Let $\{a(n)\}$ be a sequence satisfies $a(1)=1$, $a(2)=2$, and $a(n)=a(n-1)+a(\lfloor\ln(n)\rfloor)$ for $n\geq 3$.
According to the definition, it seems that $a(n)=\Omega(n\ln n\ln\ln n\ln \ln \ln n ...\ln^{(k)}n)$ for any constant $k$.
Does $\sum_{i=1}^{\infty}\frac{1}{a(i)}$ still diverge? If yes, how fast it diverges?
Consider the function $f(x)=1/x$ on $[1,e]$ and extend it on $[1,\infty)$ by equality $e^xf(e^x)=f(x)$ for $x\geqslant 1$. Then both functions $f(x),xf(x)$ decrease on $[1,\infty)$.
At first, I claim that $1/a_n\geqslant f(n)$. This is true for $n=1,2$, and we induct in $n$. Assume that the claim is proved for $1,\dots,n-1$; denote $k=\lfloor \log n\rfloor$. We have $$a_n=a_{n-1}+a_k\leqslant \frac1{f(n-1)}+\frac1{f(k)}\leqslant \frac1{f(n-1)}+\frac1{f(\log n)}=\frac1{f(n-1)}+\frac1{nf(n)}\leqslant \frac1{f(n)},$$ since $xf(x)$ decreases.
Next, the integral of $f$ diverges. Indeed, from the change of variables formula $$\int_{e^N}^{e^{e^N}}f(x)dx=\int_{N}^{e^N}f(e^t)e^tdt=\int_N^{e^N}f(t)dt,$$ we get a sequence of disjoint segments over which $f$ has the same integral.
Hence $\sum f(n)$ also diverges by the integral test, and so does $\sum a_n$. The growth of divergence for $\int f$ is seen from the proof, and for $\sum 1/a_n$ the growth is essentially the same, since we have suitable reverse estimates aswell.
