Fibonacci Sequence Table
| n | F(n) | F(n+1)/F(n) |
|---|---|---|
| 1 | 1 | 1.000000 |
| 2 | 1 | 2.000000 |
| 3 | 2 | 1.500000 |
| 4 | 3 | 1.666667 |
| 5 | 5 | 1.600000 |
| 6 | 8 | 1.625000 |
| 7 | 13 | 1.615385 |
| 8 | 21 | 1.619048 |
| 9 | 34 | 1.617647 |
| 10 | 55 | 1.618182 |
| 11 | 89 | 1.617978 |
| 12 | 144 | 1.618056 |
| 13 | 233 | 1.618026 |
| 14 | 377 | 1.618037 |
| 15 | 610 | 1.618033 |
| 16 | 987 | 1.618034 |
| 17 | 1597 | 1.618034 |
| 18 | 2584 | 1.618034 |
| 19 | 4181 | 1.618034 |
| 20 | 6765 | 1.618034 |
The Fibonacci sequence starts with 1, 1 and each term is the sum of the two before it: 1, 1, 2, 3, 5, 8, 13, 21... The ratio of consecutive terms converges to the golden ratio φ ≈ 1.6180339887 - an irrational number that appears in sunflower seed spirals, pinecone scales, and the nesting pattern of Romanesco broccoli.
This table shows the first 20 terms with their ratios, computed exactly - and the calculator extends to F(50) = 12,586,269,025 using arbitrary-precision arithmetic that handles numbers beyond the 32-bit integer limit.
How to use
- Enter how many terms to see (1-50); the table shows the sequence with consecutive ratios.
- Watch the ratio column: by term 10 the ratio is already within 0.001 of the golden ratio φ.
- The prime factorisation and golden ratio columns explain why the sequence appears in nature.
Frequently asked questions
Why does the ratio converge to the golden ratio?
The limit of F(n+1)/F(n) satisfies the same recurrence as the sequence itself: r = 1 + 1/r, giving r² = r + 1 and r = (1+√5)/2. The convergence is fast - by F(20)/F(19) the ratio matches φ to six decimal places.
Where does the Fibonacci sequence appear in nature?
Sunflower seeds are arranged in Fibonacci-number spirals (typically 34 one way, 55 the other) because this packing maximises seed density. Pinecones, pineapples and Romanesco broccoli show the same pattern. The mechanism is efficient packing, not mysticism - the golden ratio is simply the irrational number whose continued fraction has all 1s, making it the hardest to approximate by rationals.
Can I compute F(n) without calculating all previous terms?
Yes - Binet’s formula gives F(n) = (φⁿ − ψⁿ)/√5 directly, where φ = (1+√5)/2 and ψ = (1−√5)/2. It works up to about F(70) before floating-point precision causes errors; beyond that you need the recurrence or matrix exponentiation.