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For a decreasing probability sequence $p(n)$, if $G(n,p(n))$ contains a Hamilton cycle with high probability, then $G(n,Cp(n))$ is pancyclic with high probability for every fixed sufficiently large $C$; three independent rounds suffice. One first exposes a Hamiltonian round and uses the independent sprinkled edges to create cycles of every shorter length.

Use sprinkling of a binomial random graph to write $G(n,p)=G_1\cup G_2$, where the rounds are independent,and $C$ is chosen large enough that $1-p=(1-p_1)(1-p_2)$. By the given theorem, $G_1$ has a Hamilton cycle $v_1{v}_2\cdots v_n{v}_1$ with high probability.

Condition on such a cycle. For each $3\le \ell \le n-1$, every chord $v_i{v}_{i+\ell -1}$ closes one of the two paths around the Hamilton cycle into a cycle of length $\ell$. There are at least $n/2$ distinct candidate chords, so the probability that $G_2$ supplies none is at mostA union bound over the fewer than $n$ lengths shows that all these cycles occur simultaneously with high probability when $c>4$. The Hamilton cycle itself supplies length $n$, so $G$ is pancyclic.

The sharp Hamilton cycle threshold for the Erdős-Rényi model says thatonly above the window $np=\log n+\log \log n+o(1)$. The hypothesis therefore places $p(n)$ above that window. The Hamiltonicity-to-pancyclicity sprinkling principle then says that three independent $G(n,p(n))$ rounds contain every cycle graph $C_{\ell }$, $3\le \ell \le n$, with high probability: one round supplies a Hamilton cycle, while the other two supply the chords and short-cycle edges used to obtain all intermediate lengths.

The union of the three rounds has individual edge probabilityBy the standard monotone coupling, it is a subgraph of $G(n,3p(n))$. Since being pancyclic is an increasing graph property, the required probability tends to one. If $3p(n)>1$, interpret the latter parameter as $\min \{3p(n),1\}$, in which case the conclusion is immediate.