What is the 30th Fibonacci number?

F(30) = 832,040. It comes from adding the two terms before it: F(29) + F(28) = 514,229 + 317,811 = 832,040. It has 6 digits, and the terms either side are 514,229 and 1,346,269.

Is 832,040 a prime number?

No — 832,040 is composite. It has to be: F(2) = 1 and F(3) = 2 and F(5) = 5 and F(6) = 8 and F(10) = 55 and F(15) = 610 divide it, because 2 and 3 and 5 and 6 and 10 and 15 divide 30.

What is F(30) divided by F(29)?

832,040 ÷ 514,229 = 1.618033989. The golden ratio is φ = 1.618033989, so this is off by 1.69e-12 — below it, since the ratios alternate on either side as they close in.

What are the Lucas number and running total at index 30?

L(30) = 1,860,498, which is F(29) + F(31) = 514,229 + 1,346,269. The running total F(1) + … + F(30) is 2,178,308, which equals F(32) − 1 = 2,178,309 − 1 — an identity that holds at every index, not just this one. The squares up to here sum to F(30) × F(31) = 1,120,149,658,760.

What is Cassini’s identity?

F(n)² − F(n−1)F(n+1) = (−1)ⁿ⁻¹ — the square of any term differs from the product of its neighbours by exactly one, alternating in sign. It is the basis of a well-known dissection puzzle in which a square appears to be rearranged into a rectangle of a different area; the missing unit hides along a thin sliver, and the sliver is exactly this identity made visible.

Where is the sequence actually used?

In more places than the folklore suggests, and mostly unglamorous ones. Fibonacci heaps give the fastest known asymptotic bounds for several graph algorithms. Fibonacci search divides a sorted array using these numbers rather than halves. Fibonacci coding produces a self-synchronising variable-length encoding, and the Zeckendorf representation is what makes it work. Agile teams estimate in Fibonacci numbers because the widening gaps discourage false precision.

Why do the ratios alternate around φ?

Because of the second root. Binet’s formula writes F(n) as a combination of φⁿ and ψⁿ, where ψ = (1 − √5) ÷ 2 is about −0.618. Since ψ is negative and smaller than 1 in size, its powers shrink towards zero while flipping sign at every step — and that flipping is exactly what pushes each successive ratio to the other side of φ. It also explains the speed: the error is governed by ψⁿ, which decays geometrically.

How is the golden ratio involved?

Divide any term by the one before it and the answer approaches φ = 1.618033988749895, alternating above and below as it goes. The error shrinks by roughly a factor of φ² at each step, so convergence is quick: by the fortieth term the ratio matches φ to every digit an ordinary computer number can hold. Any sequence built by adding the last two terms does this, whatever it starts from.

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.

The 30th Fibonacci Number

The 30th Fibonacci number is 832,040. Index 30 is a milestone term whose value 832,040 is divisible by F(3), F(5), F(6), F(10) and F(15), the index 30 having more small divisors than any of its neighbours. It follows from the rule the whole sequence runs on: add the two terms before it, so F(30) = F...

BY POSITION

INDEX n

WORTH TRYING

F(30)

832,040

RATIO TO F(29)

1.618034

6 digits · previous 514,229 · next 1,346,269 · φ is 1.618034, so this is off by 1.7e-12

Digits

Lucas L(n)

1,860,498

Sum of F(1)…F(n)

2,178,308

Ratio to previous

1.618033989

Exact in a double?

yes

Index n

AROUND F(30)

F(n)

DIGITS

196,418

317,811

514,229

832,040

1,346,269

2,178,309

3,524,578

STEPS

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

F(30) = F(29) + F(28) = 514,229 + 317,811

6 digits, small enough to hold exactly in ordinary arithmetic

F(30) = 832,040

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.

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(30) = 832,040. This page opens with index 30 loaded, so nothing needs typing.
2
The STEPS panel shows where it comes from: F(29) + F(28) = 514,229 + 317,811 = 832,040.
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.