TITLE: Can a positive binary quadratic form represent 14 consecutive numbers?
QUESTION [59 upvotes]: NEW CONJECTURE: There is no general upper bound.
Wadim Zudilin suggested that I make this a separate question. This follows
representability of consecutive integers by a binary quadratic form
where most of the people who gave answers are worn out after arguing over indefinite forms and inhomogeneous polynomials. Some real effort went into this, perhaps it will not be seen as a duplicate question.
So the question is, can a positive definite integral binary quadratic form
$$ f(x,y) = a x^2 + b x y + c y^2 $$
represent 13 consecutive numbers?
My record so far is 8: the form $$6x^2+5xy+14y^2 $$
 represents the 8 consecutive numbers from 716,234 to 716,241. Here we have discriminant $ \Delta = -311,$ and 2,3,5,7 are all residues $\pmod {311}.$ I do not think it remotely coincidental that $$6x^2+xy+13 y^2 $$ represents the 7 consecutive numbers from 716,235 to 716,241. 
I have a number of observations. There is a congruence obstacle $\pmod 8$ unless, with 
$ f(x,y) = a x^2 + b x y + c y^2 $  and $\Delta = b^2 - 4 a c,$ we have $\Delta \equiv 1 \pmod 8,$ or 
$ | \Delta | \equiv 7 \pmod 8.$ If a prime $p | \Delta,$ then the form is restricted to either all quadratic residues or all nonresidues $ \pmod p$ among numbers not divisible by $p.$ 
In what could be a red herring, I have been emphasizing $\Delta = -p$ where $p \equiv 7 \pmod 8$ is prime, and where there is a very long string of consecutive quadratic residues $\pmod p.$ Note that this means only a single genus with the same $\Delta = -p,$ and any form is restricted to residues. I did not anticipate that long strings of represented numbers would not start at 1 or any predictable place and would be fairly large. As target numbers grow, the probability of not being represented by any form of the discriminant grows ( if prime $q \parallel n$ with $(-p| q) = -1$), but as the number of prime factors $r$ with $(-p| r) = 1$ grows so does the probability that many forms represent the number if any do.  Finally, on the influence of taking another $\Delta$ with even more consecutive residues, the trouble seems to be that the class number grows as well. So everywhere there are trade-offs.
EDIT, Monday 10 May. I had an idea that the large values represented by any individual form ought to be isolated. That was naive. Legendre showed that for a prime $q \equiv 7 \pmod 8$ there exists a solution to $u^2 - q v^2 = 2,$ and therefore infinitely many solutions. This means that the form $x^2 + q y^2$ represents the  triple of consecutive
numbers $q v^2, 1 + q v^2, u^2$ and then represents $4 + q v^2$ after perhaps skipping $3 + q v^2$.
Taking $q = 8 k - 1,$ the form $ x^2 + x y + 2 k y^2$ has no restrictions $\pmod 8,$ while an explicit formula shows that it represents every number represented by   $x^2 + q y^2.$ Put together, if $8k-1 = q$ is prime, then $ x^2 + x y + 2 k y^2$ represents infinitely many  triples. If, in addition, $ ( 3 | q) = 1,$ it seems plausible to expect infinitely many  quintuples. It should be admitted that the recipe given seems not to be a particularly good way to jump from length 3 to length 5, although strings of length 5 beginning with some $q t^2$ appear plentiful.
EDIT, Tuesday 11 May. I have found a string of 9, the form is $6 x^2 + x y + 13 y^2$ and the numbers start at $1786879113 = 3 \cdot 173 \cdot 193 \cdot 17839$ and end with 
$1786879121$ which is prime. As to checking, I have a separate program that shows me the particular $x,y$ for representing a target number by a positive binary form. Then I checked those pairs using my programmable calculator, which has exact arithmetic up to $10^{10}.$ 
EDIT, Saturday 15 May. I have found a string of 10, the form is $9 x^2 + 5 x y + 14 y^2$ and the numbers start at $866988565 = 5 \cdot 23 \cdot 7539031$ and end with 
$866988574 = 2   \cdot  433494287.$
EDIT, Thursday 17 June. Wadim Zudilin has been running one of my programs on a fast computer. We finally have a string of 11, the form being $ 3 x^2 + x y + 26 y^2$ of discriminant $-311.$ The integrally represented numbers start at  897105813710 and end at   897105813720. Note that the maximum possible for this discriminant is 11. So we now have this conjecture: For discriminants $\Delta$ with absolute values in this sequence
http://www.oeis.org/A000229
some form represents a set of $N$ consecutive integers, where $N$ is the first quadratic nonresidue. As a result, we conjecture that there is no upper bound on the number of consecutive integers that can be represented by a positive quadratic form.

REPLY [4 votes]: Here's a heuristic that suggests why arbitrarily large strings of consecutive 
numbers should be representable by some binary quadratic form.  For simplicity 
consider a prime $p$ that is $3\pmod 4$ so that $-p$ is a fundamental discriminant.  Suppose that $p$ has been chosen in such a way that all the primes $\le k$ are 
quadratic residues $\pmod p$.  Suppose there are $h$ quadratic forms of discriminant $-p$ (and note that there is only one genus).  
Recall that a number $n$ is represented by some form of discriminant $-p$ if 
every prime factor $\ell$ of $n$ satisfies $\chi(\ell)=1$.  As discussed in my 
answer to Achieving consecutive integers as norms from a quadratic field we should expect to find many strings of $k$ consecutive numbers, each of which is representable by some form of discriminant $-p$.  
In my answer to that question, I focused on such strings of (almost) prime numbers, 
but the same Hardy-Littlewood heuristics would predict lots of strings $n+1$, $\ldots$, 
$n+k$ where each $n+j$ is divisible only by primes that are quadratic residues 
$\pmod p$, and each $n+j$ has a typical number of prime factors.  Under the 
restriction that all prime factors of $n$ are quadratic residues $\pmod p$, if $n$ is 
large then typically it will have about $\frac 12 \log \log n$ such prime factors.  Moreover we may expect these prime factors to be roughly equally distributed 
in the class group (which is of fixed size $h$).  Thus since there are $2^{\omega(n)}$ 
factorizations of $n$ as a product of two ideals, we would expect that typically there are many such factorizations with one of the factors lying in a prescribed 
ideal class.  
Summarizing one would expect that there are many strings $n+j$ 
($1\le j\le k$) with each $n+j$ composed of about $\frac 12\log \log n$ primes 
that are all quadratic residues $\pmod p$, and (typically) each such $n+j$ would 
be represented by every form of discriminant $-p$.  
It should be possible with a little effort to turn this into a precise 
Hardy-Littlewood type conjecture, but I don't see any hope of a proof. 
The heuristic described above is probably classical.  One place where this 
heuristic is described is a paper of Blomer and Granville: see pages 9 and 10 of 
http://www.dms.umontreal.ca/~andrew/PDF/quadraticforms.pdf