Binet's Formula Calculator (Fibonacci sequence)
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/10/2026
Binet's formula gives the nth term of the Fibonacci sequence directly without calculating the preceding terms: F(n) = (φⁿ − ψⁿ) ÷ √5, where φ is the golden ratio. For n=10, this formula gives exactly 55 — the same value as the classic iterative calculation (1,1,2,3,5,8,13,21,34,55).
Explanation
The Fibonacci sequence, defined by a simple recurrence (each term is the sum of the two preceding ones), has a remarkable and at first surprising property: there exists a CLOSED formula, Binet's formula, that gives any term directly without having to calculate all the terms before it. This formula is written F(n)=(φⁿ−ψⁿ)/√5, where φ=(1+√5)/2 is the golden ratio already on this site, and ψ=(1−√5)/2 its conjugate (the second root of the same equation x²=x+1 that defines φ). The most astonishing result of this formula is that it systematically mixes powers of two IRRATIONAL numbers (φ and ψ, both involving √5), and yet their difference, once divided by √5, always gives exactly an INTEGER — the irrational parts canceling perfectly at each step of the calculation. This formula directly complements the Fibonacci sequence calculator already published, which computes the same result by the usual ITERATIVE method (successively adding each pair of terms): both methods give exactly the same result, but Binet's formula has the conceptual advantage of requiring no calculation loop, any term being obtainable directly without knowing those before it. In numerical practice, however, this advantage has a limit: since φⁿ becomes a very large number while ψⁿ (with |ψ|<1) becomes tiny, the floating-point rounding of this formula loses its exact precision earlier than the integer iterative calculation, which is why this calculator is capped at n=70 rather than the more generous cap of the iterative calculator.
Example: the tenth Fibonacci term
Inputs
Index sought: n=10.
Calculation
F(10) = (φ¹⁰ − ψ¹⁰) ÷ √5 ≈ (122.9919 − (−0.0081...)) ÷ 2.236 ≈ 55.
Result
The tenth term of the Fibonacci sequence is exactly 55, the same value as the classic iterative calculation 1,1,2,3,5,8,13,21,34,55.
Frequently asked questions
How can an integer result come out of a formula full of square roots?
Because φ and ψ are the two roots of the same quadratic equation x²−x−1=0, which gives them very particular complementary algebraic properties: when φⁿ−ψⁿ is expanded using the binomial theorem, all terms containing an odd power of √5 cancel exactly between the two expansions, leaving only terms in even powers of √5, which then simplify exactly with the division by √5 itself. The final result is guaranteed to be an integer by this algebraic construction, never by numerical coincidence.
Why is the input cap lower than that of the iterative Fibonacci calculator?
Because Binet's formula handles irrational numbers represented in floating point (with finite precision), while the iterative calculation directly handles exact integers, with no loss of precision as long as the result stays below 2⁵³. The rounding of φⁿ in floating point introduces an error that becomes enough to distort the final result from n=71 with this formula, whereas the iterative calculation stays exact up to n=78 — two different limits for two different calculation methods, each as rigorous as the other within its respective domain of validity.
Does this formula have any use beyond mathematical curiosity?
Yes: it allows instant calculation of a high-index term without having to unroll the whole sequence from the start, a real advantage in algorithm analysis and theoretical computer science, where the performance of a CONSTANT-TIME calculation (Binet's formula, a single operation) is precisely compared against a LINEAR-TIME calculation (iteration, one operation per term) for the same mathematical problem — a classic teaching example of algorithmic complexity.