Circle-to-circle contacts on a hexagonal grid.

In OEIS sequences A047932 and A263135, coins are placed in a "spiral" on the faces or vertices of a hexagonal grid respectively, and the number of coin-to-coin contacts are counted. The following images illustrate these two constructions respectively:

Examples

The following images illustrate $$A047932(11) = 21$$ and $$A263135(22) = 27$$. That is, when $$11$$ coins are placed in a spiral on the faces of a hexagonal grid, there are 21 coin-to-coin connections. When $$22$$ coins are placed in a spiral on the vertices of a hexagonal grid, there are $$27$$ coin-to-coin connections.

Asymptotically, $$A047932(n) \sim \frac{6}{2}n$$ and $$A263135(n) \sim \frac{3}{2}n$$, because non-boundary coins have 6 and 3 neighbors respectively. Thus it's natural to ask how $$A047932(n)$$ compares with $$A263135(2n)$$.

An observation

Numerically, it appears that $$A263135(2n) - A047932(n) = \left\lceil \sqrt{3n - \frac 34} - \frac 12 \right\rceil = A216256(n),$$ but I'm not sure how to prove it. Obviously, A263135 and A047932 have combinatorial interpretations—the ones above! But what do I mean by a combinatorial interpretation of $$A216256(n)$$? One combinatorial pattern is the run-lengths, which are the positive integers with the odd numbers repeated. $$\underbrace{1}_1, \underbrace{2}_1, \underbrace{3, 3}_2, \underbrace{4, 4, 4}_3, \underbrace{5, 5, 5}_3, \underbrace{6, 6, 6, 6}_4, \underbrace{7, 7, 7, 7, 7}_5, \underbrace{8, 8, 8, 8, 8}_5, \underbrace{9, 9, 9, 9, 9, 9}_6, \dots$$

What is a combinatorial proof of the conjectured equation above? Or, if that's too hard, is there a way to derive a formula for $$A263135(2n)$$ and use the fact that we know that $$A047932(n) = \lfloor 3n - \sqrt{12n-3}\rfloor$$?