What is the 4th Fibonacci number?

F(4) = 3. It comes from adding the two terms before it: F(3) + F(2) = 2 + 1 = 3. It has 1 digit, and the terms either side are 2 and 5.

Yes — 3 is a Fibonacci prime. It is among the smallest Fibonacci primes. Whether the sequence contains infinitely many primes is still unproved.

What is F(4) divided by F(3)?

3 ÷ 2 = 1.500000000. The golden ratio is φ = 1.618033989, so this is off by 1.18e-1 — below it, since the ratios alternate on either side as they close in.

What are the Lucas number and running total at index 4?

L(4) = 7, which is F(3) + F(5) = 2 + 5. The running total F(1) + … + F(4) is 7, which equals F(6) − 1 = 8 − 1 — an identity that holds at every index, not just this one. The squares up to here sum to F(4) × F(5) = 15.

Why do large Fibonacci numbers need special arithmetic?

Because ordinary computer numbers are exact only up to 9,007,199,254,740,991, and the sequence passes that at the 79th term. F(78) is 8,944,394,323,791,464 and fits; F(79) is 14,472,334,024,676,221 and does not. A page using ordinary arithmetic does not warn you — it returns something plausible that is wrong in its final digits, and stays wrong from then on.

Which Fibonacci numbers divide which?

F(d) divides F(n) exactly when d divides n. So every third term is even, every fourth is divisible by 3, every fifth by 5, and so on. The stronger statement is that gcd(F(m), F(n)) = F(gcd(m,n)), which is one of the tidiest facts in the whole subject and explains at a stroke why the divisibility pattern is so regular.

Are there infinitely many Fibonacci primes?

Nobody knows, and it is one of the more embarrassing open problems in the subject given how elementary it sounds. Fibonacci primes are known — 2, 3, 5, 13, 89, 233, 1597 and so on — and very large ones have been found, but no proof exists that the list never ends. One thing is settled: apart from F(4) = 3, a Fibonacci prime can only occur at a prime index, though the reverse fails, as F(19) = 4,181 = 37 × 113 shows.

What are the only square and cube Fibonacci numbers?

Above 1 there is exactly one of each: 144 is the only perfect square and 8 the only perfect cube. Both facts look like they should be easy and are not — the square case was settled in 1964 and the general question of Fibonacci perfect powers was only closed in 2006. It is a fair illustration of how a sequence with a one-line rule can hide genuinely hard problems.

What is the Zeckendorf representation?

Every positive whole number is a sum of Fibonacci numbers with no two of them adjacent in the sequence, and there is exactly one such sum. For 100 it is 89 + 8 + 3. The greedy method — repeatedly take the largest that fits — always lands on this unique representation, which is a stronger guarantee than greedy algorithms usually give.

The 4th Fibonacci Number

The 4th Fibonacci number is 3. Index 4 is the first term that is not 1 or 2, and one of only four Fibonacci numbers that are also triangular. It follows from the rule the whole sequence runs on: add the two terms before it, so F(4) = F(3) + F(2) = 2 + 1 = 3. Divided by the term before it, 3 ÷ 2 come...

BY POSITION

INDEX n

WORTH TRYING

F(4)

RATIO TO F(3)

1.500000

1 digit · previous 2 · next 5 · φ is 1.618034, so this is off by 1.2e-1

Digits

Lucas L(n)

Sum of F(1)…F(n)

Ratio to previous

1.500000000

Exact in a double?

yes

Index n

AROUND F(4)

F(n)

DIGITS

STEPS

Every term is the sum of the two before it, starting from F(0) = 0 and F(1) = 1

F(4) = F(3) + F(2) = 2 + 1

1 digit, small enough to hold exactly in ordinary arithmetic

F(4) = 3

GOLDEN SPIRAL

The squares build the sequence. Each new square takes the whole long side of what is already there, so its side is the sum of the two before it — which is the Fibonacci rule stated in geometry rather than arithmetic. The rectangle is always F(k) by F(k+1), and the quarter arcs join end to end into the spiral. It approaches the golden spiral without ever being one: this is built from circular arcs, and the true golden spiral is a smooth logarithmic curve. Square 4 is picked out in gold.

THE RATIO CLOSING ON φ

n = 3

2.00000000

n = 5

1.66666667

n = 8

1.61538462

n = 12

1.61797753

n = 20

1.61803396

n = 40

1.61803399

Bar length is the distance from φ = 1.618033989, so shorter is closer. The ratio alternates above and below it and the error falls by roughly a factor of φ² each step, which is why n = 40 already matches φ to the last digit a double can hold.

Exact, not floating point. F(78) is the last term that fits exactly in ordinary double arithmetic; F(79) is 14,472,334,024,676,221, which is past the limit. Everything here is computed with exact big integers, so F(500) is right to the final digit rather than rounded. Indices run to 1,000.

Live simulation · drawn to scale · updates as you type

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HOW TO USE

1
The answer is already here: F(4) = 3. This page opens with index 4 loaded, so nothing needs typing.
2
The STEPS panel shows where it comes from: F(3) + F(2) = 2 + 1 = 3.
3
The six tiles carry the derived quantities — digit count, the Lucas number, the running total, the ratio to the previous term, and whether the value still fits exactly in ordinary computer arithmetic.
4
Change the index to carry on, or switch to IS IT ONE? to test whether some other number appears in the sequence at all.

THE FORMULAS

The ruleF(n) = F(n−1) + F(n−2)

The seedsF(0) = 0, F(1) = 1

Golden ratioφ = (1 + √5) ÷ 2

Binet's formulaF(n) = (φⁿ − ψⁿ) ÷ √5

Ratio limitF(n) ÷ F(n−1) → φ

Sum of the first nF(1) + … + F(n) = F(n+2) − 1

Sum of squaresF(1)² + … + F(n)² = F(n)·F(n+1)

Cassini identityF(n)² − F(n−1)F(n+1) = (−1)ⁿ⁻¹

Greatest common divisorgcd(F(m), F(n)) = F(gcd(m,n))

Lucas numbersL(n) = F(n−1) + F(n+1)

Fibonacci test5N² ± 4 is a perfect square

Zeckendorfa unique sum of non-consecutive terms

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Last updated: July 29, 2026 · Formula verified · Exact big-integer arithmetic, so terms past F(78) are right to the final digit · Eagle-eyed accuracy for every calculation.