How this instrument works
The Fibonacci sequence is built from the plainest rule in mathematics that isn't outright repetition: each term is the sum of the two before it, with F₁ = 1 and F₂ = 1 fixing the start. Leonardo of Pisa, later called Fibonacci, popularized it in 1202 in Liber Abaci through a puzzle about breeding rabbits, though the same counting pattern shows up centuries earlier in Sanskrit poetry, where scholars were tallying the ways long and short syllables combine.
Binet's formula reaches any term directly, without walking through the ones before it. It comes from treating the recurrence Fₙ = Fₙ₋₁ + Fₙ₋₂ as an algebraic equation: substituting a trial solution xⁿ produces x² = x + 1, whose two roots are φ = (1+√5)/2 and ψ = (1−√5)/2. Every Fibonacci number is a mix of these two roots raised to the nth power, and the √5 in the denominator is exactly what turns that mix back into a whole number every time.
Because ψ is a fraction between −1 and 0, ψⁿ shrinks toward zero as n grows and barely nudges the result — by n = 10 it contributes less than 0.01 to the total before rounding. That is also why the ratio of consecutive terms, Fₙ₊₁ ⁄ Fₙ, homes in on φ ≈ 1.618034 rather than settling anywhere else: it is the same golden ratio that turns up in pinecone spirals and sunflower seed heads, arising from this identical recurrence rather than being fitted to it after the fact.
- Enter the position you want in the n (term index, F₁=1, F₂=1, ...) field — 1 for the first term, 2 for the second, and so on.
- Leave n at 0 to see F₀ = 0, the sequence's often-overlooked starting point before F₁.
- Read the result in the Fₙ field — the exact Fibonacci number at that position, already rounded to a whole number.
- Raise n gradually and watch Fₙ grow: past the first dozen terms it roughly multiplies by φ ≈ 1.618 with every step up.
Worked example — the 10th Fibonacci number
Set n = 10. Using φ ≈ 1.618034 and ψ ≈ −0.618034 carried to six decimals, φ¹⁰ ≈ 122.991869 and ψ¹⁰ ≈ 0.008131. Subtracting gives φ¹⁰ − ψ¹⁰ ≈ 122.983739, and dividing that by √5 ≈ 2.236068 lands on 55.000000 — which is exactly F₁₀. The tenth Fibonacci number is 55.
Building the same answer the slow way means listing all ten terms in order — 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 — and stopping once the count reaches the tenth entry. Binet's formula skips that walk entirely: it reaches 55 straight from n = 10 without ever touching F₉ or any earlier term, which starts to matter once n climbs into the hundreds and a term-by-term sum would take that many additions.
Questions
What is the Fibonacci sequence's actual definition?
It is the sequence built by F₁ = 1, F₂ = 1, and Fₙ = Fₙ₋₁ + Fₙ₋₂ for every term after — each number is simply the sum of the two before it: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, and onward. Some sources also start the list at F₀ = 0, which this sheet supports by accepting n = 0.
How does Binet's formula compute a whole number from irrational parts?
φ and ψ are both irrational, but the irrational parts cancel completely inside φⁿ − ψⁿ, leaving a clean multiple of √5 every time — dividing by √5 then produces a whole number exactly, with rounding needed only to clear tiny floating-point residue from ψⁿ. This is a real algebraic cancellation, not an approximation dressed up as one.
Why does the golden ratio show up in a formula about integers?
Because the recurrence Fₙ = Fₙ₋₁ + Fₙ₋₂ is linear, its solutions have the form xⁿ, and substituting that in produces x² − x − 1 = 0 — the golden ratio's own defining equation. Its two roots, φ and ψ, are the only pair that can generate every term of the sequence, so they are forced into the formula by the algebra itself, not chosen for convenience.
What mistake do people usually make with this formula?
Dropping the ψⁿ term because it looks negligible for large n — but skipping it early on gives visibly wrong answers: at n = 1 it would predict Fₙ ≈ φ ⁄ √5 ≈ 0.7236 instead of the true value of 1. The ψⁿ term matters for every small n and only becomes safely ignorable once n is large enough that rounding absorbs it.
How does the ratio of consecutive Fibonacci numbers relate to the golden ratio?
Fₙ₊₁ ⁄ Fₙ converges toward φ ≈ 1.618034 as n grows, though it never lands on it exactly for any finite n — 55 ⁄ 34 ≈ 1.617647 is already close, and 6,765 ⁄ 4,181 ≈ 1.618034 agrees with φ to five decimals. The convergence follows directly from Binet's formula, since ψⁿ ⁄ φⁿ vanishes as n increases, leaving the ratio governed by φ alone.
Does the formula still work at n = 0?
Yes — F₀ = (φ⁰ − ψ⁰) ⁄ √5 = (1 − 1) ⁄ √5 = 0, matching the sequence's conventional starting value that sits before F₁ = 1. This sheet accepts n = 0 as a valid term index for exactly that reason, rather than restricting entries to n = 1 and above.