In number theory, Sylvester's sequence is an integer sequence in which each term is the product of the previous terms, plus one. Its first few terms are
2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence A000058 in the OEIS).
for a number E that is approximately 1.26408473530530...[10] (sequence A076393 in the OEIS). This formula has the effect of the following algorithm:
s0 is the nearest integer to E 2; s1 is the nearest integer to E 4; s2 is the nearest integer to E 8; for sn, take E 2, square it n more times, and take the nearest integer.
This would only be a practical algorithm if we had a better way of calculating E to the requisite number of places than calculating sn and taking its repeated square root.[11]
The double-exponential growth of the Sylvester sequence is unsurprising if one compares it to the sequence of Fermat numbersFn; the Fermat numbers are usually defined by a doubly exponential formula, , but they can also be defined by a product formula very similar to that defining Sylvester's sequence:[12]
Since this sequence of partial sums (sj − 2)/(sj − 1) converges to one, the overall series forms an infinite Egyptian fraction representation of the number one:
One can find finite Egyptian fraction representations of one, of any length, by truncating this series and subtracting one from the last denominator:
The sum of the first k terms of the infinite series provides the closest possible underestimate of 1 by any k-term Egyptian fraction.[2] For example, the first four terms add to 1805/1806, and therefore any Egyptian fraction for a number in the open interval (1805/1806, 1) requires at least five terms.
It is possible to interpret the Sylvester sequence as the result of a greedy algorithm for Egyptian fractions, that at each step chooses the smallest possible denominator that makes the partial sum of the series be less than one.[15]
Uniqueness of quickly growing series with rational sums
As Sylvester himself observed, Sylvester's sequence seems to be unique in having such quickly growing values, while simultaneously having a series of reciprocals that converges to a rational number. This sequence provides an example showing that double-exponential growth is not enough to cause an integer sequence to be an irrationality sequence.[16]
To make this more precise, it follows from results of Badea (1993) that, if a sequence of integers grows quickly enough that
and if the series
converges to a rational number A, then, for all n after some point, this sequence must be defined by the same recurrence
that can be used to define Sylvester's sequence.[17]
If i < j, it follows from the definition that sj ≡ 1 (mod si ). Therefore, every two numbers in Sylvester's sequence are relatively prime. The sequence can be used to prove that there are infinitely many prime numbers, as any prime can divide at most one number in the sequence. More strongly, no prime factor of a number in the sequence can be congruent to 5 modulo 6, and the sequence can be used to prove that there are infinitely many primes congruent to 7 modulo 12.[20]
Unsolved problem in mathematics:
Are all the terms in Sylvester's sequence squarefree?
Much remains unknown about the factorization of the numbers in Sylvester's sequence. For instance, it is not known if all numbers in the sequence are squarefree, although all the known terms are.[21]
As Vardi (1991) describes, it is easy to determine which Sylvester number (if any) a given prime p divides: simply compute the recurrence defining the numbers modulo p until finding either a number that is congruent to zero (mod p) or finding a repeated modulus.[3] Using this technique he found that 1166 out of the first three million primes are divisors of Sylvester numbers,[22] and that none of these primes has a square that divides a Sylvester number.
The set of primes that can occur as factors of Sylvester numbers is of density zero in the set of all primes:[23] indeed, the number of such primes less than x is .[24]
The following table shows known factorizations of these numbers (except the first four, which are all prime):[4]
Znám's problem concerns sets of numbers such that each number in the set divides but is not equal to the product of all the other numbers, plus one. Without the inequality requirement, the values in Sylvester's sequence would solve the problem; with that requirement, it has other solutions derived from recurrences similar to the one defining Sylvester's sequence. Solutions to Znám's problem have applications to the classification of surface singularities (Brenton and Hill 1988) and to the theory of nondeterministic finite automata.[26]
Curtiss (1922) describes an application of the closest approximations to one by k-term sums of unit fractions, in lower-bounding the number of divisors of any perfect number, and Miller (1919) uses the same property to upper bound the size of certain groups.[27]
Brown, D. J. (1979). A lower bound for on-line one-dimensional bin packing algorithms. Tech. Rep. R-864. Coordinated Science Lab., Univ. of Illinois, Urbana-Champaign.
Erdős, Paul; Graham, Ronald L. (1980). Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique, No. 28, Univ. de Genève. MR0592420.
Odoni, R. W. K. (1985). "On the prime divisors of the sequence wn+1 =1+w1⋯wn". Journal of the London Mathematical Society. Series II. 32: 1–11. doi:10.1112/jlms/s2-32.1.1. Zbl0574.10020.
Rosenman, Martin; Underwood, F. (1933). "Problem 3536". American Mathematical Monthly. 40 (3): 180–181. doi:10.2307/2301036. JSTOR2301036.
Salzer, H. E. (1947). "The approximation of numbers as sums of reciprocals". American Mathematical Monthly. 54 (3): 135–142. doi:10.2307/2305906. JSTOR2305906. MR0020339.