What is the 41st Fibonacci number?

F(41) = 165,580,141. It comes from adding the two terms before it: F(40) + F(39) = 102,334,155 + 63,245,986 = 165,580,141. It has 9 digits, and the terms either side are 102,334,155 and 267,914,296.

Is 165,580,141 a prime number?

No — 165,580,141 is composite. That is worth noting, because the index 41 is prime, and a prime index is necessary for a Fibonacci prime above F(4) but nowhere near sufficient. F(19) = 4,181 = 37 × 113 is the first case where the pattern breaks.

What is F(41) divided by F(40)?

165,580,141 ÷ 102,334,155 = 1.618033989. The golden ratio is φ = 1.618033989, so this is identical to φ as far as ordinary computer arithmetic can tell — below it, since the ratios alternate on either side as they close in.

What are the Lucas number and running total at index 41?

L(41) = 370,248,451, which is F(40) + F(42) = 102,334,155 + 267,914,296. The running total F(1) + … + F(41) is 433,494,436, which equals F(43) − 1 = 433,494,437 − 1 — an identity that holds at every index, not just this one. The squares up to here sum to F(41) × F(42) = 44,361,286,907,595,736.

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.

How can I check whether a number is in the sequence?

There is an exact test: N is a Fibonacci number if and only if 5N² + 4 or 5N² − 4 is a perfect square. Nothing outside the sequence passes and nothing inside it fails. It beats comparing against φⁿ ÷ √5 because it stays in whole numbers, so it does not degrade for large values the way a floating-point comparison would.

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.

What happens at negative indices?

The rule runs backwards perfectly well. Rearranging F(n) = F(n−1) + F(n−2) gives F(n−2) = F(n) − F(n−1), which extends the sequence to negative indices: F(−1) = 1, F(−2) = −1, F(−3) = 2, F(−4) = −3. The pattern is F(−n) = (−1)ⁿ⁺¹ F(n), so the magnitudes mirror the positive side and the signs alternate. This page covers the non-negative side.

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.

The 41st Fibonacci Number

The 41st Fibonacci number is 165,580,141. Index 41 is a prime index with a composite value, and the first term above a hundred and fifty million. It follows from the rule the whole sequence runs on: add the two terms before it, so F(41) = F(40) + F(39) = 102,334,155 + 63,245,986 = 165,580,141. Divid...

BY POSITION

INDEX n

WORTH TRYING

F(41)

165,580,141

RATIO TO F(40)

1.618034

9 digits · previous 102,334,155 · next 267,914,296 · matches φ to every digit a double can hold

Digits

Lucas L(n)

370,248,451

Sum of F(1)…F(n)

433,494,436

Ratio to previous

1.618033989

Exact in a double?

yes

Index n

AROUND F(41)

F(n)

DIGITS

39,088,169

63,245,986

102,334,155

165,580,141

267,914,296

433,494,437

701,408,733

STEPS

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

F(41) = F(40) + F(39) = 102,334,155 + 63,245,986

9 digits, small enough to hold exactly in ordinary arithmetic

F(41) = 165,580,141

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(41) = 165,580,141. This page opens with index 41 loaded, so nothing needs typing.
2
The STEPS panel shows where it comes from: F(40) + F(39) = 102,334,155 + 63,245,986 = 165,580,141.
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.