A checked finding

Your pattern, reframed: N − 1 = (k+1)p.

You noticed that some Goldbach pairs have the bigger prime equal to the smaller one times something, plus one. That turns out to be a real number-theory statement — just not a universal one. Every figure below was computed and verified up to ten million.

01The reframing

Your condition is a Goldbach pair with an extra shape:

N = p + q, q = k·p + 1 (p, q prime)

Substitute the second into the first:

N = p + (k·p + 1) = (k+1)·p + 1

the clean form

N − 1 = (k+1)·p

So your question is exactly: does N − 1 have a prime factor p such that N − p is also prime?

02The numbers (verified)

Counting the even numbers that have at least one such pair, over growing ranges:

range of even N	have the property	share
4 – 100	17 / 49	34.7%
4 – 1,000	198 / 499	39.7%
4 – 10,000	1,914 / 4,999	38.3%
4 – 100,000	17,750 / 49,999	35.5%
4 – 1,000,000	163,078 / 499,999	32.6%
4 – 3,000,000	470,824 / 1,499,999	31.4%
4 – 10,000,000	1,506,856 / 4,999,999	30.1%

so it is not universalIt holds for a big minority — never all. And the share slowly falls as N grows (≈38% at ten thousand, ≈33% at a million). There is even a run of 32 consecutive even numbers with no such pair: 649,200 through 649,262.

Two quick examples:

works · 100

100 = 3 + 97, 97 = 3·32 + 1 (and 100 − 1 = 99 = 3 · 33)

fails · 102No prime factor of 101 (which is prime) lands a prime partner — 102 has no pair of this form at all.

03The multiplier can never be an odd prime

For N > 4 both p and q are odd, so q − 1 is even. Since p is odd, k = (q − 1) / p must be even. So if you ask for k itself to be prime, the only option is:

k = 2, i.e. q = 2p + 1

a known structure appearsThose are exactly the Sophie Germain prime pairs — p and 2p + 1 both prime. Your condition has one of the classic prime patterns hiding inside it.

04Tied to the factorization of N − 1

When k + 1 is also prime, N − 1 = p·r is a product of two primes — so each prime factor can give its own pair. Take 120:

119 = 7 · 17 120 = 7 + 113, 113 = 7·16 + 1 120 = 17 + 103, 103 = 17·6 + 1

Both prime factors of N − 1 produce a valid pair. Your Goldbach representation is wired directly to the factorization of N − 1.

An infinite family (so it's not just small-number luck)

Fix p = 3. The condition becomes q = 3k + 1, i.e. q ≡ 1 (mod 3). Every such prime q gives a valid even number N = q + 3. By Dirichlet's theorem there are infinitely many primes q ≡ 1 (mod 3) — so infinitely many N satisfy your pattern.

05Why it drifts down — and where it ends

For each candidate prime p you need a prime q = N − p — and a standard Hardy–Littlewood / Cramér heuristic makes that likelihood about C₂ / ln N per prime factor p of N − 1, where C₂ ≈ 0.660 is the twin-prime constant. By Erdős–Kac, N − 1 has roughly ln ln N distinct prime factors to try. Put together, the chance of failing for a given N is about

P(fail) ≈ exp( − C₂ · ω(N−1) / ln N ) ≈ exp( − C₂ · lnln N / ln N )

so the true limit is 0, not 38%Since ln ln N / ln N → 0, that exponent goes to 0, so P(fail) → 1 and the density → 0 — just extraordinarily slowly. The often-quoted ≈38% is only the value near N ≈ 10⁴; extend the count and it keeps sliding: 38.3% → 30.1% by ten million, and still falling.

◆The honest verdict

did you find something?Yes — a legitimate observation, not a new theorem. The equivalence N − 1 = (k+1)p is real and gives the idea genuine structure, linking a Goldbach pair of N to a factorization of N − 1, and touching Sophie Germain primes and Dirichlet. But since it holds for only ~⅓ of even N (and provably not all), the original “always such a pair” form is disproved.

The precise question it leaves — for which even N does some prime p | (N − 1) also make N − p prime? — turns out to be governed by known heuristics: no special name, a Hardy–Littlewood-type density that decays to zero as N → ∞. Related raw views: the comet, first primes.

asked the experts — and got an answerI posted this on Math Stack Exchange — “Is this property of even numbers studied/named?” The response confirmed it: no special name, and the density follows a Hardy–Littlewood / Erdős–Kac heuristic. (One nuance — the answer quotes ≈38%, but that's a finite-N value; both the formula and the extended count above show the density actually decays toward 0.)