What is the 1st Fibonacci number?

F(1) = 1. It comes from adding the two terms before it: F(1) is one of the two seed values, not a sum. It has 1 digit, and the terms either side are 0 and 1.

What is F(1) divided by F(0)?

F(1) has no meaningful ratio to the term before it, because F(0) is zero.

What are the Lucas number and running total at index 1?

L(1) = 1, which is F(0) + F(2) = 0 + 1. The running total F(1) + … + F(1) is 1, which equals F(3) − 1 = 2 − 1 — an identity that holds at every index, not just this one. The squares up to here sum to F(1) × F(2) = 1.

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.

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.

How do you get a term without listing them all?

Binet's formula gives F(n) = (φⁿ − ψⁿ) ÷ √5 directly, but it relies on irrational numbers and rounds badly for large n. The method used here is fast doubling, which builds F(2k) and F(2k+1) from F(k) and F(k+1) and so reaches the thousandth term in about ten steps instead of a thousand — and it never leaves the whole numbers, so nothing is approximated.

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.

The 1st Fibonacci Number

The 1st Fibonacci number is 1. Index 1 is the first term the sequence actually produces, and one of two consecutive 1s. Together with F(2) it is the seed everything else grows from. It is one of the two values the sequence is seeded with rather than a sum of anything, which is why the rule only begi...

BY POSITION

INDEX n

WORTH TRYING

F(1)

RATIO TO F(0)

—

1 digit · previous 0 · next 1

Digits

Lucas L(n)

Sum of F(1)…F(n)

Ratio to previous

—

Exact in a double?

yes

Index n

AROUND F(1)

F(n)

DIGITS

STEPS

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

F(1) is one of the two seed values, not a sum

1 digit, small enough to hold exactly in ordinary arithmetic

F(1) = 1

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 1 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.

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HOW TO USE

1
The answer is already here: F(1) = 1. This page opens with index 1 loaded, so nothing needs typing.
2
F(1) is a seed rather than a sum, so the STEPS panel says so instead of showing an addition.
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.