Primes of the form $\frac{n^2-n+4}{2}$ satisfy Hardy-Littlewood analogue?

Let $n,a,b$ be positive integers with $a<b$. Consider primes of the form $f(n)=\dfrac{n^2-n+4}{2}$. Let $C(a,b)$ denote the amount of primes of the form $f(n)$ between (and including) $f(a)$ and $f(b)$. Now it appears that $C(a,b) < C(1,b-a+1)+1$. This observation is an analogue of the second Hardy-Littlewood conjecture.( http://en.wikipedia.org/wiki/Second_Hardy%E2%80%93Littlewood_conjecture)

If we assume there are an infinite amount of primes of the form $f(n)$ and/or the original second Hardy-Littlewood conjecture is true , can we prove that $C(a,b) < C(1,b-a+1)+1$?

Is it possible to say anything about $C(a,b) < C(1,b-a+1)+1$ without assuming an infinite amount of primes of the form $f(n)$ and/or the original second Hardy-Littlewood conjecture is true ?

Is there a counterexample known ?

Notice that a finite amount of primes $f(n)$ is also potentially consistant with $C(a,b) < C(1,b-a+1)+1$.

I was thinking about using a weaker form of the second Hardy-Littlewood namely $\pi(2,x+1)+1>\pi(y,y+x)$ where $\pi(x,y)$ means counting primes from $x$ till $y$. But with no success so far.

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I don't knnow. These primes are tabulated at oeis.org/A055472 – Gerry Myerson Feb 17 '13 at 22:51