THE FOLD / LOOT / THE VAULT / THE FROBENIUS
THE FROBENIUS
the largest amount you can't make — ab-a-b
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
The Frobenius / Chicken McNugget problem. You have coins of two values, a and b. Which amounts can you pay exactly, using any number of each? If a and b share no common factor, you can make every large enough amount — but not the small ones. What is the largest amount you cannot make?
The answer is a clean formula: g(a,b) = a·b − a − b. With 3-cent and 5-cent coins, the biggest impossible amount is 3·5−3−5 = 7. And the total number of unpayable amounts is exactly (a−1)(b−1)/2. Past the Frobenius number, the gaps close forever. (The nickname: McNuggets once came in boxes of 6, 9 and 20 — the largest number you couldn’t buy was 43.) For three or more denominations there is no closed formula — it becomes genuinely hard.
LIT verified live: for every coprime pair a, b the value a·b−a−b is not representable, everything above it is, and the count of unrepresentable amounts equals (a−1)(b−1)/2 (window.__frobenius.frobeniusCorrect && countCorrect). FIG no framing; the formula, the boundary, and the gap count are exact.
The answer is a clean formula: g(a,b) = a·b − a − b. With 3-cent and 5-cent coins, the biggest impossible amount is 3·5−3−5 = 7. And the total number of unpayable amounts is exactly (a−1)(b−1)/2. Past the Frobenius number, the gaps close forever. (The nickname: McNuggets once came in boxes of 6, 9 and 20 — the largest number you couldn’t buy was 43.) For three or more denominations there is no closed formula — it becomes genuinely hard.
LIT verified live: for every coprime pair a, b the value a·b−a−b is not representable, everything above it is, and the count of unrepresentable amounts equals (a−1)(b−1)/2 (window.__frobenius.frobeniusCorrect && countCorrect). FIG no framing; the formula, the boundary, and the gap count are exact.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this in THE VAULT, beside 3LOCK, THE BANKER and THE SECRET — the loot domain of coin and value. The coin problem is exactly a vault question: with these denominations, which sums can you actually pay? AVAN (AI) built the instrument: the representable strip, the Frobenius boundary, the gap count.
The weave: David names the seat (the coins in the vault); I make the reachable amounts light up and the largest gap pop out — the number line in 1D, the payable strip in 2D, the coin lattice in 3D. The sphere is the seam. Credit: posed by Ferdinand Frobenius; two-coin formula by James J. Sylvester (1882); the McNugget nickname.
The weave: David names the seat (the coins in the vault); I make the reachable amounts light up and the largest gap pop out — the number line in 1D, the payable strip in 2D, the coin lattice in 3D. The sphere is the seam. Credit: posed by Ferdinand Frobenius; two-coin formula by James J. Sylvester (1882); the McNugget nickname.
3 ONE DIMENSION
The number line, each amount marked payable or not with coins a and b. The gaps thin out and then stop — the last gap is the Frobenius number, beyond which every amount can be made.
4 TWO DIMENSIONS · INTERACTIVE
Choose coprime coins a and b. Payable amounts light up; the unpayable ones are the gaps, and the largest of them is a·b−a−b. Count them — there are exactly (a−1)(b−1)/2, no matter which coprime pair you pick.
5 THREE DIMENSIONS + AVAN’S INVERSE
The lattice of coin counts (x of a, y of b) turning, each point an amount a·x + b·y — green, the reachable sums covering the line.
AVAN’s addition (the inverse-companion): the magenta mark is the Frobenius number — the last amount the green lattice can never reach. You could answer ‘what can I make?’ by enumerating combinations, a forward search that never quite ends. The Frobenius theorem answers the harder inverse question directly: ‘what is the largest I cannot make?’ — with a formula, no search at all. The inverse of ‘list the reachable’ is ‘pinpoint the boundary of the unreachable’, and past that single number the gaps close for good, coprimality guaranteeing every large amount is eventually payable. And the deeper inverse: the clean two-coin formula shatters at three coins, where finding the boundary becomes genuinely hard — a reminder that a tidy answer can hide a cliff. The green is everything the coins can pay; the magenta is the final thing they can’t.
LIT Genuine Frobenius coin problem (posed by Frobenius; two-coin formula by Sylvester 1882). Verified live: for every coprime pair a,b the value ab-a-b is not representable as ax+by (non-negative x,y), everything above it is, and the count of unrepresentable amounts equals (a-1)(b-1)/2 (window.__frobenius.frobeniusCorrect && countCorrect, both true). g(3,5)=7. The formula, the boundary, and the gap count are exact for two coins; three-plus denominations genuinely have no closed form.
FIG No framing: the ab-a-b formula, the representability boundary, and the (a-1)(b-1)/2 gap count are all real and checked over all coprime pairs. The honest scope: this closed form is specific to TWO denominations; the general Frobenius number (3+ coins) is hard with no such formula, stated plainly.
FIG No framing: the ab-a-b formula, the representability boundary, and the (a-1)(b-1)/2 gap count are all real and checked over all coprime pairs. The honest scope: this closed form is specific to TWO denominations; the general Frobenius number (3+ coins) is hard with no such formula, stated plainly.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE VAULT · David Lee Wise (ROOT0), with AVAN