Fibonacci sequence calculator
The 10th Fibonacci number is 55 in the Fibonacci sequence calculator: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, each term the sum of the two before it.
F(10)
- Recursive formula Each term is the sum of the two before it, starting from F(1) = 1 and F(2) = 1. F(0) = 0 is the same rule run one step back.
- F(10) = F(9) + F(8) = 34 + 21 = 55
- Binet's formula The full formula is F(n) = (φ^n - ψ^n) / √5, with φ = (1 + √5) / 2 ≈ 1.6180339887 and ψ = (1 - √5) / 2 ≈ -0.6180339887. ψ^n shrinks toward zero (ψ^10 ≈ 0.008131), so F(n) is φ^n / √5 rounded to the nearest whole number; the exact 55 comes back with the ψ term included.
- φ^10 / √5 = 55.0036, which rounds to 55
- Sum of F(1) to F(10) The sum of the first n Fibonacci numbers is F(n + 2) - 1: here F(12) - 1.
- 143
- Ratio F(10) / F(9) The ratio of neighbors approaches the golden ratio φ = 1.6180339887; here it is off by 0.000387.
- 1.6176470588
- Digits
- 2
- Even or odd Every third Fibonacci number is even (F(3) = 2, F(6) = 8, F(9) = 34), because the pattern odd, odd, even repeats forever.
- odd
- First 10 terms
- 1, 1, 2, 3, 5, 8, 13, 21, 34, 55
| n | F(n) | F(n) / F(n - 1) |
|---|---|---|
| 1 | 1 | - |
| 2 | 1 | 1 |
| 3 | 2 | 2 |
| 4 | 3 | 1.5 |
| 5 | 5 | 1.6666667 |
| 6 | 8 | 1.6 |
| 7 | 13 | 1.625 |
| 8 | 21 | 1.6153846 |
| 9 | 34 | 1.6190476 |
| 10 | 55 | 1.6176471 |
- Every value is exact: the terms are computed in whole-number arithmetic of unlimited length, so F(1,000), with 209 digits, is right to the last one.
How the Arithmetic Sequence Calculator works
The formula for the Fibonacci sequence is a recurrence: each term is the sum of the two before it.
F(n) = F(n - 1) + F(n - 2), with F(1) = 1 and F(2) = 1
So F(3) = 1 + 1 = 2, F(4) = 1 + 2 = 3, and the sequence runs 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. Many textbooks start one step earlier with F(0) = 0, which changes nothing else: F(10) is 55 either way.
Binet's formula gives any term without listing the ones before it:
F(n) = (φ^n - ψ^n) / √5, where φ = (1 + √5) / 2 ≈ 1.6180339887 and ψ = (1 - √5) / 2 ≈ -0.6180339887
ψ is smaller than 1 in size, so ψ^n fades toward zero and F(n) is φ^n / √5 rounded to the nearest whole number. For n = 10, φ^10 ≈ 122.9919, and 122.9919 / 2.2360680 = 55.0036, which rounds to 55. In ordinary floating point that rounding holds only up to about F(70). F(100) = 354,224,848,179,261,915,075 has 21 digits, more than a double-precision number holds exactly, so exact terms need whole-number arithmetic.
The golden ratio. Divide a term by the one before and the ratio closes in on φ: 55 / 34 = 1.6176 and 89 / 55 = 1.6182, and by F(40) / F(39) the two agree to 15 decimal places. It is the same φ as in Binet's formula, and it is why consecutive Fibonacci numbers turn up as the spiral counts in sunflower heads and pine cones.
Sums and patterns. The first n terms add up to F(n + 2) - 1: 1 + 1 + 2 + ... + 55 = 143 = F(12) - 1. Every third term is even, because the pattern odd, odd, even repeats forever. And F(n) has about 0.209n digits, since log10 φ ≈ 0.20899: F(1,000) has 209.
Arithmetic Sequence Calculator FAQ
What is the 20th Fibonacci number?
6,765. It is F(19) + F(18) = 4,181 + 2,584.
What is the 50th Fibonacci number?
12,586,269,025, which has 11 digits. The sequence passes a billion at F(45) = 1,134,903,170.
Is 0 a Fibonacci number?
Yes, as F(0), in the convention that starts 0, 1, 1, 2, 3. Starting from 1, 1 instead gives the same term at every position from 1 on, so F(10) = 55 in both.