Any odd number is of form $a+b$ where $a^2+b^2$ is prime

Here are some heuristics. As Hans Engler defines, let $k(n)$ be the number of pairs $(a,b)$ with $a<b$ for which $a+b=n$ and $a^2+b^2$ is prime. In other words, $$ k(n) = \#\{ 1\le a < \tfrac n2 \colon a^2 + (n-a)^2 = 2a^2 - 2an + n^2 \text{ is prime} \}. $$ Ignoring issues of uniformity in $n$, the Bateman–Horn conjecture predicts that the number of prime values of an irreducible polynomial $f(a)$ up to $x$ is asymptotic to $$ \frac x{\log x} \prod_p \bigg( 1-\frac1p \bigg)^{-1} \bigg( 1-\frac{\sigma(p)}p \bigg), $$ where $\log$ denotes the natural logarithm and $$ \sigma(p) = \#\{ 1\le t\le p\colon f(t) \equiv 0 \pmod p \}. $$

We now calculate $\sigma(p)$ for $f(a) = 2a^2 - 2an + n^2$. Note that the discriminant of $f$ is $(-2n)^2 - 4\cdot2n^2 = -4n^2$. Therefore if $p$ does not divide $-4n^2$, the number of solutions is given by the Legendre symbol $$ \sigma(p) = 1 + \bigg (\frac{-4n^2}p\bigg) = 1 + \bigg (\frac{-1}p\bigg) = \begin{cases} 2, &\text{if } p\equiv1\pmod 4, \\ 0, &\text{if } p\equiv3\pmod 4. \end{cases} $$ Furthermore, we can check by hand that if $p=2$ then $\sigma(p)=0$, while if $p$ divides $n$ then $\sigma(p)=1$. Therefore our prediction becomes $$ k(n) \approx \frac{n/2}{\log(n/2)} \cdot 2 \prod_{\substack{p\equiv1\pmod 4 \\ p\nmid n}} \frac{p-2}{p-1} \prod_{\substack{p\equiv3\pmod 4 \\ p\nmid n}} \frac p{p-1}. $$ (We're abusing notation: those two products don't individually converge, but their product converges when the primes are taken in their natural order.) In principle that constant could be cleverly evaluated to several decimal places. But for the purposes of experiment, perhaps it's valuable to note that $k(n)$ should be approximately $n/\log n$, times some universal constant, times $$ \prod_{\substack{p\equiv1\pmod 4 \\ p\mid n}} \frac{p-1} {p-2}\prod_{\substack{p\equiv3\pmod 4 \\ p\mid n}} \frac {p-1}p; $$ and so the data can be normalized by that function of $n$ to test for consistency.

One equivalent way to formulate this conjecture is also the following:

For each natural number $n$ we have: $ \displaystyle 0 = \prod_{\underset{\gcd(a,b)=1}{2n+1=a+b}} (\Omega(a^2+b^2)-1)$

This is equivalent as to say that the polynomial $\displaystyle f_n(t) = \prod_{\underset{\gcd(a,b)=1}{2n+1=a+b}} (t-\Omega(a^2+b^2))$ has zero $1$.

where $\Omega$ counts the prime divisors with multiplicity.

COMMENT.-This is another way, maybe interesting for some people, of stating the same problem.

Given an odd natural number, $2n + 1$, there are $n$ different ways to express it as the sum of two natural $$2n+1=(2n-k)+(k+1);\space k=1,2,....,n$$ Then the problem can be stated as follows equivalently: $$\text{ For all natural 2n+1 greater than 1}\text{ at least one of the n numbers}\\\begin{cases}M_1=4n^2+1\\M_2=(2n-1)^2+2^2\\M_3=(2n-3)^2+3^3\\...........\\...........\\M_n=(n+1)^2+n^2\end{cases}\\ \text{ is a prime}$$

NOTE.- It is known that such a prime (if it exists) is necessarily of the form $p=4m+1$. Besides each $M_k$ has a factorization of the form $$M_k=\prod p_i^{\alpha_i}\prod q_j^{2\beta_j}$$ where $\alpha_i,\space \beta_j$ are non-negative integers,the primes $p_i$ and $q_j$ being of the form $4m+1$ and $4m-1$ respectively.

While larger 2n + 1 is, more likely to exist such a prime number. It would seem that the conjecture is true