A checked finding · search strategy

The worst-case hunter — the smartest place to look.

If you wanted to break Goldbach, you'd ignore the spikes and hunt the numbers with the fewest pairs. Here's an equation that generates exactly those worst cases — and the chance of finding a counterexample among them, which is the best odds possible.

the worst-case generator

N = 2^k

Powers of two have no odd prime factors, so the Goldbach boost ∏(p−1)/(p−2) hits its floor of 1 — the fewest possible pairs for their size. These sit on the very bottom edge of the comet. (The family N = 2·prime is just as low.) Hunt here and nowhere else.

k → N = 2^64

2^162^322^642^1282^5122^20482^10000

chance of finding a counterexample at N = 2^64

≈ 1 in 10^3.18×10^15

This is the best odds you can possibly get — the single most vulnerable kind of number at this size. And it's still that.

worst case (N = 2^k)

1 in 10^3.18×10^15

top band (÷2·3·5·7·11·13)

1 in 10^1.22×10^16

targeting advantage vs spikes

10^9.04×10^15× better

Verified on real powers of two (actual pair counts, then the implied fail chance):

N = 2^k	actual pairs	fail chance
2^6	5	1 in 10^2
2^7	3	1 in 10^1
2^8	8	1 in 10^3
2^9	11	1 in 10^5
2^10	22	1 in 10^10
2^11	25	1 in 10^11
2^12	53	1 in 10^23
2^13	76	1 in 10^33
2^14	151	1 in 10^66
2^15	244	1 in 10^106
2^16	435	1 in 10^189
2^17	749	1 in 10^325
2^18	1,314	1 in 10^571
2^19	2,367	1 in 10^1,028
2^20	4,239	1 in 10^1,841
2^21	7,471	1 in 10^3,245

the verdictEven aimed at the single weakest number of its size, the chance of finding a counterexample is 1 in 10^3.18×10^15. Smart targeting buys you an astronomical relative edge over the spikes — and it is still, absolutely, never. The only way a hunt here ever succeeds is if primes secretly correlate against these specific numbers, which is exactly the unproven gap. So this is the best possible place to look — and it's hopeless, which together is why Goldbach is believed but open.

The equation N = 2^k streams worst-case even numbers; the fail chance uses the measured worst-case density ≈ 0.78·N/(lnN)² pairs, then P(fail) ≈ e^(−pairs). Drag k up: the exponent X grows without bound, so 1-in-10^X races to 1-in-10^(10^3000) by k = 10000. (Heuristic, not proof.)

Why 2^k? The number of Goldbach pairs is ≈ 2C₂ · N/(lnN)² · ∏(p−1)/(p−2) over odd primes dividing N. That product is a boost that's bigger the more small factors N has — so it's smallest (equal to 1) when N has none, i.e. a power of two. Fewest pairs ⇒ highest failure chance ⇒ the best target. See the bands this creates on the comet, the growth on lucky primes, and where the whole thing stalls on the frontier.