Prime Generating Polynomials(Hardy-Littlewood constant)

Jacobson found polynomial of the form x^2 + x + c with largest Hardy-
Littlewood constant C(D)~5.36 for the least D=1-4c=-
13598858514212472187,where

 

(1)C(D)=Product(1-(D/p)/p-1),

 

p is prime>2,(D/p) denotes the Legendre symbol

(Math.Comp., 2003, 72, 499-519).

In this paper Jacobson and Williams found polynomial of the same form with
larger C(D)~5.65 and 72-digit D. But in this case D is not the least of
possible magnitudes of D.

But if wanted not least D, there is simple idea to find stronger results.

For the first time D=5(mod8).Then according to (1) we have D=2(mod3), D=2
or 3(mod5) and so on. Thus we generate two sets: s1=(5,2,2,…) and
s2=(8,3,5,…,pmax). Since all numbers from set s2 are coprime, then
according to Chinese Remainder Theorem there exist D such that (D/p)=-1 for
all pmax+1>p>2.

Using the idea, were found the next results with m-digit D`s

(Table 1):

Table1.

C(D)______m______pmax
6.01477…__276____659
7.01289…__1007___2377
8.01071…__3234___7559
9.00134…__10006__23173
10.00130…_31680__73291
11.00010…_98670__227651

For example, for pmax=659 we have:
D=262508778659464115096185323435905718621126859589721245815217424331\58852200686713735757701335690265711480595015201939640121478280530259\08516600030845284564979968834904572433379987818950634053387670481251\59694554959443856960907671060549826406564583370499654905232954370723070397.

Whether C(D) may be infinitely large or not?

The proposed procedure allows simple to answer on the question. Let us
consider C(q):

(2)C(q)=Product(1+1/p-1),

where the product is taken over the primes q=>p>2. Clear that with growing
q, C(q) infinitely increase. Well known that in proposed procedure for any
member of set s2 there exist at least one candidate for set s1 and then we
can always find D for any pmax. If q=pmax than C(D)>=C(q). And consequently
C(D) may be infinitely
large.

I would like to know any information about Hardy-Littlewood constant
estimations.