is now at the head of the list.

The 7th element after that is the term for 5x11; again, the list that is currently being generated is already showing 11 as being the prime after next. Looking back at the previous lists, though, we see that this process is not being applied trivially; in the list resulting from the elimination of the two-times table, the first elimination was indeed the term for 2², and then for 2x3, and 2x5, but then before the elimination of the one for 2x7, there was one for the 2³ term. Later, for the definition of the Babylonian function, there is a need for a test whether n is a power-of-a-prime, and this seems connected, here. This glitch occurs later and later in later lists, as larger and larger primes are being handled.

For the next version of the list, we start as usual by taking off the head of the list, and adding it to the list of prime numbers. The next element to be eliminated, 7², is found at the 12th position in the list.

11 13 17 19 23 28 31 37 41 43 47 53 59 ... 

After that, the algorithm would need to eliminate the terms for 7x11, 7x13, and so on, up to 43x7², until the first glitch involving the removal of the 7³ term. A later glitch will certainly occur at the 7⁴ term, but before that there are also the potential glitches for 11x7², 13x7² to consider.

There seem to be parallels between these glitches in the desceptively smooth process, like finding unexpected harmonics with Lissajou figures.

Clusters of primes

Occasionally, prime numbers occur in clusters, or at least in pairs. Examples include 5 and 7, 11 and 13, 17 and 19. Indeed, by necessity, the number pair must straddle a number from the 6-times table (so that neither of the members of the pair is divisible by 2 or 3). However, though a necessary condition, it is not a sufficient one (since this has ignored the effect of divisibility by other prime numbers).

In effect, we have a partial application of the sieve of Eratosthenes (performed for the first two primes only). Another partial application of the sieve of Eratosthenes is provided by the chosen number base in which to represent the natural numbers; in the decimal system, beyond the first few, all of the prime numbers have 1, 3, 7 or 9 as their last digit; this follows, since all numbers ending in 0 or 5, or in 0, 2, 4, 6 or 8 have been eliminated as being divisible by 5 or 2 (that make up the value 10 of the chosen number base).

Perfect numbers and Mersenne primes

Each of the natural numbers can be decomposed into their prime factors:

(i=1,∞)Π piai

where most of the values of a_i are zero. In the case of the prime numbers, every one of the values of a_i is zero, except for one that has the value 1.

This notation is traditionally used when looking for the highest common factor (HCF) of two natural numbers, or the lowest common denominator (LCD) when adding two rational fractions together. The lowest common multiple (LCM) of two natural numbers is also used, but is simply the result of diividing the product of the two by their HCF.

A perfect number, N, is one for which "the sum of the factors of N"=N (where 1 is included as one of the factors, but N is excluded). For convenience, we start by considering those natural numbers for which:

N = 2n.q

where q is a prime number greater than 2. If this is a Mersenne prime,

q = 2r-1

where r is prime, and known non-prime values of q, such as the Lissajou-like glitch that occurs for r=11, are avoided. For N to be a perfect number:

2.(the sum of the factors of N)
 = 1.(1+2r+2r+2r+...+2n) +
   q.(1+2r+2r+2r+...+2n)
 = (q+1).(2n+1-1)
Therefore:
2.2n.(2n-1) = 2n.(2n+1-1)
So:
(2n-1) / (2n+1-1) = 2r-n-1
Therefore n=r-1 and so:
N = 2r-1.(2r-1)

This generates the following pairs of (r=Mersenne generator, N=perfect number): (2,6) (3,28) (5,496) (7,8128) (11,invalid) (13,33550336). By inspection, other varients on 2ⁿ.q, such as choosing a larger prime number for the first term, or raising the second prime number, q, to some power greater than 1, or multiplying by further prime number terms, is going to be out of the range for generating any sum of the factors of N with a value of 2N.

Primes in the vicinity of powers of two

2ⁿ is not divisible by 3 (or by any prime number greater than 2). Exactly one of 2ⁿ-1 or 2ⁿ+1 must be divisible by 3. Indeed, it is 2^even-1 or 2^odd+1 that are divisible by 3. In general, 2ⁿ±1 is divisible by p, for n=(p-1).m (where p is a prime number, and m is just the scaling factor). For other n, 2ⁿ%p takes on not more than each of the values, 1,2,3,..,(p-1), but not 0 (of course).

22-1 is divisible by 3, as is any 2even-1.
24-1 is divisible by 5, as is 28-1, 212-1.
26-1 is divisible by 7, as is 212-1.
210-1 is divisible by 11.

We will define the modulus division operator, %, to give its result in the range -|y/2|<x%y≤|y/2|. In the case of y being prime and greater than 2, this amounts to –(y-1)/2 < x%y <= +(y-1)/2. 2ⁿ⁺¹%p has twice the value of 2ⁿ%p. 2⁴%5 = +1, 2⁵%5 = +2, 2⁶%5 = +4 (that is, -1), 2⁷%5 = -2, 2⁸%5 = -4 (that is, +1). 2⁶%7 = +1, 2¹²%7 = +1, and, in between, values of (p-2). In particular, we are interested in the middle ones (remembering that 2⁰%p = +1 for all prime p>2). 2⁹%7 = +1,  and 2³%7 = +1. 2¹⁰%11 = +1,  and 2²⁰%11 = +1. 2¹⁵%11 = -1,  and 2⁵%11 = -1. Why does 7 behave differently to the others? The mid-point is +1, not -1. 2⁶%7 = +1,  2⁷%7 = +2, 2⁸%7 = +4 (that is, -3),  2⁹%7 = -6 (that is, +1), 2¹⁰%7 = +2,  and 2¹¹%7 = +4 (that is, -3), 2¹²%7 = -6 (that is, +1). Therefore, -1, -2 and +3 are not represented.

(Chart, from my log-book 29-Sep-1998, to be included here.)

It is a bit like recurring decimals: cycles start when the calculation comes back to an earlier state. Powers of two end up at nought-recurring. All the prime numbers recur with no more than (p-1) terms, and all recur starting with the +1 term. Therefore all pose threats to the Mersenne numbers being prime.

If p recurs in n terms, p² recurs in 2n terms, and p³ recurs in 3n terms. (Though this is not true for p=5).

If p recurs in n terms, p.2^m also recurs in n terms, offset to the right by m columns. If there is an odd number of recurring terms, red(n), there are no mirror terms, green(n). If there is an even number of recurring terms, red(n), there are green(n/2) mirror terms, except for multiples of (2^m-1).(2^p-1).

Babylonian function

Just as fac=[n] {if(n<2;1;n*fac(n-1)} and fib=[n] {if(n<2;n;fib(n-1)+fib(n-2)} so we can define babylonian=[n] {if(n<2;1;if(ispowerofprime(n): p*babylonian(n-1); babylonian(n-1))}. The value of babylonian(n) is the smallest integer that can be divided by each of the integers between 1 and n. Starting from babylonian(0) this generates the series: 1, 1, 2, 6, 12, 60, 60, 420, 840, 2520, 2520...

Mathematically, this is not a very useful function, since it is not very smooth, with a mixture of steps and plateaux. It is the reason, though, for the imperial measurement system having several of the units that it does: 12 inches in a foot, pennies in a shilling, months in a year, eggs in a dozen, hours in half a day; 60 minutes in an hour or degree, 60 seconds in a minute (of time, or of angle). These are integers that are easily divisible, even in Roman numeral representation, by many simple integers, in the days before the discovery of the number zero, and invention of positional arabic number representation (and hence the ability to perform long division operations).

As another passing thought, we can define sievebase=[n] {if(n<2;1;if(isprime(n): p*sievebase(n-1); sievebase(n-1))}. this generates the series: 1, 1, 2, 6, 6, 30, 30, 210, 210, 210, 210, 2310, and so on. Alternatively, we could define sievebase=[n] {if(i<1;1;p(n)*sievebase(n-1)}, where p(n) is the mythical function that returns the value of the n^th prime. Either way, these generated values are the number bases that would be needed to impliment the idea of using the number base for representing the integers as a sieve of Eratosthenes.

Linear spiral

This is related to the process of converting to polar coordinates.

The corners, starting in the bottom-left corner of shell-n, are at 2n(2n+1), (2n+1)(2n+1), (2n+1)(2n+2) and (2n+2)(2n+2)-1. For n>0, only the last of these has the possibility of being prime, though this does not occur for the first four shells shown above. Other lines can be spotted, radiating out, for each of the times-tables (albeit difficult to discern in the small table shown below). Here, coloured cells show the highest prime number by which each is divisible (in this case limited to the highest prime less than the square-root of 99, namely 7). This leaves the prime numbers, here shown in white.

Logarithmic spiral

There are as many prime numbers in the range 10^k+1 to 10^k+2, as there are in the range 10^k to 10^k+1. The Riemann zeta function.

zeta(n) = 1 + 1/2n + 1/3n + 1/4n + …
                        = (2n/(2n-1)).(3n/(3n-1)).(5n/(5n-1)).(7n/(7n-1)). (11n/(11n-1))...

It is tempting to wonder at possible connections of this in quantum mechanics: the periodic table is based partly on the sequence of odd numbers (1 3 5 7...), whose sum can also be expressed in terms of the perfect squares. There is a suggestion that the long-sought proof of the Riemann hypothesis, with the zeros of the zeta function for prime numbers all lying on the vertical line 0.5+n.i (NS, 22-Mar-2008, p40) might even be found first in the energy states of a suitably chosen quantum system, such as a large atom or molecule (NS, 11-Nov-2000, p32). As a further thought, there is a possible connection, too, with proving the Schanuel conjecture (NS, 21-Jul-2007, p38).

Integer solutions for Pythagorus

The most well-known integer solution for Pythagorus is the 3,4,5 triangle; but there are others, such as 5,12,13. Since a²=c²-b², we define b=c-k. This means that a=√(2kc-k²). Setting c=km, this becomes a=k√(2m-1). When k is even, this has integer solutions for c=5k/2, 5k, 17k/2, 13k, 37k/2, 25k, 65k/2, 41k, 101k/2, and so on; when k is odd, the integer solutions are for c=5k, 13k, 25k, 41k, and so on.

Considering the D₁ Babbage difference column for the perfect squares (1, 3, 5, 7, 9...) we can define Sj to be the sequence of integers, each one the sum of j successive elements in the D₁ sequence. Thus, S2 is (4, 8, 12, 16, 20...) and S3 is (9, 15, 21, 27, 33, 39...) and S1 is simply D₁. For Pythagorus, we are interested in cases where j²+2.m.j² is also a perfect square. This generates triangles whose sides are ((n²-1)/2, n, (n²+1)/2). That is, the triangles (0,1,1) (4,3,5) (12,5,13) (24,7,25) (40,9,41), in each case an even-numbered side, an odd-numbered side and an odd-numbered hypotenuse; with areas, respectively, of 0, 6, 30, 84, 180.

Towards Fermat’s Last Theorem

Fermat’s Last Theorem was finally proved in 1995. There is no intention to go to any such lengths, here, but merely to dabble in a few related ideas. These centre round the equation: aⁿ+bⁿ=cⁿ and to ask if this equation has any integer solutions (a, b and c all integers) for any given value of n (also an integer).

From the previous paragraphs, we know that for n=2, a²+b²=c² has many integer solutions. For n=1, we can see that a¹+b¹=c¹ only has integer solutions when a=0 or b=0, and then has an infinite number of them (c can then take on any integer value). For n=0, there are no interesting solutions to a⁰+b⁰=c⁰, since it is only true if 2=1, or if some of a, b and c are zero.

For n=3, we can repeat the process for the integer solutions to Pythagorus, to explore a³=c³-(c-k)³. For k=1, a³=3c²-3c+1, so c=(1/2)±√((1/4)-(1/3)(1-a³)). Starting from the integer values of a³=0, 1, 8, 27, 64, 125,218, 343, we seek integer solutions of 2m+1=√(4n+1). This can only occur for every third value of a³, namely a³=1, 64, 343, and so on, yielding values of √(4n+1) that are 1, 85, 457, respectively. Apart from the degenerate solution of the first of these, subsequent terms do not appear to be perfect squares (at least not so far).

Towards Collatz and the wondrous numbers

From any natural number, we can trace the chain of subsequent numbers that are visited when applying the Collatz algorithm: if n is even, "divide by 2", else "multiply by 3 and add 1". For example, starting at n=9 would give: "9 28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 4 2 1 4 2 1...". Since the "multiply by 3 and add 1" rule is only ever applied once in isolation, whereas the "divide by 2" rule can be applied repeatedly, the above sequence can be encoded as [0 2 1 1 2 3 4 2 2 2 2 2 2...] meaning "divide by two 0 times before applying 3n+1, then twice before applying 3n+1, then once before applying 3n+1, and so on. Just as with recurring decimals, it is useful to have a notation to represent the recurring loop at the end: [0 2 1 1 2 3 4 (2)].

It is noted that there is no halting condition in the Collatz algorithm, therefore all sequences must end in a recurring loop. The question therefore becomes whether there are any other loops, other than the "4 2 1" sequence. The [(2)] loop requires n=(3.n+1)/2/2, which is indeed satisfied when n=1. The question is, are there any other loops possible? Hofstadter 1980) implicitly illustrates how the most important natural number still tends to be overlooked, and which comes down to n=n/2ⁱ, which is indeed satisfied when n=0, and which completes the parallels with the recurring decimals like 1.333(3)recurring, 0.(142857)recurring and 1.5(0)recurring. Other than that, every loop must begin with a 3n+1 step (preceeded by a token divide by 2⁰ for the example, above, when starting at 9). After that, every 3n+1 step must be followed by at least one divide by 2 step, since the 3n+1 always generates an even number. A loop round two values would require n=(3.n+1)/2, which is satisfied when n=-1 (the loop: -1 -2 -1 -2 -1 -2...). A loop round j+1 values would require n=(3.n+1)/2^j, which would require n=-1/(2^j-3), which only has integer values for the two cases (j=3 or 2) just considered.

Another way of forming a loop round four values would require n=(3.((3.n+1)/2)+1)/2, which is also satisfied when n=-1 (namely twice round the loop mentioned above). This can be generalised to n=(3.((3.n+1)/2ⁱ)+1)/2^j, and then on to the cases for n=(3.((3.(3.n+1)/2ⁱ)+1)/2^j)+1)/2^k, and so on, looking for solutions that are not negative or fractional. The following examples can be listed.

[(1)]→ -1, [(2)]→1, [(3)]→1/5, [(4)]→1/13, [(5)]→1/29
[(x x)]=[(x)] and [(y x)]=[(x y)] neither of which had been a forgone conclusion,
   and [(2 1)]→ -5, [(3 1)]→11/7, [(4 1)]→19/23
[(x x x)]=[(x)] and all rotations of [(x y x)]=[(x x y)],
   and [(1 1 2)]→ -19/11, [(1 2 2)]→23/5

This can be generalised.

[(x)] : 2x.n = 3n+1
[(x y)] : 2y.n = 3((3n+1)/2x)+1 so 2x+y.n = 9n+3+2x
[(x y z)] : 2x+y+z.n = 27n+9+3.2x+2x+y
[(x y z w)] : 2x+y+z+w.n = 34n+33+32.2x+31.2x+y+30.2x+y+z
and so on

This can be regrouped.

[(x y)] : 2x.(n.2y-1) = 3.(3n+1)
[(x y z)] : 2x.(n.2y+z-2y-3) = 9.(3n+1)
[(x y z w)] : 2x.(n.2y+z+w-2y+z-3.2y-9)=27.(3n+1)
and so on, to the case of
[(x1 x2 x3 ... xi)] : 2x1.A = 3i-1.(3n+1)

In each case, the product of an even number and an odd number equals the product of an odd number and what must therefore be an even number. Thus, n must be an odd number; but this then means that (3n+1) must equal 2^x; which in turn means that A must equal 3^i-1. In each case, this comes down to n=1 and each of the x, y, z, w terms being 2. As seen earlier, [(2 2 2)]=[(2 2)]=[(2)] in this notation, so all cases come down to the loop of (4,2,1)recurring.

Circle of fifths on the well-tempered keyboard

There are many systems that consist of two rules: one that does most of the interesting work, and one that just strips out the mundane stuff (perhaps including the partial garbage collection administrative work not shared equally between the S and K combinators of a Non-von-Neumann-style Functional Language Processor). It also has parallels with the way that a five-sided die can be implemented using a normal six-sided one, and merely stating, "if you throw a six, then throw again." With Collatz and the Wondrous numbers, above, it is the 3n+1 rule that seems to do the interesting work, and the n/2 is just applied repeatedly to strip out all the factors of 2. The same thing happened in the test for perfect squares: one rule to test if a binary number ending in ...01 is a perfect square, another to dismiss those ending with ...10 or ...11, and another just to shift out the double zeros at the ends of those ending in ...00.

This also happens on the Western musical scale, notably in the circle of fifths modulation of key. Each augmentation by a perfect fifth is really a jump to the third harmonic of the current fundamental, and a jump back by an octave. However, this still results in the key changes becoming higher up on the stave, and often, but not always, requires a jump back by a further octave. The sequence of modulations might be expressed as: [3/2 3/2/2 3/2 3/2/2 3/2 3/2/2 3/2/2 3/2 3/2/2 3/2 3/2/2 3/2/2]. It is not a coincidence that this sequence mirrors the TTsTTTs pattern of the major scale in Western music, and the pattern of black and white notes on the keyboard.

Treating the 3n step as being related to the 3n+1 step for Collatz and the Wondrous numbers, this sequence can be summarised in the notation that was used above as: [(1 2 1 2 1 2 2 1 2 1 2 2)] but is only approximately a complete cycle, to the extent that 3¹² is close in value to 2¹⁹ (slightly out by a margin of just over 1%, which is about a sixth of the percentage difference between adjacent semitones).

Parabolæ-within-parabolæ embroidery on the times tables

A primary-school exercise, to get embroidery-orientated children involved in mathematics, and mathematics-orientated children involved in embroidery, consisted of stitching wool through the holes of a suitably prepared card. The card had twelve holes equally spaced along the x-axis, and twelve along the y-axis, plus one at the intersection of the two axes at the origin. The wool was to be stitched between the (12,0) hole and the (0,0) hole, then between !11,0) and (0,1), then (10,0) and (0,2), then (9,0) and (0,3), and so on, until (0,0) and (0,12).

The result is the familar parabole shape emerges from the nest of tangents. If the times-tables chess board is printed on the card, so that the x and y-axes pass through the centres of the zero-times-table row and zero-times-table colummn, we can list contents of the cells that each of the tangents pass through.

Thread	Gradient	Cells traversed
(0,12) to (0,0)	-∞	0
(0,11) to (1,0)	-11	0 0
(0,10) to (2,0)	-5	0 5 0
(0,9) to (3,0)	-3	0 6 6 0
(0,8) to (4,0)	-2	0 6 8 6 0
(0,7) to (5,0)	-1.4
(0,6) to (6,0)	-1	0 5 8 9 8 5 0

In general, a thread from (0,12-i) to (i,0) has a gradient of -(12-i)/i and is integer for the integer factors of 12, having a value of 1 less than the complementary factor. The sequence of values of the cells traversed is stangely symmetrical, despite the assymetric orientation of the thread.

The next member of the babylonian sequence is 60, yielding the following table.

Thread	Gradient	Cells traversed
(0,60) to (0,0)	-∞	0
(0,59) to (1,0)	-59	0 0
(0,58) to (2,0)	-29	0 29 0
(0,57) to (3,0)	-19	0 38 38 0
(0,56) to (4,0)	-14	0 42 56 42 0
(0,55) to (5,0)	-11	0 44 66 66 44 0
(0,54) to (6,0)	-9	0 45 72 81 72 45 0
(0,50) to (10,0)	-5	0 45 80 105 120 125 120 105 80 45 0
(0,48) to (12,0)	-4	0 44 80 108 128 140 144 140 128 108 80 44 0
(0,45) to (15,0)	-3	0 42 38...
(0,40) to (20,0)	-2	0 38 72...
(0,30) to (30,0)	-1	0 19 56...

In general, the thread from (0,n-i) to (1i,0) has a gradient of -(n-i)/i, and passes first through a cell whose vale is 0, then one whose value is !n-i)-(n-i)/i, then one whose value is 2.(!n-i)-2.(n-i)/i), and so on, reaching a peak at one whose value is (n.i-i²)/4, and then symmetrically down in values to a final cell whose value is 0. For example:

Thread	Gradient	Cells traversed
(0,52) to (8,0)	-6.5	0 45.5 78 97.5 104 97.5 78 45.5 0

Bragg contour layers

Cartessian coordinates ressemble a simple orthogonal crystal. (Really it is the other way round, that a simple orthogonal crystal is laid out on a Cartessian coordinate system.) The zeroth layer of the crystal runs from (0,0) to (∞,0), the first layer after that runs from (0,1) to (∞,1), the second from (0,2) to (∞,2), and so on. These layers are traced out by the moves of a type-0 chess piece.

A type-1 chess piece traces out the most obvious diagonal layer. The first runs from (1,0) to (0,1), the second from (2,0) through (1,1) to (0,2), the third from (3,0) to (2,1) to (1,2) to (0,3), not forgetting the zeroth layer at (0,0). The contents of those cells are simply the product of the ordinate and abscissa: "0" for the zeroth layer, "0 0" for the first, "0 1 0" for the second, "0 2 2 0" for the third, "0 3 4 3 0" for the fourth, "0 4 6 6 4 0" for the fifth, "0 5 8 9 8 5 0 for the sixth. In general, for the nth layer, it runs from 0 all the way up to a peak at (n/2)², and then symmetrically back down to 0. In the case of odd values of n, this leads to a peak between two actual cells, just as in the way for a black-cell bishop being able to make two fractional moves to land on a white cell.

For a type-2 piece, the first layer is from (1,0) to (0,2), the second from (2,0) to (1,2) to (0,4), the third from (3,0) to (2,2) to (1,4) to (0,6). Each of the layers have the same number of cells in them as for the corresponding type-1 layer, but just with each value multipled by two. In general, then, the nth type-i layer starts at (n,0), reaches a peak valued cell of (i/(n-2)/2)², and ending at a final cell (0,i.n).

There is an identical set of layers formed by swapping the ordinate and abscissa, starting at (i.n,0), reaching a peak valued cell of i.(n/2)², and ending at a final cell (0,n).

Eratosthenes divergence trajectories

Choosing one of the non-prime cells (in any of the m-times tables, for m greater than 1) a line can be drawn back to its corresponding cell in the one-times table, and the values of the other cells that the line passes through can be traced.

These lines are running across the Bragg-like layers, in a similar way to the way that magnetic field lines run across iso-magnetic field contours. They run down the GRAD-like paths, and converge on a point in the one-times table, yielding a DIV-like star network.

The cell (m,n) xontains a non-prime for all m greater tan 1 and n greater than 1. It is possible that either, or both, m and n are non-prime; but even without this, we can trace four lines: (m,n) to (m.n,1) and to (1,m.n); and (n,m) to (m.n,1) and to (1,m.n). These, respectively, have the gradients: -1/m, -n, -1/n, and -m.

A power of a prime number, such as 32, appears of course in the one-times table (the one-times column), and in the 32-times table (in the one-times row). It also appears in the 16, 8, 4 and 2-times tables, with gradients back to its occurence in the one-times column of -2; -4, -8 and -16, respectively; and gradients to its occurence in the one-times row of -1/16, -1/8, -1/4 and -1/2, respectively. The repective lines are:

- (1,32) is a degenerate case
- (2,16) (1,32) having values 32 and 32, respectively
- (4,8) (3,16) (2,24) (1,32) having values 32, 48, 48, 32, respectively
- (8,4) (7,8) (6,12) (5,16) (4,20) (3,24) (2,28) (1,32)
     having values 32, 56, 72, 80, 80, 72, 56, 32, respectively
- (16,2) (16,2) (15,4) (14,6) (13,8) (12,10) (11,12) (10,14) (9,16)
     (8,18) (7,20) (6,22) (5,24) (4,26) (3,28) (2,30) (1,32)
     having values 32, 60, 84, 104, 120, 132, 140, 144, 144, 140, 132, 120, 104, 84, 60, 32
- (32,1) (31,2) (30,3) (29,4) (28,5) (27,6) (26,7) (25,8)
     (24,9) (23,10) (22,11) (21,12) (20,13) (19,14) (18,15) (17,16)
     (16,17) (15,18) (14,19) (13,20) (12,21) (11,22) (10,23) (9,24)
     (8,25) (7,26) (6,27) (5,28) (4,29) (3,30) (2,31) (1,32)
     having values 32, 62, 90, 116, 140, 176, 182, 200, 216, 230, 242, 252, 260, 266, 270, 272, 
        272, 270, 266, 260, 252, 242, 230, 216, 200, 182, 176, 140, 116, 90, 62, 32

Each parabola (quadratic) is easily calculated using Babbage differences. Based on the factorisation of pⁿ, in this case 2ⁿ, each parabola starts at (2^n-i,2ⁱ), reaches its peak values in between (2^n-1+1,2^n-1) and (2^n-1,2^n-1+1), and on to (2ⁱ,2^n-i) and (0, (2ⁿ+2^n-i). Indeed, the extension (by one step in both directions, to the 0th row and 0th colummn) of each of the lines can be noted:

- (2,0) (1,32) (0,64)
- (3,0) (2,16) (1,32) (0,48)
- (5,0 (4,8) (3,16) (2,24) (1,32) (0,40)
- (9,0) (8,4) (7,8) (6,12) (5,16) (4,20) (3,24) (2,28) (1,32) (0,36)
- (17,0) (16,2) (16,2) (15,4) (14,6) (13,8) (12,10) (11,12) (10,14) (9,16)
     (8,18) (7,20) (6,22) (5,24) (4,26) (3,28) (2,30) (1,32) (0,34)
- (33,0) (32,1) (31,2) (30,3) (29,4) (28,5) (27,6) (26,7) (25,8)
     (24,9) (23,10) (22,11) (21,12) (20,13) (19,14) (18,15) (17,16)
     (16,17) (15,18) (14,19) (13,20) (12,21) (11,22) (10,23) (9,24)
     (8,25) (7,26) (6,27) (5,28) (4,29) (3,30) (2,31) (1,32) (0,33)

To be tied in with HCF, unexpected alignments in factors, or in the expansion of the sieve of Eratosthenes, rational fractions, notes of the diatonic music scale, Lissajou figures, unexpected resonnance.

If these were to be considered the fall-lines, they must pass contours that are perpendicular to them, namely with gradients of m, 1/n, n, and 1/m, respectively. But then, returning to the analogy with Bragg-type analysis, the X-rays are not constrainted to falling perpendicularly on each layer of the crystal.

It is not clear, at present, whether any of this is leading anywhere useful, but at least it has been enjoyable dabbling just for the fun of it, to see where it might lead (to be continued).

Philosophical analogy

The Difference Engine hints at a possible philosophical analogy. If an error occurs in a calculation, either due to mechanical failure, or an inherent rounding error, it manifests as an apparent divergent wave of errors in all of the differences columns, all the way out to the D_∞ column, and building up not just into future rows of the calculation, but also seemingly retrospectively into past rows.

Glitches in the desceptively smooth process, like finding unexpected harmonics with Lissajou figures.

With the Mersenne primes, occasional non-prime values are thrown up, phase-locked loop or Lissajou-like resonnance, that occurs for r=11. Chaotic patterns in the vicinity of prime numbers posing threats to the Mersenne numbers being prime.