Fibonacci analogue of A124171

[Jonathan Post]

This is a kind of composition of A124171 and the Fibonacci sequence.  But it
is not a composition or convolution that I recognize, but perhaps a
transform new to OEIS.

The function is slow-growing at first.  The smallest n such that a(n) > n
occurs when a(560) = 610.  But eventually, the superpolynomial Fibonacci
dominates the merely cubic tetrahedral numbers, and the mean value of a(n)/n
exceeds any fixed bound.

There is a slower-starting such analogue that starts with F(0) = 0 and f(1)
= 1, the triangles beginning:
0
0
0, 1
0
0, 1
0, 1, 1
0
0, 1
0, 1, 1
0, 1, 1, 2
0
0, 1
0, 1, 1
0, 1, 1, 2
0, 1, 1, 2, 3
reading by rows gives offset 0,36 and many zeroes (a whole lotta nuthin').

On 12/7/06, Jonathan Post <jvospost3 at gmail.com> wrote:
>
> 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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