A GENERALIZATION OF THE PRIME PAIRS / HARDY-LITTLEWOOD / DICKSON / SCHINZEL AND SIERPINSKI CONJECTURES

by

STEPHEN CHILTON^Ack

There exist an infinity of Ns such that both N and N+2 are prime.

Hardy and Littlewood generalized this to "k-tuples" of the form N, N+a₁, N+a₂, ... N+a_k, conjecturing / claiming that

For any given (a₁, ... a_k) there are an infinite number of Ns for which the k-tuple N, N+a₁, N+a₂, ... N+a_k will be all prime (unless there is a trivial divisibility condition⁽¹⁾that ensures that one of these will be nonprime).

The prime pairs conjecture is a special case of this for the 1-tuple (2).⁽²⁾

Dickson's Conjecture, which is discussed in more detail here in Chris Caldwell's informative nest of pages on prime numbers, extends the conjecture to a k-tuple of linear equations with integer coefficients, i.e., a₁N + b₁, a₂N + b₂, … a_kN + b_k.⁽³⁾ Dickson's formulation requires that there be no prime number p that divides the product of these linear equations for all N.

In 1958, Schinzel and Sierpinski proposed "Hypothesis H", generalizing Dickson's Conjecture to k-tuples of polynomials with positive leading integers where, just as in Dickson's formulation, there is no prime number p that divides the product of these polynomials for all N. Let me call this last restriction the "divisible product condition".

I wish to further generalize these conjectures by substituting a "suitable reduction condition" for the divisible product condition. For ease of presentation I will state the new condition it in terms of the Hardy-Littlewood Conjecture, but I really mean it to be extended in the obvious way to Dickson's Conjecture and to Schinzel's and Sierpinski's Hypothesis H.

For any k-tuple (a₁, ... a_k), there are an infinity of Ns for which every "suitably reduced" member of the combination is prime

where

"suitably reduced" = "divided by the divisor to which the 'trivial divisibility condition' refers".

[Spring 1999: I AM INDEBTED TO PROFESSOR WILLIAM ADAMS OF THE DEPARTMENT OF MATHEMATICS, UNIVERSITY OF MARYLAND, FOR THE OBSERVATION THAT THE "GENERALIZATIONS" I PROPOSE DO NOT GO BEYOND DICKSON'S CONJECTURE, EVEN THOUGH THEY APPEAR TO. I'M LEAVING THIS PAGE UP TO CONVEY THIS INFORMATION AND THIS APPRECIATION.]

A specific example may be needed to explain the concept of "suitably reduced". Consider the quadruple with offsets 0, 1, 3 and 5: (N, N+1, N+3, N+5). It is apparent (and can be verified by examining the product of these four factors) that either one or three of the numbers must be divisible by 2 and one or two of the numbers must be divisible by 3, so that this quadruple does not satisfy the divisible product condition. With the suitable reduction condition, however, we need not give up yet. Let's consider the class of Ns such that N/2 and (N+1)/3 are integers. In that case, the suitable reduction condition permits N to be "suitably reduced" by division by 2, and N+1 to be "suitably reduced" by division by 3. The specific conjecture for this quadruple (and this choice of suitable reductions)⁽³⁾ then becomes,

There are an infinity of Ns for which N/2, (N+1)/3, N+3 and N+5 are all prime.

Here are some of these values of N:

• N=2, yielding the quadruple (2,3,5,7), which is "suitably reduced" to the quadruple of primes (1,1,5,7), 1 being accepted as prime in this context.
• N=14, yielding the quadruple (14,15,17,19), which is "suitably reduced" to the quadruple of primes (7,5,17,19).
• N=38, yielding the quadruple (38,39,41,43), which is "suitably reduced" to the quadruple of primes (19,13,41,43).
• N=878, 1046, 1226, 2126, 2306, 3326, and so on, presumably ad infinitum.

We must also consider the cases where other elements are divisible by 2 and by 3. This leads us into one other consideration, however. Consider, as before, the quadruple (N, N+1, N+3, N+5). If N is divisible by 2, then none of the other numbers are divisible by 2. However, if N+1, N+3, and N+5 are all divisible by 2, then at least one of them is divisible by 4: either N+3 is divisible by 4, or else N+1 and N+5 are. "Suitable reduction", therefore, demands not just that we reduce it not just by the factor(s) referred to in the divisible product condition; we must consider each pattern of factors and continue permissible reduction until no further reduction is possible.

Consider the case, then, where N+1 is divisible by 4, in which case N+5 will also be, and N+3 will be divisible by 2 (but not 4), and N will not be divisible by 2. If (N+1) and (N+5) are both divisible by 4, then at least one of them must be divisible by 8; let us suppose it is (N+1). And now let us also suppose that we take the case where N is divisible by 3, in which case N+3 will also be divisible by 3, and N+1 and N+5 will not be. Combining these choices and reduction, our hypothesis becomes,

There are an infinity of Ns for which N/3, (N+1)/8, (N+3)/6 and (N+5)/4 are all prime.

Here are some of these values of N:

• N=15, yielding the quadruple (15,16,18,20), which is "suitably reduced" to the quadruple of primes (5,2,3,5).
• N=39, yielding the quadruple (39,40,42,44), which is suitably reduced to the quadruple of primes (13,5,7,11)
• N=183, yielding the quadruple (183,184,186,188), which is suitable reduced to the quadruple of primes (61,23,31,47).
• N=8391, 13863, 16311, 20343, 28983, 32583, and so on, presumably ad infinitum.

The general conjecture is that the same is true of any k-tuple with any pattern of suitable reductions.⁽⁴⁾ The Hardy-Littlewood, Dickson, and Schinzel & Sierpinski formulas would then seem to be modifiable by a replacement of the ms in some way by "suitably reduced" ms.

A further generalization is,

Given any k-tuple of offsets (a₁, ... a_k) and any k+1-tuple of integer divisors (d₀, d₁, ... d_k), There are an infinity of Ns for which the k+1 quantities (N/d₀), ((N+a₁)/d₁), ... ((N+a_k)/d_k), "suitably reduced" (for each specific "suitable reduction"), are all prime (and thus also integers, obviously).

For example, let k=3, a = (1, 3, 5), and d = (1, 1, 2, 5). It is straightforward to show that these conditions are inconsistent: no integer N can satisfy the condition that both (N/3) and (N+5)/5 be both prime, since if N is divisible by both the relative primes 3 and 5, then N/3 must be divisible by 5 and thus cannot be prime.

On the other hand, let k=3, a = (11, 17, 22), and d = (1, 1, 1, 2). It is apparent that either N and N+22 must be even or N+11 and N+17 must be even. Since (N+22)/2 must be an integer, N must be even and thus we are entitled to reduce N and N+22 by 2. Because of the divisor d₄=2, we are entitled to further reduce N+22 again by 2. At least one of the numbers N, N+11, N+17, and N+22 must be divisible by 3. There are three different possible suitable reductions: that N is divisible by 3, that N+11 and N+17 are divisible by 3, or that N+22 is divisible by 3. The generalization holds for any choice; let us choose the second of the two: that N+11 and N+17 are divisible by 3. So our specific suitable reduction would be to divide N by 2, (N+11) by 3, (N+17) by 3, and (N+22) by 4. This is satisfied by N=22, 166, 310, ... ad infinitum?

While there is but slight merit in thus plausibly extending these previous conjectures, I do so with the thought that the creation of as general a (true) conjecture as possible will direct us to an appropriate method of solution. (I say "us", when really I mean "you".)

So some questions for those of you reading this: Do the extensions arising from the suitable reduction condition maintain the plausibility of the previous conjectures? If so, would it be worth sending this along as a note to some mathematics journal? (Which?) Is there someone else you know of who would be interested in this kind of question? Please write me here to let me know.

FOOTNOTES

* I appreciate Dr. Thomas N. Herzog and Prof. James C. Owings, Jr. for their useful references.

1. For example, the 3-tuple N, N+1, N+2 will assuredly have members divisible by 2 and by 3, and if N is divisible by 2, then either N or N+2 will be divisible by 4.

2. In fact, the Hardy-Littlewood Conjecture holds that the density of such prime k-tuples is asymptotic to ... well, a complicated formula that can be found here.

3. Of course, all the leading coefficients must be positive. I am indebted to Professor Caldwell for the reference to Dickson's Conjecture and the web page discussing it. I find it curious that Dickson's Conjecture appeared in 1904, eighteen years previous to 1922, when (I believe) Hardy & Littlewood's article appeared.

4. Just for a lark, I looked at the octuple generated by (2, 3, 5, 7, 11, 13, 17, 19), implying the offsets (0, 1, 3, 5, 9, 11, 15, 17). Analysis reveals that one possible set of suitable reductions is the octuple N/2, (N+1)/3, (N+3)/5, (N+5), (N+9), (N+11), (N+15), (N+17). Below 1,686,733,372, there are eight Ns for which this octuple are prime (not including the generating octuple N=2, which requires no reductions at all): 79,682; 6,213,782; 18,362,282; 129,664,802; 236,197,382; 256,392,662; 1,174,132,562; and 1,643,537,162 . And presumably this series continues ad infinitum.

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