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(This is more like a comment with images) Here are some simulations of the values $c = c(p)$ using the grid of size $1000\times1000$ and $500$ steps together with some fitting curves. $\hspace{8em}$ The data clearly deviates from the polynomial $2p^2$, and although the above plot may seem to suggest that $c(p)$ assumes a nice closed form, I believe that ...

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Seems fine but it is easier to note that $$n+1\leq 2n$$ for $n\geq 1$ whence $$(n+1)^3\leq (2n)^3=8n^3$$ for $n\geq 1$ as desired.

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By the most recent bound on Linnik's Theorem, there is an absolute constant $c$ such that for every prime $q < cp_n^{1/5}$, there is a prime $p < p_n$ such that $p \equiv 1 \pmod{q}$. Your least common multiple is therefore divisible by all primes below $cp_n^{1/5}$. The prime number theorem implies that the product of all primes below $cp_n^{1/5}$ is ... 2 Your differential equation has closed-form solutions \eqalign{f \left( t \right) &=c_{{1}}{t}^{1/2-a/2}{{ J}_{-\sqrt {{a}^{2}-2\,a +4\,b+1}}\left(2\,{\frac {\sqrt {b\epsilon}}{\sqrt {t}}}\right)}\cr &+c_{{2 }}{t}^{1/2-a/2}{{ Y}_{-\sqrt {{a}^{2}-2\,a+4\,b+1}}\left(2\,{\frac {\sqrt {b\epsilon}}{\sqrt {t}}}\right)}} whereJ$and$Y$are Bessel ... 1 Here is a key lemma (which you can and should prove for yourself) that helps to resolve so many questions of this type. I'll phrase it in terms of functions of$x$where$x\to\infty$, although it holds for other domains as well. Lemma: Let$f,g\colon [1,\infty)\to[2,\infty)$be functions. If$\log f(x) = o(\log g(x))$as$x\to\infty$, then$f(x) = o(g(x))$... 1$\log((1/n) + n^2 )$If we just focus on the term inside of the logarithm.$ let x = 1/n + n^2$As n grows, we notice that the 1/n term effectively becomes zero and the overpowering term is$n^2$We can now state that$\log((1/n) + n^2 )$has growth$log(n^2)$when n gets large. We know from logarithms that$log(a^b) = b* log(a)$So$log(n^2)$can be ... 1 Yes it does. Let$t_N$denote the algorithm time. Let$f_N = \max\{t_M/M^2:M\le N\}$. Suppose$f_N$diverges. Then$t_N$fails to be$O(\sqrt{f_N} N^2)$. Therefore$f_N$does not diverge. Since$f_N$is a non-decreasing sequence, it must be bounded. Set$C= \sup f_N$. Then$t_N \le C N^2\$.

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