# The Prime Sequence Problem

The Prime Sequence Problem [Research]. This perplexing matter has been reset as an exercise in the linear algebra of a Euclidean four dimensional hyperspace and spaces of descending dimension and a finite discrete version of the Legendre Transformation, where intercepts are expressed as a function of gradients.
Since mathematics is a perfectly logical subject, the distribution of the prime numbers on the real axis [and imaginary axis] cannot be a matter for statistics, but subject to the rules of logic and therefore explicable. This contribution has been called the above title because the purpose here is to explain the prime sequence. Logically this can be achieved, but only if the necessary information needed to solve this problem is at hand. If some known mathematical principle has been overlooked or if a new principle relating to the solution has not yet been discovered, then this deficit would need to be made good, but the development appears to be on the right lines. All previous attempts to solve this problem have failed, probably because it is almost certainly impossible to find the solution in one dimension - along the real axis - not withstanding the use of complex, two dimensional numbers. The solution might be discovered in spaces of a higher dimension such as in 4D spaces.
The prime numbers fall into two exclusive classes:[a] two [2] in a class of its own containing just one member, itself. and [b] the infinite class: [... - 5, - 3, 3, 5, 7, 11, ...] that are a subclass of the uneven integers: [ ... - 3, - 1, 1, 3, 5, 7, 9, 11, ... ]. It is convenient to consider negative and positive primes comprehensively. Here, the primes: [3, 5, 7, 11, ... ] have been assigned the ordinals: [1, 2, 3, 4, ... ], corresponding to p[1] = 3, p[2] = 5, ... p[4] = 11 etc. The complementary composites are in the sequence: [ ... -9, 9, 15, 21, 25, 27, 33, 35, 39, ... ] Any uneven integer can be expressed as a Binary Remainder Equation [B.R.E]. Moreover while there is a minimum number of terms in a B.R.E., there is no maximum number of terms. There can be any finite number of terms in a Binary Remainder Equation. [dropping caps. from now on]. Even integers, zero included and the prime number two [2], cannot be represented as binary remainder equations. There are elementary relationships connecting even and uneven integers that lead to equations in Euclidean spaces some of whose solutions are ordered sets of prime numbers. One such relationship is that an even number of uneven integers when added together result in an even integer summation, noting that zero is an even integer.
There are finite dimensional Euclidean spaces of even dimension containing solution points to equations whose terms, which are even in number may be prime numbers [except two [2]] up to the final term or final terms, such that the algebraic sum of all the terms is zero. Adjacent or consecutive solution points in the hyperspaces may be joined by lines to form an arc of a polygon. The segments [polygonal sides] of the polygonal arc have a length and direction that is specified by direction ratios. In four dimensional space which is the first non-degenerate instance there are four components of each point in solution space. A neighboring point has four components, these being ordered quadruples, very often of primes. The direction ratios of which there are four for each segment of the polygonal arc constitute ordered quadruples in a direction ratio 4D space that is related to the primitive solution space. The direction ratios can be squared, so removing irrationals from the direction ratios and creating another related 4D space of direction ratio squares. The primitive solution space, the direction ratio space and the direction ratio squares space contain polygonal arcs. The sides of these arcs can be produced to intersect with the planes formed by the axes of co-ordinates. This will be seen to effectively reduce the dimension of the space in which the intersection points exist. These intercepts and their direction ratios contain all the information about the primitive polygonal arc. This is like discrete versions of the Legendre transformation. Unlike in the Legendre transformation, it is necessary to use direction ratios or direction cosines rather than gradients in spaces of dimension greater than 2D. The intercepts may be found in each Euclidean space. Some direction ratios are worked out below so helping to make the above statements clearer. A starting solution point can be assigned to the above equations that are described below.
The polygonal arc, starting from perhaps three solution points as datum solutions of Equation 0.002. [see below] continues in 4D space in such a way that the next point defining the successor segment does not have all of its co-ordinates freely assignable. This is because the successor prime depends exclusively on all of the antecedent primes. If the prime sequence is written: [p[1], p[2], ... p[k - 1], p[k], p[k + 1], ... , then p[k + 1] depends on the p[1], ... p[k] exclusively [noting that in this context, p[1] = 3, never two]. The polygonal arc has sides and chords of which there is for any [k] value, a combinatorial number of such chords. Associated with each chord there are four direction ratios and up to four intersections, but at least one intersection with the planes defined by the four axes of co-ordinates. These matters imply equations connecting the primes: [p1], ... p[k]], all assumed to be known and p[k + 1], the unknown successor prime. The foregoing applies only in the situation where the solution points in 4D space are adjacent and in the sequence: [3, 5, 7, -15], [5, 7, 11, -23], [7, 11, 13, -31], ... [p[k - 1], p[k], p[k + 1], - [p[k - 1] + p[k] + p[k + 1]], the first three solution points being the datum points.
There are other ways of setting out the sequence of solution points, such as: [p[1], p[2], p[3], p[4]], ... [p[k - 2], p[k - 1], p[k], p[k + 1]], where both p[k] and p[k + 1] are unknown primes that are subject to two conditions , one being imposed by Equation 0.002., that is: -[p[k] + p[k + 1]] = [p[k - 2] + p[k - 1]] and the other condition, difficult to find, is that p[k] is the least prime that will satisfy Equation 0.002. so that p[k + 1] is a prime and not a composite, such any in the list of composites above. Further down, equations formulated from 4D conditions such as the chords, sides, direction ratios and intercepts of segments [sides] of the 4D polygonal arc, above may make it possible to compute the successor prime. If not, then further information would be needed but only derivable from the antecedent primes.
It might look much better to discard Equation 0.002., and write, for example, the sequence of 4D points as: [3, 5, 7, 11], [5, 7, 11, 13], ... , but the constraint supplied by Equation 0.002. is then missing, making the resulting equations possibly unsolvable. Other constraints would involve the squares, cubes and fourth powers of the solution point co-ordinates that satisfy Equation 0.002.
In 4D space this could be [3, 5, 7, - 15] but it may be better to work with two prime sequences separately: [a] [4.k + 1] primes such as 5, 13, 17, 29 etc. and [b] [4.k + 3] primes such as 3, 7, 11, 19, 23 etc. and then synthesize the two sequences. In that case, a starting point would be [5, 13, 17, - 35] in 4D hyperspace.
The object is to find operations that can be done on uneven integers to produce more uneven integers and then express all of these uneven integers as binary remainder equations [B.R.E.]'s and thence to discover a pattern in the resulting array of binary remainders.
Derivation of a binary remainder equation: Given Q[N, n] = [2.n - 1] where the integer [n] takes values: [ ... - 2, - 1, 0, 1, 2, 3, ... ], equations that are a special kind of division by two [2] can be found easily. Q[N, n] = 2.Q[N - 1] + r[N - 1], counting down from N to zero by integral steps. Q[N - 1] is an uneven integer provided that the remainder r[N - 1] is given by: r[N - 1] = [i]^{1 + Q[N, n]}, where [i] satisfies: i^{2} = - 1. The remainder is either [- 1] or [+ 1], but never zero. The integer N comes from the inequalities: 2^{N} < Q[N, n] < 2^{N + 1} . Here, the value of [n] has to be consistent with these inequalities and takes upper and lower limit values, and some intermediate values. The value of n[minimum] is given by: [2.n[min] - 1] = 2^{N} + 1. and the value of n[maximum] is given by: [2.n[max] - 1] = 2^{N + 1} - 1. From this we find that n[min] = [2^{N - 1} + 1] and n[max] = 2^{N}. There are: 2^{N - 1} intermediate values of Q[N] in the interval defined by the inequalities above. The next iteration equation is: Q[N - 1] = 2.Q[N - 2] + r[N - 2] Again, Q[N - 2] is an uneven integer provided that the remainder is given by: r[N - 2] = [i]^{1 + Q[N - 1] The final iteration is: Q[1] = 2.Q[0] + r[0]. The dividend, Q[1] is identifiable with three [3], Q[0] is identifiable with unity [1] and the remainder. r[0] is identifiable with unity [1]. Taking the antecedent to Q[N, n], it is convenient to write Q[N + 1] = 2.Q[N] + r[N] and set r[N] = 1 [dropping the integer [n] in the definition of Q[N, n]]. These rather fancy division algorithm equations in which the divisor is two [2], can be rearranged thus:
Q[N - 1] = [1/2].[Q[N] - r[N - 1]]
Q[n] = Q[N, n] as defined.
Q[N - 1] = [1/2].[Q[N, n] - i^{1 + Q[N]}]
Q[j - 1] = [1/2].[Q[j] - i^{1 + Q[j]}]
Q[N - j] = 1/2.[Q[N - [j - 1]] + i^{1 + Q[N - [j - 1]]}]
Q[0] = [1/2].[Q[1] - i{1 + Q[1]}] ...............................................................................................
Generally, Q[j - 1] = [1/2].[Q[j] - r[j - 1]] and r[j - 1] = i^{1 + Q[j]}
Example: Q[5] = 53, then Q[4] = [1/2].[53 - i^{1 + 53}] = 27, Q[3] = [1/2].[27 - i^{1 + 27}] = 13, Q[2] = [1/2].[13 - i^{1 + 13}] = 7, Q[1] = [1/2].[7 - i^{1 + 7} = 3, Q[0] = [1/2].[3 - i^{1 + 3}] = 1. The remainders are: r[5] = + 1, r[4] = i^{1 + 53} = - 1, r[3] = i^{1 + 27} = + 1, r[2] = i^{1 + 13} = - 1, r[1] = i^{1 + 7} = + 1 and r[0] = i^{1 + 3} = + 1. Putting the values of r[1] to r[N - 1] into the binary remainder equation [see below] means putting r[1], .. , r[4] into the B.R.E.:
53 = 2^{5} + 2^{4} + r[1].2^{3} + r[2].2^{2} + r[3].2^{1} + r[4] and that is:
53 = 32 + 16 + [+ 1.8] + [- 1].4 + [+ 1].2 + [- 1] which is true. In terms of putting s = - 1 and t = + 1, 53 = 32t + 16t + 8t + 4s + 2t + s or formally: 53 = [t, t, t, s, t, s], an ordered set of five remainders where the exponents of two [2] are implied.
The above quotients can subsequently be expressed in terms of their remainders in binary remainder equations as has been done in the example above of Q[5] = 53.
There is a useful result: 2^{k} = 2^{N} - 2^{N - 1} - ... - 2^{k} Proof: simply add 2^{k} to each side of this equation. Then add 2^{k + 1] to the resulting equation. Continue until the result becomes apparent by finally reversing the process.
Any binary number can be written implicitly as for example: 1101. This is means: 1101 = [+ 1].2^{3} + [+ 1].2^{2} + [0].2^{1} + [+ 1].2^{0}] That is: 8 + 4 + 0 + 1 = 13[decimal]. Thirteen can also be written as: 13 = 8 + [+ 4] + [+ 2] + [- 1]. This is the binary remainder equation for the number thirteen [13]. Generally: Q[N] = b[N].2^{N} + b[N - 1].2^{N - 1} + ... + b[1].2^{1} + b[0].2^{0} The binary coefficient: b[N] = 1, being the most significant digit and: b[0] = 1, because Q[N] is an uneven integer. By comparison with the iterated division algorithm equations, we can write: b[k] = [1/2].[1 + r[N - k]] or b[N - j] = [1/2].[1 + r[j]] Substitution for b[k] or b[N - j] into the explicit general equation Q[N]= b[N].2^{N} + ... ... + b[0].2^{0} gives the B.R.E. for Q[N] that has the least number of terms needed to express Q[N]. Writing: Q[N] = 2^{N].[1/2].[1+ r[0]] + 2^{N - 1}.[1/2].[1 + r[1]] + ... + 2^{N - [N - 1]].[1/2].[1+ r[N - 1]] + 2^{N - [N]].[1/2].[1 + r[N]]
Two halves appear that cancel with a minus one from the G.P summation: [2^{N} - 1]
This equation separates into a geometrical progression in two [2] and since r[N] and r[0] are each unity, this result follows:
Q[N] = 2^{N} + 2^{N - 1} + 2^{N - 2}.r[1] + 2^{N - 3}.r[2] + ... + r[N - 1]
The above is the minimum binary remainder equation for Q[N].
Using: Q[N] = 2.Q[N - 1] + r[N - 1], by rearrangement: Q[N - 1] = [1/2].Q[N] - r[N - 1]].
Substitution for Q[N] in the above equation gives Q[N - 1] in terms of the remainders: [r[1], ... ... r[N - 2]]. A substitution of 2^{N} - 2^{N - 1} to replace + 2^{N - 1} is required. This iterative process can be continued until we have:
1 [unity] = 2^{N} - 2^{N - 1} .... - 2^[1] - 1. This is the final B.R.E.
Negative Q[N]'s, that is: - Q[N] can be expressed as B.R.E.'s too.
There are some elementary relationships amongst even and uneven integers [often overlooked].
Let E represent an even integer [leaving a remainder of zero in ordinary division by two [2]] and U represent an uneven integer [leaving a remainder of unity [1] in ordinary division by two], then:
U[1] + U[2] + ... + U[2.k] = E[2.k + 1]
In view of the above relationships connecting even and uneven integers, we can write:
2^{L}.P[L] = 2^{1}.p[1] + 2^{2}.p[2] + ... + 2^{K}.p[K], where P[L] and p[1], ... p[K] are any primes whatsoever.
We can also write:
2^{L}.P[L] = p[1}^{n[1]} + p[2]^{n[2]} + ... + p[2.k]^{n[2.k]} , noting the even number of terms being added on the right to give an even summation. The indices [positive integers]: n[1], ... ... n[2.k] are all greater than zero.
The simplest version of the above is: 2^{L}.P[L] = p[1] + p[2]. Equation 0.01.
The next simplest version is:
2^{L}.P[L] = p[1] + p[2] + p[3] + p[4] Equation 0.02.
EXAMPLE [1] In Equation 0.01. as an example set L = 3 and P[L] = 37, a [4.k + 1] prime and give p[1] the values: [3, 5, 7, 11, ... ]. Values for p[2], for the want of a better way, can be found by trial and error.
The results are conveniently expressed as an ordered sets of two primes: [p[1, 1], p[2, 1]], [p[1, 2], p[2, 2]], ... [p[1, j], p[2, j]], where [j] can be increased indefinitely.
The pairs of values represent points in two dimensional Euclidean space and if P[L] is allowed to vary, a three dimensional space of prime solutions.
In Equation 0.02., the [p[1, j], p[2, j], p[3, j], p[4, j]] are points in a four dimensional Euclidean space. If P[L] is allowed to vary, then there is a five dimensional prime space of solutions.
In Equations 0.01. and 0.02. it is convenient to keep L and P[L] constant and allow p[1] in Equation 0.01. to increase in the normal prime progression: [3, 5, 7, 11, ... ].
The Euclidean distance between successive solutions can in all cases be found:
d[1, 2] = [Sigma[p[1, [j + 1]] - p[1, j]]^2 + ... + [p[2, [j + 1]] - p[2, j]]^{2}]^{1/2}, where the square root is taken as positive or negative.
In the above example: 2^{3}.37 = 296 Some prime solutions [followed preferably by the class of uneven integer [4.k + 1] and [4.k + 3]] are:
[3, 393], [13, 283], [19, 277], [67, 229], [73, 223], [97, 199], [103, 193], [139, 157] ... .
Including the negative primes, there are probably an infinite number of solutions.
The distances are respectively: [2]^{1/2}.2.5, [2]^{1/2}.2.3, [2]^{1/2}.16.3, [2]^{1/2}.2.3, [2]^{1/2}.2.5, [2]^{1/2}.2.3, [2]^{1/2}.16.3, [2]^{1/2}.2.3, [2]^{1/2}.4.9. The eighth case being an exception.
Given a set of [n] solutions to Equation 0.01. and/or Equation 0.02. or any other higher even dimension version, there are C[n, 2] ways of joining any two points that represent solutions. This gives n!/[2![n - 2]!] distances that can be permuted in [n!/[2![n - 2]!]! different ways.
Moreover, since addition is commutative and Equation 0.01. is symmetrical in p[1] and p[2], any one solution can be considered as fixed, and in the other [n - 1] solutions, the order of the two primes can be permuted in 2^{n - 1} different ways. This applies to higher order equations but with [2.k]^{n - 1} variations. This is a further ramification that may not have any significance and can be deferred.
[Categorization of the primes, P[L], a constant, p[1, j]] and p[2, j]] into [4.k + 1] and [4.k + 3]'s and L a constant has not yet been done.] All the primes need then be written as B.R.E's. It is convenient to write a B.R.E. with positive signs throughout and replace the general remainders, [r[j]] with [- 1] replaced by s and [+ 1] replaced by t.
Looking at Equation 0.02. again, let L = minus infinity or noting that zero is an even number, in which case [relabeling]:
[Let p[0] = 2, in an entirely separate class]
And let p[1] = 3, p[2] = 5, p[3] = 7, the only triplet and q[j] is any uneven integer that may be prime or composite. Unless specified as definitely 11, 13, ..., p[4], p[5], ... can represent any prime.
p[1] + q[1] = 0
The solution point: [p[1], q[1]] is in a 2D space.
q[1] = - p[1] Equation 0.001.
Equation 0.001. is degenerate.
p[1] + p[2] + p[3] + q[1] = 0
Equation 0.002.
[p[1], p[2], p[3], q[1]] is a solution point in 4D space.
[p[1], p[2], p[k], p[j]] might also be a solution in 4D space.
p[1, j] + p[2, j] + p[3, j] + p[4, j] = 0
The above solution set is true probably in an infinite number of instances. Note that: [p[1, j], p[2, j] and p[3, j]] are points in 3D space.
Analogous Equations 0.003. and 0.004. would be:
p[1] + p[2] + p[3] + p[4] + p[5] + q[1] = 0 Equation 0.003.
Noting that: [p[1], p[2], .. p[5], q[1]] is a solution point in 6D space. Likewise is : [p[1], ... p[4], p[5], p[6]] a solution point in 6D space.
P[1] + p[2] + ... + p[8] = 0 Equation 0.004.
Noting that: [[p1], p[2], ... p[8]] is a point in 8D space.
A starting solution point in 8D space could be: [3, 5, 7, 11, 13, 17, 19, - 75].
The quadruple: [p[1], p[2], p[3], p[4]] is a solution point in a four dimensional [4D] space. Similar remarks apply to Equations 0.003., 0.004. and to equations of higher [even] dimension [[2.n]D], with integer [n] taking values: [2, 3, 4, ... ]
This can be continued indefinitely for equations representing [2.n] dimensional spaces, integer: [n] taking values: [1, 2, 3, ... ]. The question and matters cited below would apply for each above space of a prescribed dimension.
For example in Equation 0.002., The prime p[1] is given the starting values: [3, 5, 7, 11, ... ]. p[1] is increased in the ascending order starting with p[1] = 5, The primes p[2] and p[3] are found by trial and error, taking the least value of p[2]. Similar applies in Equations 0.003. and 0.004. .........................
It may be expeditious to have the initial datum solution points: [3, 5, 7, -15], where q[1] = - 15 and separately: [3, 5, 11, - 19], where p[4] = 11 and p[5] = - 19. ...............................................................................................
Returning to Equation 0.002.
Some solutions of Equation 0.002., taking P[0] = 3 are: [1] [5, 11, - 19], [2] [5, 23, - 31], [3] [7, 13, - 23], [4] [7, 19, - 29]. There are C[4, 2], which is six distances in 3D space for the four solutions above. A sufficiency of solutions in the proper ascending order need be computed as primary solution points so as to be able to compute further derived results. The inter-solution point distances are just one of several results derivable from the primary solution points. Other derivable results are contained in the discussion below.
An interesting question is whether or not the lines joining the solution points intersect at other points. If the answer is yes, then more solution points might be found and the process iterated. This applies to all of the above equations where there are prime solutions that define points in Euclidean spaces.
Another interesting matter is to find the intercepts on the axes of the planes formed by selecting three solution points at a time in the case of Equation 0.002. and in the the cases of Equations 0.003. and 0.004., the intercepts of the hyperplanes on the relevant axes. In the case of Equation 0.003. five solution points would presumably define a hyperplane in 5D hyperspace and in the case of Equation 0.004., seven solution points would define a hyperplane in 7D space. [statements in this paragraph subject to verification]
Yet another interesting matter is that three [primary] solution points define a plane and three such independent planes, formed of separate independent [non-shared] points intersect at another point, a secondary point. These secondary points give rise to more intersecting planes and tertiary points. This process could continue indefinitely. The line produced by the intersection of any two planes intersects with the planes formed by the co-ordinate axes at some point.
A particular case is where four solution points define a tetrahedron. The four planes defining the tetrahedron, in the case of 3D space, give rise to four sets of three intercepts on the axes. The three intercepts produced by the four planes define four points in three space which give another tetrahedron defined by four planes. Apparently, this process of finding the intercepts of a tetrahedron can be continued indefinitely.
The edges of the tetrahedrons produced intersect with the planes formed by the axes at other points which have co-ordinates.
The points of intersection on the axial planes form more tetrahedrons whose edges produce further points of intersection on the axial planes [This statement requires verification by examples].
Primary solution points can be joined to form triangles and polygons in higher dimensional spaces, 3D, 5D, 7D, ... .
A series of consecutive primary solution points, joined by straight lines constitute a polygonal arc in a Euclidean space. Each segment of the arc, produced, has intercepts of effectively of one dimension less than the dimension of the primary solution points [see paragraph below] at the produced ends of the segment. Traveling up the arc gives one set of intercepts and traveling down the arc gives the same set of intercepts. Referring to the paragraph below, when the subsequently derived intercepts are in a 2D space, that is having two co-ordinates at each point in the 2D space, this constitutes a derived polygonal arc lying in a plane in 2D space. At this stage, we can find the intercepts on the two co-ordinate axes, say X and Y axes, of each segment of the polygonal arc and find the gradient in the usual way. Now we have the gradients as a discrete function of the corresponding intercepts. There are two intercepts and a gradient and its inverse - one intercept on the X axis and another intercept on the Y axis. This is a discrete version of the Legendre Transform. The actual function is however known, since the derived solution points were used to find the gradients and intercepts in the first place. If this function is regular, not erratic, then the prime sequence could be found in principle by working back to the primary solution points and inferring the next segment of the polygonal arc in the relevant higher dimension space. The final gradients and intercepts on the X and Y axes in the 2D space are likely to be rational numbers or recurring decimals, not algebraic irrationals. Once the discrete function or functions and their intercept and gradient form in 2D have been found, the successive differences of the gradients and the differences of the corresponding intercepts would produce a higher order discrete function analogous to higher order Legendre derivatives. This process continued would in the limit give a continuous function and its corresponding Legendre transform. This would be a function capable of being differentiated and integrated in the ordinary sense. The foregoing statements need verification by actual examples.
It is useful to consider one of the primes: p[0] as constant, so lowering by unity the dimension of the solution points of the equations: 0.002. or 0.003. or 0.004. from equations respectively having solution points of four, six and eight dimensions, since it is easier to deal with spaces of a lower dimension.
Given an [n] dimensional space, any two solution points in this space can be joined with a straight line produced so as to intersect in several places on the [axial] planes formed by the axes. The dimension of the intersection points is effectively [n - 1], since on the axial planes, one of the other values of the variables along its axis is zero. Selecting the intersection points, two at a time, [presumably defining a line in [n] dimensions] gives lines in an [n - 1] dimensional space. These lines produced, intersect with the axial planes in the [n - 1] space and the intersection points have an effective dimension of [n - 2]. This process can be continued until lines joining points in 2D space give intersection points of 1D space on the 2D axes, that is ordinary numbers.
In practice, the even numbered dimensional version of Equations 0.002., 0.003. and 0.004. would need to be considered. This means working with 4D, 6D and 8D spaces of primary solution points and selecting the appropriate ascending sets of assigned primes in each equation so that the successor set of assigned primes is adjacent to its antecedent set, so making the consecutive primary solution points the minimum distance between one another in the space or hyperspace of even dimension.
There is another development of Equations 0.002., 0.003., 0.004. and successor equations having solution points in higher [2.n] dimensional spaces. There are several different processes, any one of which might answer the problem implied by the title. These are best illustrated by examples.
[1] Write down a solution to equation 0.002.: [3, 5, p[2],p[3]] One solution is: [3, 5, 11, - 19]. Now permute the co-ordinates. There are 4!,which is twenty-four permutations of the ordered quadruple. Some of these are: [3, 5, 11, -19], [3, 5, - 19, 11], [3, - 19, 5, 11] and twenty-one more. These twenty-four permutations constitute twenty-four points in 4D space and form a regular shape in 4D space. The number of edges of this, presumably, polyhedron is given by: C[24, 2], which is: 276 in number. Each of these lines produced have places of intersection on the axial planes: [x[0] = 0, ... x[3] = 0]. This eventually, by the process described above gives polygonal arc in 2D, this being a finite discrete function. Next a solution is taken for Equation 0.003. which for example is: [3, 5, 7, 11, p[4], p[5]]. A solution might be: [3, 5, 7, 11, 17, - 43]. The ordered set of six has 6!, which is 720 permutations in 6D space. There are C[720, 2], which is 258840 edges to the regular shape formed by 720 permuted solution points in 6D space giving many intersections with the axial planes as before. In this process the number of edges and their intersections with the axial planes quickly becomes astronomical as the dimension of the solution points spaces increases by two at a step.
In place of finding the intersection co-ordinates of the produced edges on the axial planes of the hyperspace, we could simply project each point on to all of the axial planes in sequence and so reduce the dimension by unity, without any further permuting of the reduced co-ordinates.
[2] [Much simpler] Taking Equation 0.002. we could put for p[0] and p[1] in the array: [p[0], p[1], p[2], p[3]] the succession: [3, 5], [5, 7], [7, 11], [11, 13] and solve for p[2] and p[3] taking the lest prime p[3] that with p[4] will give a zero summation. Then we could find the 276 intersections as above and finally end with a 1D polygonal arc representing a finite discrete function in every case and then see how each differs from its successor.
In place of finding intersections of produced edges we could just simply project the twenty-four permutations of the four solution points on to the 4D axial planes, as above and have a succession of polygonal arcs ending with a set of polygonal arcs 2D space. There may be more than one polygonal arc giving more than one finite discrete function [described by ordinates or by intercepts and gradients] because there is more than one axial plane to start off with.
Variations of the above:
[a] Categorize the primes as [4.k + 1] and [4.k + 3] integers separately in the solution sets, as far as possible, for if p[0] and p[1] are [4.k + 1] primes, p[2] and p[3] may not both be [4.k + 1] primes. [b] For example in process [2] in stead of writing p[0], p[1] as starters in Equation 0.002. and then finding the least p[2] that together with - p[3] gives a zero summation, write instead: p[0], p[1], p[2] and q[3], where uneven integer q[3] may or may not be prime, to yield a zero summation.
What is needed is to discover a regular discrete finite function or regular discrete finite functions derived, as above, from well ordered primary point solutions, such that the successor prime or some successor primes can be computed using ordinary conventional elementary algebra. Having and using all of the above information including the B.R.E.'s may make it finally possible to deduce how the prime sequence works. This is a structural problem.
From here on, it is a matter of working out examples to see how matters develop and transpire. Subsequently, some of the above statements about the even dimensional spaces and the corresponding even dimensional equations 0.002. - 0.004. might need modification.
A good place to start is with Equation 0.002.0 and consecutive solution points in 4D space and find some polygonal arcs in 2D and the corresponding finite discrete functions in their specific forms. Some other consecutive points that are related to the solution points in hyperspace, of Equation 0.002., are: [q[1], q[2], q[3], q[4]] ... , and [p[j], p[k], p[j + 1], p[k + 1]] ...
[a] The first form of Equation 0.002. to be considered is: p[1] + p[2] + p[3] + q[1] = 0 Given the first two solution points: [3, 5, 7, -15] and [5, 7, 11, -23], these specify a line in 4D. If four planes in 4D represent a point, subject to the equations representing the planes having consistency:
a[1, 1].x[1] + a[1, 2].x[2] + a[1, 3].x[3] + a[1, 4].x[4] = d[1] ............................................................... a[4, 1].x[1] + ... ... + a[4, 4].x[4] = d[4]
then three planes would represent a line in 4D. Two solution points, such as the above, might lie on this line, so that:
3.a[1, 1] + 5.a[1, 2] + 7.a[1, 3] - 15.a[1, 4] = d[1]
3.a[2, 1] + 5.a[2, 2] + 7.a[2, 3] - 15.a[2, 4] = d[2]
3.a[3, 1] + 5.a[3, 2] + 7.a[3, 3] - 15.a[3, 4] = d[3]
and
5.a[1, 1] + 7.a[1, 2] + 11.a[1, 3] - 23.a[1, 4] = d[1]
5.a[2, 1] + 7.a[2, 2] + 11.a[2, 3] - 23.a[2, 4] = d[2]
5.a[3, 1] + 7.a[3, 2] + 11.a[3, 3] - 23.a[3, 4] = d[3]
The above six equations representing a line in 4D space relate in some way to the direction ratios and intercepts of the line.
There may be a parametric representations of the lines in 4D, 3D and 2D spaces respectively.
This means that some of the matrix coefficients are not independent but decided by the others. There may be other constraints on the coefficients a[i, j] and constants d[1], d[2] and d[3].
Cramer's solution could be appropriate. The constants d[1], ... d[4] are somewhat undecided at present. Let's suppose that the line equation has already been been established. This is:
b[1, 1].x[1] + b[1, 2].x[2] + b[1, 3].x[3] + b[1, 4].x[4] = h[1]
b[2, 1].x[1] + b[2, 2].x[2] + b[2, 3].x[3] + b[2, 4].x[4] = h[2]
b[3, 1].x[1] + b[3, 2].x[2] + b[3, 3].x[3] + b[3, 4].x[4] = h[3]
Here, x[1], ... x[4] are variables and h[1], h[2] and h[3] might be different from the d[1], d[2] and d[3].
In turn, set x[1] = 0, then x[2] = 0, then x[3] = 0 and finally, x[4] = 0 in the three equations above. In each case, subject to consistency, we can find the co-ordinates of the places where the line intersects the planes formed by the axes. Labeling these points but recording which co-ordinate was zero, we have at most four triples [h, k, j] in 3D space. If the line in 4D was parallel to one of the 4D planes, then there would be fewer than four but at least one intersection point.
Logically, in 4D space, if four hyperplanes specify a point and three hyperplanes specify a line, then two hyperplanes would specify a plane and one hyperplane would specify a volume of hyperspace. Hyperplanes such as x[1] = 0, ... x[4] = 0 would logically each represent a volume. This does not affect the above argument. The procedure that gives results is the only procedure that can be followed.
How to find the direction ratios of a line segment: The algebraic differences of the respective co-ordinates of the two points, defining the line, are divided by the Euclidean distance between the two points. Example: Given the primitive solution [datum] points, [Points can be specified by an uppercase letter P, followed by a number in brackets, as P[37], the thirty-seventh point in a polygonal arc that contains thirty-seven points or vertices.] P[1] [3, 5, 7, - 15] and P[2] [5, 7, 11, - 23], the Euclidean distance, H is:
The direction ratios respectively are: [[22]^{1/2}]/22, [[22]^{1/2}]/22, [[22]^{1/2}]/11, - [2.[22]^{1/2}]/11, this being a point in direction ratio space.
The next datum point is: P[3] which is [7, 11, 13, -31]. The three datum points define two line segments of a polygonal arc, having one chord.
It would be interesting to set up equations so that we can solve for the next solution point P[4] which is known to be: [11, 13, 17, -41]. Analytically this point in 4D space is: [11, 13, p[6], -[11 + 13 + p[6]]], where p[6] is seventeen. Direction ratios and intercepts of sides and chord[s] produced are needed to be expressed as equations containing the known primes and composites and the prime, p[6] is required to be computed.
A better analytic expression for p[6] would be: [13 + 2.m[6, 5]], where the difference between the unknown prime ordinal six and the prime thirteen, ordinal five is twice some integer: m[6, 5]. Then, the above point in 4D space would be: [11, 13, [13 + 2.m[6, 5]], - [11 + 13 + [13 + 2.m[6, 5]]]. Before attempting to compute prime, ordinal six, it would be better to set down an ordered list of solution points in 4D and by diminishing the dimension, plot the resulting discrete finite function in 2D. This is a problem in linear algebra. ..... Thence, given the discrete values on the y axis versus the discrete point variable P[j] on the x axis in 2D space, find the intercepts as a function of the gradients for this and higher order derivatives.
The first twenty solution points in ascending order are: P[1] [3, 5, 7, - 15], [2] [5, 7, 11, - 23], [3] [7, 11, 13, - 31], [4] [11, 13, 17, - 41], [5] [13, 17, 19, - 49], [6] [17, 19, 23, - 59], [7], [19, 23, 29, - 71], [8] [23, 29, 31, - 83], [9] [29, 31, 37, - 97], [10] [31, 37, 41, - 109], [11] [37, 41, 43, - 121], [12] [41, 43, 47, - 131], [13] [43, 47, 53, - 143], [14] [47, 53, 59, - 159], [15] [53, 59, 61, - 173], [16] [59, 61, 67, - 187], [17] [61, 67, 71, - 199], [18] [67, 71, 73, - 211], [19] [71, 73, 79, - 223], [20] [73, 79, 83, - 235]. The above twenty solution points constitute a polygonal arc of nineteen segments in 4D space. These arcs, however long, never become a closed polygon.
If the above can be done then the next solution point would be known, producing three chords, as distinct from the sides of the three sided polygonal arc. One of these chords, the major chord starts from point [1] and ends on point [4], specified by: [[1], [4]]. The other chords [minor] are specified by: [[1], [3]] and [[2], [4]]. ...
[The squares of the direction ratios for the line from points [1] and [2] are respectively: [1/22, 1/22, 2/11, 8/11]
The next adjacent point in primitive solution space would give four more direction ratios, constituting another point in direction ratio space. These two points make a joining line, which has itself direction ratios and intersection points on the axis planes. An associated polygonal arc can be constructed in direction ratio space.
SHAWWPG19410425 (talk) 21:11, 25 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 10:04, 27 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 11:25, 27 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 13:52, 28 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 17:53, 29 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 21:25, 29 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 22:18, 30 November 2012 (U.T.C.)SHAWWPG19410425 (talk) 12:32, 1 December 2012 (UTC)