Sylvester’s sequence begins 2, 3, 7, 43, 1807, 3263443, 10650056950807, …. Its first terms can look like the start of a prime-number sequence, but 1807 is already composite. The more durable property is different: every two distinct Sylvester terms are coprime. No prime factor that appears in one term can appear in another.
Why the Terms Are Pairwise Coprime
For any earlier term si and later term sj, the definition of the sequence makes sj exactly 1 more than a product containing si as a factor. Therefore
sj ≡ 1 (mod si)
and so gcd(si, sj) = 1. This holds for every pair with i ≠ j, not only for neighboring terms.
Two Equivalent Rules Generate the Sequence
Using the index s0 = 2, Sylvester’s sequence can be defined by multiplying every earlier term and adding 1.
Product Form
s0 = 2
sn = 1 + s0s1…sn−1 for n ≥ 1
The first few steps make the rule visible:
- s1 = 2 + 1 = 3
- s2 = (2 × 3) + 1 = 7
- s3 = (2 × 3 × 7) + 1 = 43
- s4 = (2 × 3 × 7 × 43) + 1 = 1807
The same sequence also satisfies a quadratic recurrence. Since
sn − 1 = s0s1…sn−1,
the next term can be written as
sn+1 = 1 + sn(sn − 1) = sn² − sn + 1.
This second form is much faster for generating a later term because it needs only the term immediately before it.
The Remainder-One Proof of Pairwise Coprimality
A collection of integers is pairwise coprime when every two different members have greatest common divisor 1. Sylvester’s sequence satisfies the stronger pairwise condition, not merely the weaker statement that consecutive terms are coprime.
Take two terms si and sj with i < j. The later term is
sj = 1 + s0s1…sj−1.
The product on the right contains si, so that entire product is divisible by si. Adding 1 leaves remainder 1:
sj ≡ 1 (mod si).
If a number d divided both si and sj, it would also have to divide the remainder 1. The only positive divisor of 1 is 1, hence
gcd(si, sj) = 1.
The argument does not depend on how far apart the indices are. For example, it applies just as well to s0 and s6 as it does to s5 and s6.
The Pattern in Actual Terms
For 43, the product form gives
43 = 2 × 3 × 7 + 1.
Therefore 43 leaves remainder 1 when divided by 2, 3, or 7.
For 1807,
1807 = 2 × 3 × 7 × 43 + 1.
So 1807 leaves remainder 1 modulo 2, 3, 7, and 43. The same mechanism continues indefinitely.
| gcd | 2 | 3 | 7 | 43 | 1807 |
|---|---|---|---|---|---|
| 2 | 2 | 1 | 1 | 1 | 1 |
| 3 | 1 | 3 | 1 | 1 | 1 |
| 7 | 1 | 1 | 7 | 1 | 1 |
| 43 | 1 | 1 | 1 | 43 | 1 |
| 1807 | 1 | 1 | 1 | 1 | 1807 |
Pairwise Coprime Does Not Mean Every Term Is Prime
Primality describes one integer: a prime has exactly two positive divisors, 1 and itself. Coprimality describes a relationship between integers: two numbers are coprime when they share no positive divisor greater than 1.
That distinction matters immediately in Sylvester’s sequence. The term 1807 is composite because
1807 = 13 × 139,
yet it is coprime to 2, 3, 7, and 43. A composite term can therefore fit the pairwise-coprime rule perfectly.
| Index | Term | Prime Status | Prime Factorization |
|---|---|---|---|
| 0 | 2 | Prime | 2 |
| 1 | 3 | Prime | 3 |
| 2 | 7 | Prime | 7 |
| 3 | 43 | Prime | 43 |
| 4 | 1807 | Composite | 13 × 139 |
| 5 | 3263443 | Prime | 3263443 |
| 6 | 10650056950807 | Composite | 547 × 607 × 1033 × 31051 |
Prime and Coprime Answer Different Questions
A Sylvester term does not have to be prime. The theorem says that two different Sylvester terms cannot share a prime factor.
For an individual term within the supported input range, the prime number checker can test whether that one value is prime or composite. That test is separate from proving that two sequence terms are coprime.
Prime Factors Cannot Reappear in a Later Term
Pairwise coprimality has an immediate factorization consequence. If a prime p divides one Sylvester term, p cannot divide any other Sylvester term.
The factorization of 1807 introduces 13 and 139. Neither prime can occur in the factorization of any other term in the sequence. The same applies to the four prime factors of s6:
- 547
- 607
- 1033
- 31051
Each belongs to s6 and to no other Sylvester term.
An Euclid-Style Proof That There Are Infinitely Many Primes
Every Sylvester term is greater than 1, so it has at least one prime divisor. Because different terms are coprime, a prime divisor used by one term cannot be reused by another. An infinite sequence of Sylvester terms therefore forces infinitely many distinct prime divisors.
This mirrors the product-plus-one mechanism in Euclid’s proof of the infinitude of primes. The new number itself does not have to be prime. It only needs a prime divisor that was absent from the earlier product.
The Same Product Structure Produces an Egyptian Fraction Identity
The reciprocal pattern
1/2 + 1/3 + 1/7 + 1/43 + 1/1807 + …
is not separate from the recurrence. It comes from the same identity
sn+1 − 1 = sn(sn − 1).
This gives
1/sn = 1/(sn − 1) − 1/(sn+1 − 1).
The terms telescope when added. For the first n reciprocals,
1/s0 + 1/s1 + … + 1/sn−1 = 1 − 1/(sn − 1).
For example,
1/2 + 1/3 + 1/7 = 41/42 = 1 − 1/42,
and the next Sylvester term is 43. As n increases, sn grows without bound, so the remaining fraction 1/(sn − 1) approaches 0. The infinite reciprocal sum is therefore 1.
Why the Terms Become Huge So Quickly
The recurrence
sn+1 = sn² − sn + 1
is close to repeated squaring once sn is large. That is why the number of decimal digits rises so quickly. From s0 through s9, the digit counts are
1, 1, 1, 2, 4, 7, 14, 27, 53, 105.
The exact terms illustrate the jump:
- s5 = 3263443 — 7 digits
- s6 = 10650056950807 — 14 digits
- s7 = 113423713055421844361000443 — 27 digits
- s8 = 12864938683278671740537145998360961546653259485195807 — 53 digits
The coprimality proof does not become harder as the numbers grow. Factorization does. Knowing that sn shares no prime factor with earlier terms says nothing about how difficult it will be to identify all of sn’s own prime factors.
Which Primes Can Divide Sylvester Terms?
Pairwise coprimality controls whether a prime factor can be reused. There is also a modular restriction on which primes can occur in the first place.
For a prime divisor p > 3 of a Sylvester term,
p ≡ 1 (mod 6).
The early factorizations fit this rule: 7, 13, 43, 139, 547, 607, 1033, and 31051 are all congruent to 1 modulo 6.
One route to the restriction starts by rewriting the recurrence as
4sn = (2sn−1 − 1)² + 3.
If p divides sn, this makes −3 a quadratic residue modulo p. For primes larger than 3, the resulting quadratic-residue condition forces p into the residue class 1 modulo 6.
Pairwise Coprime Does Not Settle Square-Freeness
A positive integer is square-free when no square of a prime divides it. Pairwise coprimality prevents a prime from appearing in two different Sylvester terms, but it does not prevent a prime from appearing more than once inside a single term.
For instance, a hypothetical factorization containing p² in one Sylvester term would not contradict gcd(si, sj) = 1 for i ≠ j. It would only show that one term is not square-free.
Open Problem
Checked on August 25, 2026. No squareful Sylvester term is known, but there is still no proof that every term in the sequence is square-free and no known counterexample.
A separate result rules out an entire Sylvester term being a perfect power. That does not resolve the square-free question: an integer can contain a squared prime factor without itself being a perfect power.
A 2026 Extension of the Divisibility Structure
A newer result concerns differences between shifted Sylvester terms. Fix k ≥ 0 and define
bn = sn+k − sk for n ≥ 1.
The sequence b1, b2, … is a strong divisibility sequence, meaning
gcd(bm, bn) = bgcd(m,n).
In particular, if n divides m, then bn divides bm. This is a different statement from the pairwise-coprime property of the original Sylvester terms: it describes divisibility among structured differences sn+k − sk, not shared factors between si and sj.