Three new recent discoveries.

#### First,

it turns out that you can generate a complete list of primes in the following fashion.

Using %n% where %p + q = 2n%, let %n = 2% and find any Goldbach partitions %p, q%; if %max(q in bb Q)% is not in set %bb P%, add that %q% to %bb P%. This will generate a set %bb P% such that %bb P% contains every prime up through %2n-3%, in order. Furthermore, the corresponding values of %p% will be small. I haven't determined exactly, but I am confident of %p <= sqrt(2n)% as an upper bound.

###### Sketch of proof

Let %q% be some prime not yet in %bb P%. The first possible inclusion of %q% in a Goldbach pair will be %3 + q = 2n%, and as soon as %n = (q + 3)/2%, you will have that %q%, and it will be the largest such %q% for that %n%.

There are repeats (corresponding to prime gaps) and even backtracks of the largest %q% (deserving further investigation), but they necessarily consist of primes already in %bb P%.

#### Second,

the Cappallo-Goldbach Conjecture, refining the domain of %p% and %q% to make a stronger assertion:

%sf"For any even " N > 2,sf" there exist primes " p, q sf" such that "% %p + q = N sf" and " sqrt(N) <= p <= N//2 <= q <= N - sqrt(N).%

If true, this bound is an improvement on the original %2 <= p <= N//2 <= q <= N-2%.

Or, more simply: there will always be a valid %p% such that %p >= sqrt(N)%.

Note also that there need not be a %p < sqrt(N)%; the first such counter-example is found at %N = 98%, yielding %19 + 79% as the smallest of three partitions.

I have not proven why this should be, but confirmed it empirically through %N = 5000%. I find it suggestive that %N - sqrt(N) = sqrt(N)(sqrt(N) - 1)%, which is reminiscent of the binomial expressions which can also be used to identify primes.

For instance, it appears likely (see Lehmer's totient problem) that for any prime %p%, it must be possible to factor %p - 1% into an arbitrary number of terms of the form %(k^n - k^(n-1))%, where %k% is a unique prime and %n >= 1, n in NN%.

This also likely connects to results such as the %i%th row of Pascal's Triangle consisting only of multiples of %i " iff " i% is prime: or equivalently, that %p in bbb P iff {p:" " p | ((p),(i)) :}, 0 < i < p}%.

#### Third,

**the set of all Goldbach numbers have a bijective mapping to the set of all semiprimes.**

The numbers going across the top represent our %n%. Visually, if one drops a vertical line to meet the diagonal, and then travels on a down/left diagonal, any intersections crossed represent Goldbach pairs.

For example, the line dropped from %8% hits the intersection of the horizontal %11% line and the vertical %5% line, meaning that %5 + 11 = 2(8)% is a solution; the same applies to %3 + 13 = 2(8)%. Since every semiprime can be represented as intersections on this sort of diagram, and every Goldbach pair will likewise be generated once, when intersecting a semiprime, the mapping becomes apparent.

Alternatively, consider that any semiprime is the product of two specific primes, and conversely, any two primes form a unique semiprime. Likewise, any Goldbach pair consists of two specific primes, and while their sum may not be unique among Goldbach pairs, that pair will still only be used a single time in the set of all Goldbach pairs, providing our one-to-one map. As addition and multiplication are both commutative, ordering is irrelevant; it suffices to either always lead by the smaller/larger of the two operands, or allow both orderings when taking the sum or product.

###### Analysis

(Under construction.)

For starters, it is known that there are infinite primes, thus infinite semiprimes, which in turn provides another proof of infinite Goldbach pairs. I also reason that since any given even number has a finite number of Goldbach pairs, there must be an infinite number of Goldbach sums—that is, even numbers with at least one Goldbach pair. If there were only a finite number of these sums, each with a finite number of Goldbach pairs, we would lose our bijection. Unfortunately, this is necessary but not sufficient to prove Goldbach's Conjecture, as the infinitude of sums says nothing about their frequency within %NN%.

There should be other low-hanging fruit here. For example, since it is ...

It is interesting that while the product of two primes is a unique integer, the sum of a Goldbach pair is typically not (e.g. %3+13=5+11=16%). However, because of our one-to-one mapping, we know that the number of semiprimes must equal the number of Goldbach sums.

Let's look at bounding %p% and %q%. If we choose some %C% such that %p <= q <= C%, we yield %(pi^2(C)+pi(C))/2% semiprimes / Goldbach pairs.

(Ignoring %2%s for the moment...)

Let %C=7%.

- %3 * 3 = 9%
- %3 * 5 = 15%
- %3 * 7 = 21%
- %5 * 5 = 25%
- %5 * 7 = 35%
- %7 * 7 = 49%
- %3 + 3 = 2(3) = 6%
- %3 + 5 = 2(4) = 8%
- %3 + 7 = 2(5) = 10%
- %5 + 5 = 2(5) = 10%
- %5 + 7 = 2(6) = 12%
- %7 + 7 = 2(7) = 14%

Let %C = 17%. We obtain the following semiprimes:

- %3 * 3 = 9%
- %3 * 5 = 15%
- %3 * 7 = 21%
- %3 * 11 = 33%
- %3 * 13 = 39%
- %3 * 17 = 51%
- %5 * 5 = 25%
- %5 * 7 = 35%
- %5 * 11 = 55%
- %5 * 13 = 65%
- %5 * 17 = 85%
- %7 * 7 = 49%
- %7 * 11 = 77%
- %7 * 13 = 91%
- %7 * 17 = 119%
- %11 * 11 = 121%
- %11 * 13 = 143%
- %11 * 17 = 187%
- %13 * 13 = 169%
- %13 * 17 = 221%
- %17 * 17 = 289%

All semiprimes will be unique, and they will be bounded above by %C^2% and below by %3^2%.

We obtain the following Goldbach partitions:

- %3 + 3 = 2(3) = 6%
- %3 + 5 = 2(4) = 8%
- %3 + 7 = 2(5) = 10%
- %3 + 11 = 2(7) = 14%
- %3 + 13 = 2(8) = 16%
- %3 + 17 = 2(10) = 20%
- %5 + 5 = 2(5) = 10%
- %5 + 7 = 2(6) = 12%
- %5 + 11 = 2(8) = 16%
- %5 + 13 = 2(9) = 18%
- %5 + 17 = 2(11) = 22%
- %7 + 7 = 2(7) = 14%
- %7 + 11 = 2(9) = 18%
- %7 + 13 = 2(10) = 20%
- %7 + 17 = 2(12) = 24%
- %11 + 11 = 2(11) = 22%
- %11 + 13 = 2(12) = 24%
- %11 + 17 = 2(14) = 28%
- %13 + 13 = 2(13) = 26%
- %13 + 17 = 2(15) = 30%
- %17 + 17 = 2(17) = 34%

All Goldbach pairs will be unique, and their sums bounded above by %2C% and below by %2*3%.

I have been working through this without knowing where I was going, but now that I re-read this last part, it strikes me that the Goldbach bounds are the derivatives of the semiprime bounds with respect to the bound.

#### Reflection

I believe the first observation is essentially trivial number manipulation and contains no new useful truth. I have not yet formed an opinion about the utility of the second.