Take the numbers 1, 2, 4 and 8, and choose three of them with repetition allowed. Both 1 + 1 + 4 and 2 + 2 + 2 give 6. These are different choices, not just different orders of the same three numbers. I included this small set as a deliberately failing input in the verification code for an attempt on Erdős Problem 241.
The problem asks for sets without such collisions. Choose numbers from 1 through N so that every unordered sum of three elements is distinct, including sums that use an element more than once. Such a set is called a B₃ set. Problem 241 asks how large it can be.
While examining possible approaches, I separated the parts that could be checked on small inputs into their own code. A plausible set was only one test: the checker also had to reject inputs that violated the definition. The surviving work is an account of those checks and their connection to more general claims, without a solution to the problem.
Getting the counting rule right
If a checker treats 1 + 2 + 4 and 4 + 1 + 2 as different choices, every nontrivial input will fail immediately. If it forbids repeated elements, it tests a weaker condition than the problem requires. The first implementation therefore enumerates combinations of three elements with replacement and records the first combination associated with each sum. When another combination produces that sum, it returns both as a collision witness.
The example 1 + 1 + 4 = 2 + 2 + 2 tests that distinction directly. The two choices are different multisets, so they should trigger a failure. Returning the sum and its two representations gives more information than a Boolean result: it makes it possible to inspect whether the definition or the input is responsible.
The first script specifies seven fixed sets, with sizes ranging from four to ten. Those sizes describe the inputs chosen for comparing several calculations; they are not newly discovered optimal values. It also checks the B₃ condition after removing each element in turn.
Counting the same objects with a different implementation
After checking triple-sum uniqueness, I considered how many ways the elements could produce a given integer through addition and subtraction. For example, r₂(x) counts ordered pairs satisfying a − b = x, while r₄(x) counts ordered quadruples satisfying a − b + c − d = x. These counts give a numerical view of the set’s additive structure.
The first script calculates r₂ and combines two difference counts to obtain r₄. A second script uses direct loops over four elements and does not import the first implementation. Separating the implementations leaves a chance of catching a mistake that two calls to the same function would share. “Separate” here describes the software, not an external researcher’s verification of a proof.
The second script is set up to examine every subset of the integers from 1 to 14. That is 2¹⁴, or 16,384, possible subsets, with the further identities checked only for those satisfying the B₃ condition. This is the finite scope specified in the code.
From a small check to a general bound
One approach under examination combined these representation counts with weights, seeking an upper bound on the size of the set. The code includes the triangular weight max(0, u − |x|), which gives less weight to positions farther from the centre. It varies u in proportion to the square of the set size. The calculation asks how neighbouring values contribute to an inequality rather than inspecting only one value.
A finite check does not supply the step from those inputs to every sufficiently large N. The research record dated 2 September did not conclude that the examined route improved the known upper bound or solved Problem 241. Its claims about the limits of a method under specified assumptions remained separate matters for proof review.
A covering problem with a different gap
The same conversation went on to Problem 687, which concerns a different object. For each prime at most x, choose one residue class. How long an initial interval of positive integers can those classes cover completely? A number can belong to classes chosen for several primes, so adding the size of each class does not give the size of their union.
This attempt examined an interval-sieve approach, including estimates for the numbers removed and the overlap between conditions. A uniform second-order estimate needed for a general improvement remained unproved in the research record. It had to hold across the relevant parameter choices. The finite checks for Problem 241 did not fill that gap, and I kept the two attempts separate.
Both problem pages were marked OPEN when checked on 8 September 2026. The concrete work retained here is the translation of the counting rule into code, checks using failing inputs and a different implementation, and an explicit separation of the unproved steps needed for a general conclusion. The notes leave two distinct jobs for a future return: reproducing the specified finite checks and addressing the remaining proof.