A364132(28) = 75 and A364153(17) = 26, each found by an exhaustive computer search and re-run by a second, separately written program. A computer search, not a proof.
We found two new numbers for an old puzzle by Paul Erdős. This post shows what the puzzle asks, what we found, and how we checked it.
The puzzle
Pick some whole numbers. Line them up. Now add up every run of numbers that sit next to each other. A run can be one number, two numbers, or the whole line.
The rule: every run must add up to a different total.
Here is a small example from the OEIS page for this puzzle: the six numbers 1, 2, 4, 5, 8, 10. Their 21 run totals are all different.
The top row is the six numbers. Each row below adds one more neighbour: row 2 holds the totals of two numbers side by side, and the bottom cell is the whole line. All 21 totals are different.Drawn by the blog build from the terms above. The build recomputes all 21 sums and refuses to publish if two are equal.
Erdős and Harzheim asked how many numbers you can fit this way, using only numbers from 1 up to some limit (Erdős Problem #357).
The OEIS turns this into two lists:
A364132: the numbers must go up, smallest to largest. Each entry is the smallest top number that lets you fit n of them. (oeis.org/A364132)
A364153: the same question, but the numbers can be in any order. (oeis.org/A364153)
What we found
A364132(28) = 75. You can fit 28 rising numbers using only 1 to 75. You cannot do it with 1 to 74.
A364153(17) = 26. You can fit 17 numbers, in any order, using only 1 to 26. You cannot do it with 1 to 25.
The OEIS lists the first 22 terms of A364132 and the first 13 of A364153. The atlas had already filled in the terms between those and today's.
Here are the 28 numbers that fit inside 1 to 75:
They make 406 run totals, and no two are the same. Here are all 406, in the same triangle as the small example (hover a cell to read its total):
The 28 numbers from the A364132(28) receipt, with every run total below them.Drawn by the blog build from the terms above. The build recomputes all 406 sums and refuses to publish if two are equal.
This is the row for #357 on the Erdős atlas gap ledger, as the atlas shows it today:
A364132(30) — least n such that {1..n} contains an increasing 30-term sequence all of whose consecutive-segment sums are distinct (inverse of the Erdős–Harzheim f(n)); a(29) = 79 (certificates/erdos-357)[ 80 , ? ]
Finding one sequence that works is the easy half. It shows the answer is at most 75.
The hard half is showing nothing smaller works. For that, the computer tried every possible way at each smaller limit, and found none. That is called an exhaustive search.
Then a second program, written separately, ran the same search. It got the same answers. For A364153(17), a full re-run of the whole certificate, every search by both programs, also passed (result on PR #186).
To be clear about the limits:
It's a computer search, not a proof.
Both programs use the same basic method. They are separate code, not separate math.
It says nothing about the big question Erdős asked: how fast these lists grow forever.
The computing was done by ProjectForty2's AI research agent. The search programs were written with AI research agents. Kevin Russell directed the work.
Everything is public. You can read it, rerun it, and tell us if we got something wrong.
