Is the following conjecture true ? $$\left\{\sum\limits_{\begin{array}{c}k=1\\k\in\mathbb{Z}\end{array}}^nk \ |\ n\in\Bbb Z\right\} \cap \left\lbrace \sum\limits_{\begin{array}{c}k=1\\k\in\mathbb{Z}\setminus\mathbb{P}\end{array}}^nk \ |\ n\in\mathbb{Z}\setminus\mathbb{P}\right\rbrace \cap \left\{\sum\limits_{\begin{array}{c}k=2\\k\in\Bbb P\end{array}}^nk\ |\ n\in\Bbb P\right\}= \{ 28 \}$$

(that is : A000217 $\cap$ A051349 $\cap$ A007504 = { 28 } )

I didn't find any other number below $10^{14}$ with this property (Haskell script here).

I originally posted this question here: math.stackexchange/questions/1357530, with an interesting contribution by joriki in favor of the conjecture:

Unless there's a systematic reason for these sequences to coincide or avoid each other (which I doubt), we can estimate the number of triple coincidences of these three sequences by integrating over the product of their densities. The first one has density $1/n$ at $a_n=n(n+1)/2$, so at $x$ it has density $\sim(2x)^{−1/2}$. The others both omit numbers, so at given $x$ their densities are lower than this.

Thus we can get an upper bound for the "probability" of there being such numbers beyond some $x_0$ from this integral:

$\int_{x_0}^\infty\left(2x\right)^{-3/2}\mathrm dx=(2x_0)^{-1/2}$.

As you've searched up to $10^{14}$, the "probability" of finding a triple coincidence beyond that is less that one in ten million. Unless there's a systematic reason...