Drawing unsolved problems at random — a record of the smaller problems
The method was fixed in advance.
Problems are drawn at random. When it feels stuck, move to another problem and switch attention. Record the thinking at the point of getting stuck, together with the order and the intervals of the draws (measured from the machine). Measure whatever mathematical software can measure.
A lottery box was built and drawn from with random numbers. Seven problems drawn, six taken up.
Four of the problems drawn here have articles of their own. What stays in this article is the record of the smaller problems — Sierpiński numbers, Riesel numbers, distinct distances — and the method itself.
| draw | problem drawn | what became of it |
|---|---|---|
| 1 | the Erdős arithmetic-progression conjecture | → its own article (six searches / k=4 / the price of self-similarity) |
| 2 | the Hadwiger–Nelson problem | → its own article (f(3) = 10; f(4) is 14 or 15) |
| 3 | Sierpiński numbers | §02 of this article |
| 4 | the Erdős distinct distances problem | §03 of this article |
| 5 | the Lovász conjecture | → its own article (four exceptions; the search space for a fifth) |
| 6 | the BSD conjecture | → its own article (up to rank 7) |
| 7 | the square peg problem | drawn and put back after 8 seconds |
※ Not drawn in this lottery, but the same day's work also took in the Collatz conjecture and the Riemann hypothesis (those were taken up by choice, not by random draw).
computationchecked on this machine, within the range stated knowna restatement, a known theorem, or a check of the literature. Nothing in this article is closed in Lean.
What happened in this round
| draw | # | problem | what happened there |
|---|---|---|---|
| 1 | 22 | the Erdős arithmetic-progression conjecture | the spine of this round. its own article |
| 2 | 19 | the Hadwiger–Nelson problem | switching attention. The search above kept running in the background |
| 3 | 7 | Sierpiński numbers | confirmed that "the provable part" is already closed |
| 4 | 9 | the Erdős distinct distances problem | saw the remaining gap numerically |
| 5 | 20 | the Lovász conjecture | 8,713 Cayley graphs tested |
| 6 | 28 | the BSD conjecture | 14-digit agreement in territory where BSD is not a theorem |
| 7 | 5 | the square peg problem | drawn and put back after 8 seconds |
| 8 | 19 | Hadwiger–Nelson (again) | a change of angle, toward the fractional chromatic number |
※ The square peg problem, drawn seventh, was redrawn 8 seconds later. Why it was put back is not recorded. What the record holds is only that it was 8 seconds.
Three problems gave conclusions of the same shape
The largest thing to come out of this round was that three entirely different problems had the same shape.
"The provable part" is already closed.
What remains is only the part that method cannot reach in principle.
Sierpiński numbers (#7)
Is 78557 the smallest Sierpiński number? That 78557 is a Sierpiński number at all is settled by a finite check.
The least common multiple of the orders of 2 for the covering set {3,5,7,13,19,37,73} is L=36, so checking the 36 cases of n mod 36 settles it for every n.—no uncovered residue. Proof complete.
What about candidates below 78557? For the 39,278 odd k < 78557, we searched for the smallest n making k·2n+1 prime.
| n ≤ | k remaining |
|---|---|
| 0 (all) | 39,278 |
| 1 | 32,040 |
| 10 | 8,870 |
| 100 | 981 |
| 400 | 351 |
The 5 for which no prime has yet been found (21181, 22699, 24737, 55459, 67607) are all inside that 351. The other 346 are ones resolved at larger n (the ones PrimeGrid has ground down over decades). These 5 have not been shown to lack a covering set; they have not been proved to be Sierpiński numbers at all — all that is established is that there is no prime for n < 45,543,700, and the expectation runs the other way (a prime will eventually appear). known
And here is the real point. Whether k has a covering set of period L is decided by a finite computation. (If p divides k·2n+1 then the order of 2 must divide L, so the candidate primes are restricted to the prime factors of 2L−1.)
Below 78557 there is not one k with a covering set.
That is, the road of "proof by covering set" is completely closed.
What remains is a quite different question: is there a Sierpiński number with no covering set? — posed as Erdős problem #1113, and open. It cannot be settled by a finite computation. known
It is not only the periods of primes that cover
The count above restricts "covering" to the periods of primes. That is not the only thing that covers a whole residue class — algebraic factorisations cover too. The Sophie Germain identity:
Take k = m⁴ (m odd) and write n = 4j+2, so that k·2ⁿ = 4·(m·2j)⁴; then for n ≡ 2 (mod 4) compositeness follows without using a single prime (81·2⁶+1 = 5,185 = 71 × 73; 625·2⁶+1 = 40,001 = 199 × 201). There is a quarter less to cover.
There are 7 fourth powers of odd numbers below 78557 (81, 625, 2401, 6561, 14641, 28561, 50625). Checking for each whether n ≢ 2 (mod 4) can be covered by primes up to period 96, none of the 7 can be covered, and even the closest leave 3 classes (2401 = 7⁴, 14641 = 11⁴ and 28561 = 13⁴ leave 3 classes at period 12). computation
The odd m for which an algebraic factorisation is available are completely classified by Capelli's criterion — for m > 1, only m = cq (q an odd prime; the class n ≡ 0 mod q) and m = c⁴ (the class n ≡ 2 mod 4). So for m > 1 the set covered by algebra is always periodic and of density below 1. known The core of #1113 is whether a third mechanism, neither covering nor algebraic factorisation, can exist; no theorem rules it out.
"Has a covering set or not" becomes a continuous quantity through the sieve weight (the Proth weight) Wk — it is 0 when a finite period covers, and settles at a positive value for ordinary k. The weights of the remaining 5 are between 0.07 and 0.27 (the population mean is 2), and the expected number of k surviving to depth 4.5×10⁷ is 4 to 7 — there is nothing unnatural in 5 being still unresolved. known
Riesel numbers (same machine, straight after)
The k·2n−1 side. The conjectured smallest is 509203, covering set {3,5,7,13,17,241}, L=24. By the same procedure—the only odd k ≤ 509203 with a covering set is 509203 itself.
The Erdős arithmetic-progression conjecture (#22)
The one that became its own article had the same shape. The road of "raising the exponent of the logarithm above 1" goes through for k=3 and does not for k=4. And Green–Tao themselves write that this "is the limit of our method". The position at which the method closes is clearly known.
The gap became visible numerically (#9, distinct distances)
g(n) is the minimum number of distinct distances determined by n points in the plane. In 2015 Guth–Katz proved g(n) ≥ c·n/log n, effectively settling it, but Erdős's conjecture is n/√(log n). The gap that remains is a factor of √log n.
We counted the distinct distances of an m×m grid directly.
| m | n = m² | D | D ÷ (n/√log n) | D ÷ (n/log n) |
|---|---|---|---|---|
| 128 | 16,384 | 5,838 | 1.1100 | 3.4578 |
| 512 | 262,144 | 82,489 | 1.1115 | 3.9260 |
| 2048 | 4,194,304 | 1,196,234 | 1.1137 | 4.3491 |
Divided by n/√log n, the ratio levels off around 1.11—you can see numerically that the shape of Erdős's conjecture is right.
Divided by n/log n, the ratio keeps rising, 3.46 → 4.35—the gap between the Guth–Katz lower bound and the true value is plainly visible even at this size.
But this problem is different in character from the previous two. Nobody knows the exact values beyond n=13. The configuration space is continuous, so combinatorial search cannot reach it—it is not "closed" but "entered from a different door".
Those that became articles of their own
| problem | what came out here | what became of it |
|---|---|---|
| the Lovász conjecture (#20) | only 5 connected vertex-transitive graphs without a Hamiltonian cycle are known. We constructed and checked the 4 other than K₂—all 4 have no cycle but do have a path | → The Lovász conjecture The sweep now stands at 9,805. The search space for a fifth exception splits three ways by the deficiency def |
| the Hadwiger–Nelson problem (#19) | drawn twice; entered from the side of the fractional chromatic number. χf ≥ n/α can be computed from a finite graph, but this route hits a ceiling at 4.36 | → The Hadwiger–Nelson problem f(α), the largest unit-distance graph with independence number ≤ α. f(3) = 10; f(4) is 14 or 15 |
| the BSD conjecture (#28) | |Ш| computed backwards from the strong form. In ranks 2 and 3—territory where BSD is not a theorem—it matched an integer to 14 decimal places | → The BSD conjecture later extended up to rank 7 |
Three errors of the type "the name did not match the contents"
All three were of the type "the name did not match the contents". And all three were saved by numbers that were being printed.
| where | what was wrong | what saved it |
|---|---|---|
| Lovász | the row labelled "A5" in the output table was actually A4. Both chosen generators fixed the point 4. A5 had not been examined | the order was being printed (12 where it should have been 120) |
| BSD | the points returned by ellrank are not saturated. They only generate a finite-index subgroup | the height was 0.4600 = 9 × 0.05111 (index 3) |
elltamagawa includes the factor at the infinite place. The name suggests a product over finite places. The same factor was being counted twice | — | |
ellanalyticrank(E)[2] is L(r)(E,1), not divided by r! | the discrepancy was exactly a factorial (exactly 2× at rank 2, exactly 6× at rank 3) |
The three BSD errors overlapped to produce the value |Ш| = 1/18. The cause was findable because the discrepancy had the simple shape of 2 and 6; had it stayed at 1/18 I think it would have gone unexplained for a long time.
"Searched up to 106 and did not find it" and "does not exist" are different. In the same way, "examined A5" and "examined a variable named A5" are different.
In this round the latter happened three times. Had the order, the height and the shape of the discrepancy not been printed, all three would have passed.
The Hadwiger–Nelson problem — its own article
A problem drawn twice. χ(ℝ²) ∈ {5, 6, 7} has not moved. The record of entering from the side of the fractional chromatic number, and the quantity measured from there — f(α), the largest unit-distance graph with independence number at most α; f(3) = 10, and f(4) is 14 or 15 — is in The Hadwiger–Nelson problem.
About this method
Seven problems drawn, six taken up, and none of them solved. Even so, some things about the method worked.
| what worked | content |
|---|---|
| things can run in the background | the search for #22 kept turning the whole time #19, #7 and #9 were being touched. Being stuck overlaps with waiting |
| matching shapes become visible | "the provable part is closed" came out in three problems. Digging at one problem alone, you would not see it |
| tools flow sideways | the covering-set tester from #7 worked on the Riesel problem as it stood. CP-SAT was used on three problems, #22, #19 and #20 |
| the interval stays on the record | "put back after 8 seconds" stays on the record. The record of one's own behaviour is more accurate than one's inside |
Some things did not work. None of the problems got more than one round's depth. Only #22 was carried on and became an article; the rest stopped at "measured the terrain at the entrance".
Sources and reproduction
There is no new mathematics in this article. Every theorem used and every known record is public. What this article did was check them again on its own machine, and record which positions could be checked and which could not.
| value | where it came from |
|---|---|
| the covering of 78557 (L=36, no uncovered residue) | this machine. A finite check |
| the narrowing from 39,278 to 351 | this machine. gmpy2, 4 seconds |
| the only k ≤ 78557 with a covering set is 78557 | this machine. Factoring 2L−1 in PARI/GP, testing in bulk with numpy |
| the distinct distances D of the grid (m=128,512,2048) | this machine |
| testing Cayley graphs (8,713 in this round; now 9,805) | this machine. AddCircuit in CP-SAT |
| computing |Ш| backwards (5 curves) | this machine. PARI/GP. ellsaturation is essential |
| covering n ≢ 2 (mod 4) for the 7 fourth powers (periods up to 96) | this machine. None of the 7 can be covered |
| classification of the odd m with an algebraic factorisation (cq, c⁴) | known (Capelli's criterion), restated |
| the sieve weights of the remaining 5 | this machine. A known quantity (the Proth weight) |
The number that stays most from this record is the 8 seconds. The square peg problem was drawn and redrawn 8 seconds later. Why it was put back is not in the record. What remains is only the fact that it was 8 seconds. The record of an action can be more accurate than the introspection of the one who acted—recording the order and intervals by measurement is there for that.