Arithmetic Progressions of Primes

W. Stein^{∗}

November 29, 1998

Abstract Let 2n be an even integer. What is the longest arithmetic progression p, p + 2n, …, p + 2na with p prime? In this paper we guess some answers.

Warning: The following is just for fun! There could be a well-known conjecture which supercedes all of this.

1 Introduction

It is easy to show (see Proposition 3.1) that if

p, p + 2, p + 4

are all prime, then p = 3. Similarly, if

p, p + 6, p + 12, p + 18, p + 24

are all prime, then p = 5 and the progression is thus 5, 11, 17, 23, 29. There are probably infinitely many examples where

p, p + 6, p + 12, p + 18,

are all prime, but one would not dare think of proving this here as it is probably more difficult than the twin primes conjecture.

For each even integer 2n, let M(2n) to be the maximal length of an arithmetic progression of primes with gap 2n. Thus M(2) = 3, M(6) = 5. Let M_{∞}(2n) be the maximum length of arithmetic progressions of gap 2n which occur infinitely often. Thus one conjectures that M_{∞}(2) = 2, M_{∞}(6) = 4.

Is there an example in which M_{∞}(2n) < M(2n) − 1? Is there an upper bound on the possible values M(2n)?

∗UC Berkeley, Department of Mathematics, was@math.berkeley.edu

2 Table

The M_{?} column was computed by searching the first 100000 or so primes for an arithmetic progression of primes, having gap 2n. The M_{∞?} column was computed by doing the same search, but requiring that at least two distinct progressions were found (usually there were many more). By Proposition 3.1, the computed values of M_{?} are the actual value of M, except possibly for the last four rows in the table.

2n  2n  M_{?} M_{∞?} Examples
2  2  3  2 3
4  2^{2}  3  2 3
6  2 · 3  5  4 5
8  2^{3}  3  2 3
10  2 · 5  3  2 3
12  2^{2} · 3  5  4 5
14  2 · 7  3  2 3
16  2^{4}  2  2 3
18  2 · 3^{2}  4  4 5
20  2^{2} · 5  3  2 3
22  2 · 11  2  2 7
24  2^{3} · 3  4  4 59
26  2 · 13  2  2 3
28  2^{2} · 7  2  2 3
30  2 · 3 · 5  6  6 7, 107
32  2^{5}  2  2 5
34  2 · 17  3  2 3
36  2^{2} · 3^{2}  4  4 31
38  2 · 19  3  2 3
40  2^{3} · 5  3  2 3
42  2 · 3 · 7  5  4 5
44  2^{2} · 11  2  2 3
46  2 · 23  2  2 7
48  2^{4} · 3  5  4 5
50  2 · 5^{2}  3  2 3
210  2 · 3 · 5 · 7  10  9 199
2310  2 · 3 · 5 · 7 · 11  9  9 3823
30030  2 · 3 · 5 · 7 · 11 · 13  12 12 23143, 1498141
510510 2 · 3 · 5 · 7 · 11 · 13 · 17 13  ? 766439

3 Conclusions

Proposition 3.1. Let q be a prime. If q ∤ 2n then M(2n) ≤ q and M_{∞}(2n) ≤ q − 1. Furthermore, M(2n) = q iff q is involved in arithmetic progression of length q with gap 2n.

Proof. Reduce the arithmetic progression

p, p + 2n, …, p + 2na

modulo q. Since 2n is a unit modulo q, if the length of the arithmetic progression is ≥ q then one of the terms in the arithmetic progression is divisible by q, hence equal to q.

In the table above, there is not a single example in which q ∤ 2n, M(2n) = q and q is not the first prime in the arithmetic. This suggests

Question 3.2. Let q ∤ 2n be a prime and suppose M(2n) = q. Then must

q, q + 2n, …, q + 2n(q − 1)

all be prime?

The data also suggests the following

Conjecture 3.3. Both M(2n) and M_{∞}(2n) are ≥ 2 for all n.

The conjecture that M(2n) ≥ 2 is similar to Golbach’s conjecture. It asserts that any even integer is the difference of two primes; the conjecture that M_{∞}(2n) ≥ 2 asserts that this can be done in infinitely many ways.

It seems conceivable that M(2n!) is unbounded as a function of n.