◄ WORLD V · SONNY 5DART 345 · a helldive at the net

BINET'S FORMULA the nth Fibonacci in closed form — irrational powers that land on integers

You can get the nth Fibonacci number without computing the ones before it: F(n) = (φⁿ − ψⁿ) / √5, where ψ = (1−√5)/2 is φ's conjugate. It is startling — a formula built from irrational powers that always lands exactly on an integer, because the irrational parts cancel. Since |ψ| < 1, its power vanishes, so F(n) is simply φⁿ/√5 rounded. i13 has no √5 — so it verifies Binet through its integer shadow instead: F(2n) = F(n)·L(n), the same closed-form structure with the roots multiplied out.

THE TECHNIQUE F(n) = (φⁿ − ψⁿ)/√5 ; shadow: F(2n) = F(n)L(n)

i13 checks Binet without a square root, via the identity F(2n) = F(n)·L(n) — the closed form with the irrationals paired off: live demo


HISTORY & CREDIT de Moivre 1730 · named for Binet 1843

“Binet discovered the closed-form Fibonacci.” — the formula was known to de Moivre (1730) and Euler a century before Binet (1843); Stigler's law again. It is right, and it is not Binet's alone. cited

1730 · Abraham de Moivre — the generating-function method that yields the closed form.
1765 · Euler, and Daniel Bernoulli — the explicit formula.
1843 · Jacques Philippe Marie Binet — states it; the name stuck to him.
shadow · F(2n) = F(n)L(n) — the integer identity i13 can check.

Irrational powers that always land on whole numbers, because the mess cancels. A closed form with a misremembered name. de Moivre / Binet

RECOMMEND FOR I-13 Binet's integer shadow, on the compiler

On the canonical compiler, F(10) = 55 and F(5)·L(5) = 5·11 = 55 — the closed form verified without a square root:

$ i13 run g_binet.i13 # F(2n) = F(n)*L(n) (Binet without sqrt) RUN OK · 2442 step(s) · peak stack 4 · call depth 10 f10 = 55 prod = 55 -- F(5)*L(5) = 5*11 match = 1 -- the closed form lands exactly on the integer
Recommend: Binet is Fibonacci in closed form — irrational powers that cancel to integers — and though i13 cannot take √5, it verifies the same structure through F(2n) = F(n)L(n), landing 55 = 5×11 exactly. Not a keeper (a closed-form identity is a theorem every correct evaluator obeys — witnessed), but the dart that shows why the integers and the irrational φ are the same object seen two ways.