A Betti sequence is a map $\mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$.

A Betti sequence $b$ is realizable if there is a connected closed Kähler manifold $M$ such that $b(k)=b_k(M)$.

A Hodge diamond is a map $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$. To any Hodge diamond $h$ we associate a Betti sequence $b$ given by $b(k)=\sum_{i=0}^k h(i, k-i)$.

A Hodge diamond $h$ is realizable if there is a connected closed Kähler manifold $M$ such that $h(i, j)=h^{i, j}(M)$. Realizable Hodge diamonds have realizable Betti sequences.

A Hodge diamond is naively realizable if $h(0, 0)=1$ and there is an integer $n\geq 0$ such that

- $h(i, j)=h(j, i)=h(n-i, n-j)$ if $0\leq i, j\leq n$
- $h(i, j)\geq h(i-1, j-1)$ if $i, j\geq 1$ and $i+j\leq n$
- $h(i, j)=0$ if $i, j\geq 0$ and $\mathrm{min}(i, j)\geq n+1$

Is there a naively realizable, non-realizable Hodge diamond with a realizable Betti sequence?

For example, the naively realizable Hodge diamond with $n=2$ and $h(0, 1)=h(1, 1)=1$, $h(0, 2)=0$ is not realizable but its Betti sequence is not realizable either.