Fibonacci analogue of A124171

[Jonathan Post]

I've got a Fibonacci analogue of A124171.  This kind of analogue can be made
from almost any nice OEIS sequence. I picked Fibonacci as a core sequence
for exemplary purpose. The original is:

%I A124171
%S A124171 1,1,2,3,1,2,3,4,5,6,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
%N A124171 Sequence obtained by reading the triangles shown below by rows.
%e A124171 1
%e A124171 1
%e A124171 2 3
%e A124171 1
%e A124171 2 3
%e A124171 4 5 6
%e A124171 1
%e A124171 2 3
%e A124171 4 5 6
%e A124171 7 8 9 10

My analogue is:
%I A000000
%S A000000 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1,
1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 5, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1,
2, 3, 5, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 5, 8, ...
%N Fibonacci triangle read by rows; the triangles below read by rows.
%e A000000 1
%e A000000 1
%e A000000 1, 1
%e A000000 1
%e A000000 1, 1
%e A000000 1, 1, 2
%e A000000 1
%e A000000 1, 1
%e A000000 1, 1, 2
%e A000000 1, 1, 2, 3
%e A000000 1
%e A000000 1, 1
%e A000000 1, 1, 2
%e A000000 1, 1, 2, 3
%e A000000 1, 1, 2, 3, 5
%O 1, 10
%F For k>0, max(row(T(k))) = F(k) where T(n) = A000217(k), F(k) =
A000045(k).
%F Records of a(n) after a(1) = 1 are given by a(A000292(n)) = C(n+2,3) =
n(n+1)(n+2)/6 = F(n+1) = A000045(n+1).

What I'm missing here is the simplest form of trhe closed-form formula for
a(n) itself, and comments about the diagonals and antidiagonals.
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