## Goldbach/ Twin Prime II

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• To summarize (up to ***), or perhaps better phrase, my previous post, we established maxima and minima for numbers of integers, in any set of intervals, of
Message 1 of 6 , Aug 20 4:42 PM
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To summarize (up to ***), or perhaps better phrase, my previous post, we established maxima and minima for numbers of integers, in any set of intervals, of integers i, of specified length, for which at least one member of a set J of primes divides i. The maximum is found when the left-hand enpoint is one. The minimum is found when, for each j in J, the left-hand endpoint is as close as possible to j*min J\ {J}. It is, then, the interval [1, j*min J \ {j}) that should interest us. Consider that J is an initial segment of the sequence of primes. For each j in J \ {2}, we have, in this interval, a single multiple of j, it being j. If we think of our sequences of multiples as sequences of bricks (i.e. each brick is a factor j, in a sequence of multiples of j), it is this one, for each member of J \ {2}, that we can in principle reposition, when we rearrange our sequences of multiples, such as to stop it from taking up an extra integer, as a singleton divisor. By this I mean that, if we have columns for our bricks that are either empty or nonempty ('occupied'), this brick can in principle, and with suitable manipulation, be made simply to be one more divisor out of several in the same column; the column, of course, represents an integer. That makes, in theory, one less occupied column for each member of J \ {2}; and there are two such singletons for the prime two, making in total |J| +1.

And we can manipulate it, otherwise, so that *more* columns are occupied than if the left-hand endpoint is one, by using our 'error margin' -- ie, we squeeze in one extra multiple for each member of J.

So that is a total difference, in principle, of 2|J|+1, for any interval in our set of intervals of equal length.

***
If you take a look at Figure 1, you can see these bricks envisaged the way I have done above (here as grey boxes), for an interval [x,y]. However, here each sequence of them is doubled, by folding the number scale at its midpoint, (x+y)/2. In other words, if for j in J, the sequence of multiples of j occupies k members the interval [x, (x+y)/2], in the doubled columns, 2k are occupied by the folded sequence. You can see that the ones that are empty in the unfolded number scale are all primes, and the ones that are empty in the folded number scale must all be solutions to the Goldbach equation if x = 1 and y is odd. The only other required criterion is to do with the set of the initial segment of primes; I require that what I have thus far called J is the set P(n) of primes {p(1), p(2), ..., p(n)} whose squares do not exceed sqrt(y).

The folding doesn't work for multiples of two; nor does it happen for j in J for which j | (x+y)/2.

Consider a doubled sequence of multiples of p in P(n), as being two different _sieves_. Since the very same parameters as we have discussed above, in regard to how many columns can be occupied, apply to the set of sieves in the bottom half as apply to the set of sieves in the top half, the difference between the maximum and minimum number of columns that can be occupied for any sieve (discounting the operation of squeezing in the extra multiple for each sieve), is, at most, only one (and that's if all the other sieves don't get in the way, i.e. preclude a the creation of a singleton). It is easy to prove that this maximum total of 2|J| becomes negligible in relation to the average number of columns that are empty, among intervals of length p(n)^2, over [1, prod(P(n)]. So the proportionate discepancy from that average diminishes as n --> oo. This proves the Goldbach conjecture.

Tom
• It was suggested that somewhere in the proof of Theorem 1 there would be the error to my proof of the Goldbach (and Twin Prime) conjectures. It was only a
Message 2 of 6 , Aug 25 3:37 PM
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It was suggested that somewhere in the proof of Theorem 1 there would be the error to my proof of the Goldbach (and Twin Prime) conjectures. It was only a short proof, but I have made it yet shorter, as it occurred to me once I had written that `patch' in a post a couple of days ago, that I could cut out vast amounts of the preceding combinatorics if I only gave an explanation of that patch. I have written the explanation below, complete with a paraphrase of the most important part of the preceding combinatorics.

Notation:

/\ = set intersection.
\/ = set union.
A \ B = set A with set B substracted (relative complement of A and B).
A (+) B = symmetric difference of sets A and B (i.e. (A \ B) \/ (B \ A).
prod(K) = product of all the elements of K.

For any sets M and N of integers, let Mult(M, N) be the set of M such that at least one member of N divides M, and for n in |N let
Div(M, n) be the set of members of M that divide n.
Let T(M, n, N) = Mult(M, N \ {i}) /\ Mult(M, N).
Further, p(n) for n = 1, 2, ... will be the sequence of primes.
Let P(n) be the set comprised of the first n primes.

Theorem 1: |Mult([1, y], P(n))| + n >= |Mult([1 + i, y + i], P(n))| >= |Mult([1, y], P(n))| - n + 1.

We note that there is no i in [2, p(n)^2] for which Div(P(n), i) is and empty set, while also, for each q in P(n), there exists i such that

|Mult([1+i,y+i] {q})| - |Mult([1,y] {q})| = 1.

Consider a set K of primes such that |K| > 2, and a nonempty subset L of K, and i in [2, oo) \ J. If for any x,y, i*prod(L) > |[x, y]| while prod(L) <= |[x, y]| and i < |[x, y]|, then if

|Mult([x,y], {i)| = |Mult([x+k,y+k], {i})|,

it follows that
||Mult([x,y], {prod(L), i)| - |Mult([x+k,y+k], {prod(L), i})|| = r,
where r in {0, 1}. Indeed, since prod(L) < i*prod(L), among the members of the set H of intervals [x,y], where x > 0, for which the number of multiples of Y is s, |Mult([x, y], {prod(Y), i})| is minimal if x is the least of all the left-hand endpoints for such members. For any k in [1, oo) \ {K}, it follows that, for any interval I such that |I| = |[k * min K, y]|, if

|sum{|T([k* min K,y], k, {q})| : q in K} - sum{|T(I, k, {q})| : q in K}| = u

then
||T([k* min K, y], k, K)| - |T(I, k, K)|| <= u <= |K|.

By a simple progression, we may prove the first inequality in the statement of Theorem 1.
Let Q(n) = P(n) \ {2}. If

|Mult([2, y+1], P(n))| - |Mult([1, y], P(n))| = 1

then since Mult({1}, P(n)) is an empty set, there exists p in P(n) for which
|Mult([2, y+1], {p})| - |Mult([1, y], {p})| = 1.

Since the integer one is in the range of k, it follows by (1) that, for any i,

|Mult([1, y], P(n))| + n >= |Mult([1 + i, y + i], P(n))|.

Therefore, given any p in Q(n) and any interval I,

|T([2p, 2p + |I| - 1], p, P(n))| + n - 2 >= |T(I, p, P(n))|,

there being notably only one positive multiple, p, of p that is lower than 2p, and Div(P(n),p) = {p}.
Thus we reach the foundation of our proof: if n > 2, for any q in Q(n) \ {p} there is a subset H of P(n) for which q = min H, and for any integer i,

|Mult([pq, y], {pq})| >= |Mult([pq + i, y + i], {pq})|,

while |{k in [1, pq): Div(P(n), k) = {q}}| = 1.
The significance can be explicated as follows. Assume q > p and n > 2. Let
g: [1,y] -- > [2 + i, y + i]
be a bijection that minimises |Div(P(n), k) (+) Div(P(n), g(k))|. Suppose that there exists a subset R of [2 + i, y + i] such that

a) for each k in [pq, y], g(k) in R,

b) for each k in [pq, y], |Div(P(n), g(k))| >= |Div(P(n), k)|,

c) |T([2 + i, y + i) \ R, p, P(n))| = |T([2, pq), p, P(n))| = q - 2, and

d) |Mult([2, y], {pq})| < |Mult([2 + i, y + i], \{pq})|.

We note that, since d) is true, and since for any r,

|{m in [1, pq]: pq | m}| <= |{s in [1 + r, pq + r]: pq | m}|

it follows that for each k in [pq, y] for which Div({p,q}, k) = {p,q}, we have Div({p, q}, g(k)) = {p,q}. Therefore, since there is no k in [2, pq) \ {q} for which Div(P(n), k) = {q}, if

|Mult([2 + i, y + i], {q})| = |Mult([1, y], {q})|

then there is no g(k) for which Div(P(n), g(k)) = {p, q} and Div(P(n), k) = p. Thus, since d) is true, for some r in P(n) \{p,q}, there exists k such that Div(K, k) = {q,r} while Div(K, g(k)) = {p, q, r}. It follows that, if

|T([2 + i, y + i] \ R, p, {q})| > |T([pq, y], p, {q})|,

we have

|Mult([2 + i, y + i], {q})| > |Mult([1, y], {q})|.

Thus |Mult([2 + i, y + i], {p, q})| = |Mult([2, y], {p, q})|, rendering the inequality |T([2 + i, y + i], p, {q})| > |T([2, y], p, {q})| superfluous to our method.

http://unsolvedproblems.org/S13.pdf
Should be updated shortly.
Tom

--- In primenumbers@yahoogroups.com, "descarthes" <mgullandt@...> wrote:
>
> To summarize (up to ***), or perhaps better phrase, my previous post, we established maxima and minima for numbers of integers, in any set of intervals, of integers i, of specified length, for which at least one member of a set J of primes divides i. The maximum is found when the left-hand enpoint is one. The minimum is found when, for each j in J, the left-hand endpoint is as close as possible to j*min J\ {J}. It is, then, the interval [1, j*min J \ {j}) that should interest us. Consider that J is an initial segment of the sequence of primes. For each j in J \ {2}, we have, in this interval, a single multiple of j, it being j. If we think of our sequences of multiples as sequences of bricks (i.e. each brick is a factor j, in a sequence of multiples of j), it is this one, for each member of J \ {2}, that we can in principle reposition, when we rearrange our sequences of multiples, such as to stop it from taking up an extra integer, as a singleton divisor. By this I mean that, if we have columns for our bricks that are either empty or nonempty ('occupied'), this brick can in principle, and with suitable manipulation, be made simply to be one more divisor out of several in the same column; the column, of course, represents an integer. That makes, in theory, one less occupied column for each member of J \ {2}; and there are two such singletons for the prime two, making in total |J| +1.
>
> And we can manipulate it, otherwise, so that *more* columns are occupied than if the left-hand endpoint is one, by using our 'error margin' -- ie, we squeeze in one extra multiple for each member of J.
>
> So that is a total difference, in principle, of 2|J|+1, for any interval in our set of intervals of equal length.
>
> ***
> If you take a look at Figure 1, you can see these bricks envisaged the way I have done above (here as grey boxes), for an interval [x,y]. However, here each sequence of them is doubled, by folding the number scale at its midpoint, (x+y)/2. In other words, if for j in J, the sequence of multiples of j occupies k members the interval [x, (x+y)/2], in the doubled columns, 2k are occupied by the folded sequence. You can see that the ones that are empty in the unfolded number scale are all primes, and the ones that are empty in the folded number scale must all be solutions to the Goldbach equation if x = 1 and y is odd. The only other required criterion is to do with the set of the initial segment of primes; I require that what I have thus far called J is the set P(n) of primes {p(1), p(2), ..., p(n)} whose squares do not exceed sqrt(y).
>
> The folding doesn't work for multiples of two; nor does it happen for j in J for which j | (x+y)/2.
>
> Consider a doubled sequence of multiples of p in P(n), as being two different _sieves_. Since the very same parameters as we have discussed above, in regard to how many columns can be occupied, apply to the set of sieves in the bottom half as apply to the set of sieves in the top half, the difference between the maximum and minimum number of columns that can be occupied for any sieve (discounting the operation of squeezing in the extra multiple for each sieve), is, at most, only one (and that's if all the other sieves don't get in the way, i.e. preclude a the creation of a singleton). It is easy to prove that this maximum total of 2|J| becomes negligible in relation to the average number of columns that are empty, among intervals of length p(n)^2, over [1, prod(P(n)]. So the proportionate discepancy from that average diminishes as n --> oo. This proves the Goldbach conjecture.
>
> Tom
>
• We note that there is no i in [2, p(n)^2] for which Div(P(n) should read We note that, for any p and r in P(n) there is no i in [2, pr] for which
Message 3 of 6 , Aug 25 6:24 PM
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"We note that there is no i in [2, p(n)^2] for which Div(P(n)" should read
" We note that, for any p and r in P(n) there is no i in [2, pr] for which Div(P(n)".

BTW My reference to (1) is a reference to this equation:
"||T([k* min K, y], k, K)| - |T(I, k, K)|| <= u <= |K|".

Tom

--- In primenumbers@yahoogroups.com, "descarthes" <mgullandt@...> wrote:
>
> It was suggested that somewhere in the proof of Theorem 1 there would be the error to my proof of the Goldbach (and Twin Prime) conjectures. It was only a short proof, but I have made it yet shorter, as it occurred to me once I had written that `patch' in a post a couple of days ago, that I could cut out vast amounts of the preceding combinatorics if I only gave an explanation of that patch. I have written the explanation below, complete with a paraphrase of the most important part of the preceding combinatorics.
>
> Notation:
>
> /\ = set intersection.
> \/ = set union.
> A \ B = set A with set B substracted (relative complement of A and B).
> A (+) B = symmetric difference of sets A and B (i.e. (A \ B) \/ (B \ A).
> prod(K) = product of all the elements of K.
>
> For any sets M and N of integers, let Mult(M, N) be the set of M such that at least one member of N divides M, and for n in |N let
> Div(M, n) be the set of members of M that divide n.
> Let T(M, n, N) = Mult(M, N \ {i}) /\ Mult(M, N).
> Further, p(n) for n = 1, 2, ... will be the sequence of primes.
> Let P(n) be the set comprised of the first n primes.
>
>
> Theorem 1: |Mult([1, y], P(n))| + n >= |Mult([1 + i, y + i], P(n))| >= |Mult([1, y], P(n))| - n + 1.
>
> We note that there is no i in [2, p(n)^2] for which Div(P(n), i) is and empty set, while also, for each q in P(n), there exists i such that
>
> |Mult([1+i,y+i] {q})| - |Mult([1,y] {q})| = 1.
>
> Consider a set K of primes such that |K| > 2, and a nonempty subset L of K, and i in [2, oo) \ J. If for any x,y, i*prod(L) > |[x, y]| while prod(L) <= |[x, y]| and i < |[x, y]|, then if
>
> |Mult([x,y], {i)| = |Mult([x+k,y+k], {i})|,
>
> it follows that
> ||Mult([x,y], {prod(L), i)| - |Mult([x+k,y+k], {prod(L), i})|| = r,
> where r in {0, 1}. Indeed, since prod(L) < i*prod(L), among the members of the set H of intervals [x,y], where x > 0, for which the number of multiples of Y is s, |Mult([x, y], {prod(Y), i})| is minimal if x is the least of all the left-hand endpoints for such members. For any k in [1, oo) \ {K}, it follows that, for any interval I such that |I| = |[k * min K, y]|, if
>
> |sum{|T([k* min K,y], k, {q})| : q in K} - sum{|T(I, k, {q})| : q in K}| = u
>
> then
> ||T([k* min K, y], k, K)| - |T(I, k, K)|| <= u <= |K|.
>
> By a simple progression, we may prove the first inequality in the statement of Theorem 1.
> Let Q(n) = P(n) \ {2}. If
>
> |Mult([2, y+1], P(n))| - |Mult([1, y], P(n))| = 1
>
> then since Mult({1}, P(n)) is an empty set, there exists p in P(n) for which
> |Mult([2, y+1], {p})| - |Mult([1, y], {p})| = 1.
>
> Since the integer one is in the range of k, it follows by (1) that, for any i,
>
> |Mult([1, y], P(n))| + n >= |Mult([1 + i, y + i], P(n))|.
>
> Therefore, given any p in Q(n) and any interval I,
>
> |T([2p, 2p + |I| - 1], p, P(n))| + n - 2 >= |T(I, p, P(n))|,
>
> there being notably only one positive multiple, p, of p that is lower than 2p, and Div(P(n),p) = {p}.
> Thus we reach the foundation of our proof: if n > 2, for any q in Q(n) \ {p} there is a subset H of P(n) for which q = min H, and for any integer i,
>
> |Mult([pq, y], {pq})| >= |Mult([pq + i, y + i], {pq})|,
>
> while |{k in [1, pq): Div(P(n), k) = {q}}| = 1.
> The significance can be explicated as follows. Assume q > p and n > 2. Let
> g: [1,y] -- > [2 + i, y + i]
> be a bijection that minimises |Div(P(n), k) (+) Div(P(n), g(k))|. Suppose that there exists a subset R of [2 + i, y + i] such that
>
> a) for each k in [pq, y], g(k) in R,
>
> b) for each k in [pq, y], |Div(P(n), g(k))| >= |Div(P(n), k)|,
>
> c) |T([2 + i, y + i) \ R, p, P(n))| = |T([2, pq), p, P(n))| = q - 2, and
>
> d) |Mult([2, y], {pq})| < |Mult([2 + i, y + i], \{pq})|.
>
> We note that, since d) is true, and since for any r,
>
> |{m in [1, pq]: pq | m}| <= |{s in [1 + r, pq + r]: pq | m}|
>
> it follows that for each k in [pq, y] for which Div({p,q}, k) = {p,q}, we have Div({p, q}, g(k)) = {p,q}. Therefore, since there is no k in [2, pq) \ {q} for which Div(P(n), k) = {q}, if
>
> |Mult([2 + i, y + i], {q})| = |Mult([1, y], {q})|
>
> then there is no g(k) for which Div(P(n), g(k)) = {p, q} and Div(P(n), k) = p. Thus, since d) is true, for some r in P(n) \{p,q}, there exists k such that Div(K, k) = {q,r} while Div(K, g(k)) = {p, q, r}. It follows that, if
>
> |T([2 + i, y + i] \ R, p, {q})| > |T([pq, y], p, {q})|,
>
> we have
>
> |Mult([2 + i, y + i], {q})| > |Mult([1, y], {q})|.
>
> Thus |Mult([2 + i, y + i], {p, q})| = |Mult([2, y], {p, q})|, rendering the inequality |T([2 + i, y + i], p, {q})| > |T([2, y], p, {q})| superfluous to our method.
>
> http://unsolvedproblems.org/S13.pdf
> Should be updated shortly.
> Tom
>
> --- In primenumbers@yahoogroups.com, "descarthes" <mgullandt@> wrote:
> >
> > To summarize (up to ***), or perhaps better phrase, my previous post, we established maxima and minima for numbers of integers, in any set of intervals, of integers i, of specified length, for which at least one member of a set J of primes divides i. The maximum is found when the left-hand enpoint is one. The minimum is found when, for each j in J, the left-hand endpoint is as close as possible to j*min J\ {J}. It is, then, the interval [1, j*min J \ {j}) that should interest us. Consider that J is an initial segment of the sequence of primes. For each j in J \ {2}, we have, in this interval, a single multiple of j, it being j. If we think of our sequences of multiples as sequences of bricks (i.e. each brick is a factor j, in a sequence of multiples of j), it is this one, for each member of J \ {2}, that we can in principle reposition, when we rearrange our sequences of multiples, such as to stop it from taking up an extra integer, as a singleton divisor. By this I mean that, if we have columns for our bricks that are either empty or nonempty ('occupied'), this brick can in principle, and with suitable manipulation, be made simply to be one more divisor out of several in the same column; the column, of course, represents an integer. That makes, in theory, one less occupied column for each member of J \ {2}; and there are two such singletons for the prime two, making in total |J| +1.
> >
> > And we can manipulate it, otherwise, so that *more* columns are occupied than if the left-hand endpoint is one, by using our 'error margin' -- ie, we squeeze in one extra multiple for each member of J.
> >
> > So that is a total difference, in principle, of 2|J|+1, for any interval in our set of intervals of equal length.
> >
> > ***
> > If you take a look at Figure 1, you can see these bricks envisaged the way I have done above (here as grey boxes), for an interval [x,y]. However, here each sequence of them is doubled, by folding the number scale at its midpoint, (x+y)/2. In other words, if for j in J, the sequence of multiples of j occupies k members the interval [x, (x+y)/2], in the doubled columns, 2k are occupied by the folded sequence. You can see that the ones that are empty in the unfolded number scale are all primes, and the ones that are empty in the folded number scale must all be solutions to the Goldbach equation if x = 1 and y is odd. The only other required criterion is to do with the set of the initial segment of primes; I require that what I have thus far called J is the set P(n) of primes {p(1), p(2), ..., p(n)} whose squares do not exceed sqrt(y).
> >
> > The folding doesn't work for multiples of two; nor does it happen for j in J for which j | (x+y)/2.
> >
> > Consider a doubled sequence of multiples of p in P(n), as being two different _sieves_. Since the very same parameters as we have discussed above, in regard to how many columns can be occupied, apply to the set of sieves in the bottom half as apply to the set of sieves in the top half, the difference between the maximum and minimum number of columns that can be occupied for any sieve (discounting the operation of squeezing in the extra multiple for each sieve), is, at most, only one (and that's if all the other sieves don't get in the way, i.e. preclude a the creation of a singleton). It is easy to prove that this maximum total of 2|J| becomes negligible in relation to the average number of columns that are empty, among intervals of length p(n)^2, over [1, prod(P(n)]. So the proportionate discepancy from that average diminishes as n --> oo. This proves the Goldbach conjecture.
> >
> > Tom
> >
>
• Oh dear! That should be We note that, for any p and r in P(n) {p}, where r
Message 4 of 6 , Aug 25 6:48 PM
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Oh dear! That should be
"We note that, for any p and r in P(n) \ {p}, where r < p, there is no i in T([2, pr], p, P(n)) for which... ".

I hope there aren't any more of those....
Tom

--- In primenumbers@yahoogroups.com, "descarthes" <mgullandt@...> wrote:
>
> "We note that there is no i in [2, p(n)^2] for which Div(P(n)" should read
> " We note that, for any p and r in P(n) there is no i in [2, pr] for which Div(P(n)".
>
> BTW My reference to (1) is a reference to this equation:
> "||T([k* min K, y], k, K)| - |T(I, k, K)|| <= u <= |K|".
>
> Tom
>
> --- In primenumbers@yahoogroups.com, "descarthes" <mgullandt@> wrote:
> >
> > It was suggested that somewhere in the proof of Theorem 1 there would be the error to my proof of the Goldbach (and Twin Prime) conjectures. It was only a short proof, but I have made it yet shorter, as it occurred to me once I had written that `patch' in a post a couple of days ago, that I could cut out vast amounts of the preceding combinatorics if I only gave an explanation of that patch. I have written the explanation below, complete with a paraphrase of the most important part of the preceding combinatorics.
> >
> > Notation:
> >
> > /\ = set intersection.
> > \/ = set union.
> > A \ B = set A with set B substracted (relative complement of A and B).
> > A (+) B = symmetric difference of sets A and B (i.e. (A \ B) \/ (B \ A).
> > prod(K) = product of all the elements of K.
> >
> > For any sets M and N of integers, let Mult(M, N) be the set of M such that at least one member of N divides M, and for n in |N let
> > Div(M, n) be the set of members of M that divide n.
> > Let T(M, n, N) = Mult(M, N \ {i}) /\ Mult(M, N).
> > Further, p(n) for n = 1, 2, ... will be the sequence of primes.
> > Let P(n) be the set comprised of the first n primes.
> >
> >
> > Theorem 1: |Mult([1, y], P(n))| + n >= |Mult([1 + i, y + i], P(n))| >= |Mult([1, y], P(n))| - n + 1.
> >
> > We note that there is no i in [2, p(n)^2] for which Div(P(n), i) is and empty set, while also, for each q in P(n), there exists i such that
> >
> > |Mult([1+i,y+i] {q})| - |Mult([1,y] {q})| = 1.
> >
> > Consider a set K of primes such that |K| > 2, and a nonempty subset L of K, and i in [2, oo) \ J. If for any x,y, i*prod(L) > |[x, y]| while prod(L) <= |[x, y]| and i < |[x, y]|, then if
> >
> > |Mult([x,y], {i)| = |Mult([x+k,y+k], {i})|,
> >
> > it follows that
> > ||Mult([x,y], {prod(L), i)| - |Mult([x+k,y+k], {prod(L), i})|| = r,
> > where r in {0, 1}. Indeed, since prod(L) < i*prod(L), among the members of the set H of intervals [x,y], where x > 0, for which the number of multiples of Y is s, |Mult([x, y], {prod(Y), i})| is minimal if x is the least of all the left-hand endpoints for such members. For any k in [1, oo) \ {K}, it follows that, for any interval I such that |I| = |[k * min K, y]|, if
> >
> > |sum{|T([k* min K,y], k, {q})| : q in K} - sum{|T(I, k, {q})| : q in K}| = u
> >
> > then
> > ||T([k* min K, y], k, K)| - |T(I, k, K)|| <= u <= |K|.
> >
> > By a simple progression, we may prove the first inequality in the statement of Theorem 1.
> > Let Q(n) = P(n) \ {2}. If
> >
> > |Mult([2, y+1], P(n))| - |Mult([1, y], P(n))| = 1
> >
> > then since Mult({1}, P(n)) is an empty set, there exists p in P(n) for which
> > |Mult([2, y+1], {p})| - |Mult([1, y], {p})| = 1.
> >
> > Since the integer one is in the range of k, it follows by (1) that, for any i,
> >
> > |Mult([1, y], P(n))| + n >= |Mult([1 + i, y + i], P(n))|.
> >
> > Therefore, given any p in Q(n) and any interval I,
> >
> > |T([2p, 2p + |I| - 1], p, P(n))| + n - 2 >= |T(I, p, P(n))|,
> >
> > there being notably only one positive multiple, p, of p that is lower than 2p, and Div(P(n),p) = {p}.
> > Thus we reach the foundation of our proof: if n > 2, for any q in Q(n) \ {p} there is a subset H of P(n) for which q = min H, and for any integer i,
> >
> > |Mult([pq, y], {pq})| >= |Mult([pq + i, y + i], {pq})|,
> >
> > while |{k in [1, pq): Div(P(n), k) = {q}}| = 1.
> > The significance can be explicated as follows. Assume q > p and n > 2. Let
> > g: [1,y] -- > [2 + i, y + i]
> > be a bijection that minimises |Div(P(n), k) (+) Div(P(n), g(k))|. Suppose that there exists a subset R of [2 + i, y + i] such that
> >
> > a) for each k in [pq, y], g(k) in R,
> >
> > b) for each k in [pq, y], |Div(P(n), g(k))| >= |Div(P(n), k)|,
> >
> > c) |T([2 + i, y + i) \ R, p, P(n))| = |T([2, pq), p, P(n))| = q - 2, and
> >
> > d) |Mult([2, y], {pq})| < |Mult([2 + i, y + i], \{pq})|.
> >
> > We note that, since d) is true, and since for any r,
> >
> > |{m in [1, pq]: pq | m}| <= |{s in [1 + r, pq + r]: pq | m}|
> >
> > it follows that for each k in [pq, y] for which Div({p,q}, k) = {p,q}, we have Div({p, q}, g(k)) = {p,q}. Therefore, since there is no k in [2, pq) \ {q} for which Div(P(n), k) = {q}, if
> >
> > |Mult([2 + i, y + i], {q})| = |Mult([1, y], {q})|
> >
> > then there is no g(k) for which Div(P(n), g(k)) = {p, q} and Div(P(n), k) = p. Thus, since d) is true, for some r in P(n) \{p,q}, there exists k such that Div(K, k) = {q,r} while Div(K, g(k)) = {p, q, r}. It follows that, if
> >
> > |T([2 + i, y + i] \ R, p, {q})| > |T([pq, y], p, {q})|,
> >
> > we have
> >
> > |Mult([2 + i, y + i], {q})| > |Mult([1, y], {q})|.
> >
> > Thus |Mult([2 + i, y + i], {p, q})| = |Mult([2, y], {p, q})|, rendering the inequality |T([2 + i, y + i], p, {q})| > |T([2, y], p, {q})| superfluous to our method.
> >
> > http://unsolvedproblems.org/S13.pdf
> > Should be updated shortly.
> > Tom
> >
> > --- In primenumbers@yahoogroups.com, "descarthes" <mgullandt@> wrote:
> > >
> > > To summarize (up to ***), or perhaps better phrase, my previous post, we established maxima and minima for numbers of integers, in any set of intervals, of integers i, of specified length, for which at least one member of a set J of primes divides i. The maximum is found when the left-hand enpoint is one. The minimum is found when, for each j in J, the left-hand endpoint is as close as possible to j*min J\ {J}. It is, then, the interval [1, j*min J \ {j}) that should interest us. Consider that J is an initial segment of the sequence of primes. For each j in J \ {2}, we have, in this interval, a single multiple of j, it being j. If we think of our sequences of multiples as sequences of bricks (i.e. each brick is a factor j, in a sequence of multiples of j), it is this one, for each member of J \ {2}, that we can in principle reposition, when we rearrange our sequences of multiples, such as to stop it from taking up an extra integer, as a singleton divisor. By this I mean that, if we have columns for our bricks that are either empty or nonempty ('occupied'), this brick can in principle, and with suitable manipulation, be made simply to be one more divisor out of several in the same column; the column, of course, represents an integer. That makes, in theory, one less occupied column for each member of J \ {2}; and there are two such singletons for the prime two, making in total |J| +1.
> > >
> > > And we can manipulate it, otherwise, so that *more* columns are occupied than if the left-hand endpoint is one, by using our 'error margin' -- ie, we squeeze in one extra multiple for each member of J.
> > >
> > > So that is a total difference, in principle, of 2|J|+1, for any interval in our set of intervals of equal length.
> > >
> > > ***
> > > If you take a look at Figure 1, you can see these bricks envisaged the way I have done above (here as grey boxes), for an interval [x,y]. However, here each sequence of them is doubled, by folding the number scale at its midpoint, (x+y)/2. In other words, if for j in J, the sequence of multiples of j occupies k members the interval [x, (x+y)/2], in the doubled columns, 2k are occupied by the folded sequence. You can see that the ones that are empty in the unfolded number scale are all primes, and the ones that are empty in the folded number scale must all be solutions to the Goldbach equation if x = 1 and y is odd. The only other required criterion is to do with the set of the initial segment of primes; I require that what I have thus far called J is the set P(n) of primes {p(1), p(2), ..., p(n)} whose squares do not exceed sqrt(y).
> > >
> > > The folding doesn't work for multiples of two; nor does it happen for j in J for which j | (x+y)/2.
> > >
> > > Consider a doubled sequence of multiples of p in P(n), as being two different _sieves_. Since the very same parameters as we have discussed above, in regard to how many columns can be occupied, apply to the set of sieves in the bottom half as apply to the set of sieves in the top half, the difference between the maximum and minimum number of columns that can be occupied for any sieve (discounting the operation of squeezing in the extra multiple for each sieve), is, at most, only one (and that's if all the other sieves don't get in the way, i.e. preclude a the creation of a singleton). It is easy to prove that this maximum total of 2|J| becomes negligible in relation to the average number of columns that are empty, among intervals of length p(n)^2, over [1, prod(P(n)]. So the proportionate discepancy from that average diminishes as n --> oo. This proves the Goldbach conjecture.
> > >
> > > Tom
> > >
> >
>
• The owner of the UnsolvedProblems site no longer wants to update the site sooner than a month s time (I had made a couple of silly mistakes), so my website has
Message 5 of 6 , Aug 29 8:38 AM
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The owner of the UnsolvedProblems site no longer wants to update the site sooner than a month's time (I had made a couple of silly mistakes), so my website has the most up-to-date version:
www.foldednumbers.co.uk ,
complete with John-Derbyshire-style colloquial explanations.

Tom
• This should, now, open in Internet Explorer, albeit with unsatisfactory typesetting (I can t suss that HTML)! It s also got some minor improvements/corrections
Message 6 of 6 , Aug 30 8:25 PM
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This should, now, open in Internet Explorer, albeit with unsatisfactory typesetting (I can't suss that HTML)!
It's also got some minor improvements/corrections put in.
http://www.foldednumbers.co.uk/
The colloquial version has been a hit, if offlist replies are anything to go by.
Thanks to those who've given me positive feedback.
Tom
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