What is the 49th Fibonacci number?

F(49) = 7,778,742,049. It comes from adding the two terms before it: F(48) + F(47) = 4,807,526,976 + 2,971,215,073 = 7,778,742,049. It has 10 digits, and the terms either side are 4,807,526,976 and 12,586,269,025.

Is 7,778,742,049 a prime number?

No — 7,778,742,049 is composite. It has to be: F(7) = 13 divide it, because 7 divide 49.

What is F(49) divided by F(48)?

7,778,742,049 ÷ 4,807,526,976 = 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 49?

L(49) = 17,393,796,001, which is F(48) + F(50) = 4,807,526,976 + 12,586,269,025. The running total F(1) + … + F(49) is 20,365,011,073, which equals F(51) − 1 = 20,365,011,074 − 1 — an identity that holds at every index, not just this one. The squares up to here sum to F(49) × F(50) = 97,905,340,104,793,732,225.

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.

What is the Fibonacci sequence?

A sequence in which each term is the sum of the two before it, beginning 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. It reached European mathematics through Leonardo of Pisa in 1202, as a puzzle about breeding rabbits, though it appears in Indian prosody several centuries earlier. The rabbit setup is artificial; the sequence turned out to matter for entirely different reasons.

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.

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.

The 49th Fibonacci Number

The 49th Fibonacci number is 7,778,742,049. Index 49 is a term whose index is a square, and whose value is divisible by F(7) = 13. It follows from the rule the whole sequence runs on: add the two terms before it, so F(49) = F(48) + F(47) = 4,807,526,976 + 2,971,215,073 = 7,778,742,049. Divided by th...

BY POSITION

INDEX n

WORTH TRYING

F(49)

7,778,742,049

RATIO TO F(48)

1.618034

10 digits · previous 4,807,526,976 · next 12,586,269,025 · matches φ to every digit a double can hold

Digits

Lucas L(n)

17,393,796,001

Sum of F(1)…F(n)

20,365,011,073

Ratio to previous

1.618033989

Exact in a double?

yes

Index n

AROUND F(49)

F(n)

DIGITS

1,836,311,903

2,971,215,073

4,807,526,976

7,778,742,049

12,586,269,025

20,365,011,074

32,951,280,099

STEPS

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

F(49) = F(48) + F(47) = 4,807,526,976 + 2,971,215,073

10 digits, small enough to hold exactly in ordinary arithmetic

F(49) = 7,778,742,049

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.

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HOW TO USE

1
The answer is already here: F(49) = 7,778,742,049. This page opens with index 49 loaded, so nothing needs typing.
2
The STEPS panel shows where it comes from: F(48) + F(47) = 4,807,526,976 + 2,971,215,073 = 7,778,742,049.
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.