Here we have a monotonocally increasing sequence that predicts itself:

$1,2,2,3,\color{brown}{3},4,4,4,\color{blue}{5,5,5},6,6,6,6,...$

The $n$th term $a(n)$ predicts the number of times that term appears in the sequence. For instance, the fifth term is $3$ (brown) and thus $5$ will appear three times (blue).*

The prediction is implemented through the recursion relation

$a[n+a(a(n))]=a(n+1).$

For example, $a(5)$ is the second occurrence of $3$, and when we march $a(a(5))=a(3)=2$ steps forward from there we land at $a(7)$, which is the second occurrence of $3+1=4$.

Let us plug this into an assumed power-law asumptotic relation:

$a(n)\approx\alpha n^\beta.$

Thus

$\alpha[n+\alpha(\alpha n^\beta)^\beta]\approx\alpha n^\beta+1$

$\alpha[n+\alpha^{1+\beta}n^{\beta^2}]^\beta\approx\alpha n^\beta+1$

Take two terms of the binomial power expansion of the left side:

$\alpha [n^\beta + \alpha^{1+\beta}\beta n^{\beta^2+\beta-1}]^\beta\approx\alpha n^\beta+1$

Canceling the identical leading terms leads to

$[\alpha^{1+\beta}\beta n^{\beta^2+\beta-1}]^\beta\approx1$

from which $\beta^2+\beta-1=0$ and thus $\color{blue}{\beta=\phi-1=1/\phi}$. The corresponding value of $\alpha$ is then $\beta^{-1/(1+\beta)}=\phi^{\phi-1}$. We may elegantly express the result as

$a_n\approx(\phi n)^{\phi-1}.$

If we graph $a_n$ versus $n$ on a log-log plot and draw the straight line represented by $\alpha n^\beta$ with the values computed above, we find that the line precisely "threads the needle" through the sequence values. Try it!

*Yes, that is a shout-out to one of our favorite video channels.

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