Fibonacci Calculator
Enter which term you want and get it instantly via Binet’s formula, plus the full sequence up to that point and its running sum.
Result
| n | F(n) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 1 |
| 3 | 2 |
| 4 | 3 |
| 5 | 5 |
| 6 | 8 |
| 7 | 13 |
| 8 | 21 |
| 9 | 34 |
| 10 | 55 |
| 11 | 89 |
| 12 | 144 |
| 13 | 233 |
| 14 | 377 |
| 15 | 610 |
Two ways to reach the same number
The Fibonacci sequence is usually introduced through its recurrence — each term is the sum of the two before it, starting from 0 and 1. That’s exact and simple, but it means finding F(50) requires computing all 49 terms before it. Binet’s formula sidesteps that entirely: it expresses F(n) directly in terms of the golden ratio φ, so a single calculation reaches any term without touching the ones before it.
That a sequence built from whole-number addition has an exact formula involving an irrational number is one of the more striking small facts in elementary mathematics — the irrational parts of φⁿ and ψⁿ always cancel out to leave a clean integer.
Frequently Asked Questions
What is Binet's formula?
Binet's formula gives the nth Fibonacci number directly, without recursion: F(n) = (φⁿ − ψⁿ) / √5, where φ = (1+√5)/2 ≈ 1.6180339887 is the golden ratio and ψ = (1−√5)/2 ≈ −0.6180339887 is its conjugate. Because |ψ| < 1, the ψⁿ term shrinks toward zero as n grows, so F(n) is always essentially φⁿ/√5 rounded to the nearest integer.
Why does this calculator cap n at 70?
Binet's formula computes φⁿ and ψⁿ using ordinary double-precision floating point. Those powers keep enough significant digits for an exact rounded integer up to roughly F(70) — beyond that, tiny floating-point errors can shift the last digit or two, so the calculator stops at a term it can still guarantee is exact.
What's the running sum used for?
The sum of F(0) through F(n) has a tidy closed form of its own: it always equals F(n+2) − 1. Watching the running total climb alongside the sequence is a quick, concrete way to see that identity in action rather than just taking it on faith.
How is this different from just adding the last two numbers?
The simple recurrence F(n) = F(n−1) + F(n−2) is how the calculator actually builds the listed sequence (it's exact and easy to reason about), but it requires computing every earlier term first. Binet's formula is the interesting case: it reaches F(n) in one shot from n alone, which is why it's the headline result here.
Educational tool. Results are exact integers for every supported term (n ≤ 70); this is not a substitute for arbitrary-precision arithmetic software if you need terms beyond that range.