# 离散数学 Discrete Mathematics MATH223001/MATH223101

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We apply the lower bound in Lemma 2.5.1 to the logarithms. For a typical term, we get
$$\ln \left(\frac{m+t-k}{m-k}\right) \geq \frac{\frac{m+t-k}{m-k}-1}{\frac{m+t-k}{m-k}}=\frac{t}{m+t-k}$$

and so
\begin{aligned} \ln \left(\frac{m+t}{m}\right) &+\ln \left(\frac{m+t-1}{m-1}\right)+\cdots+\ln \left(\frac{m+1}{m-t+1}\right) \ & \geq \frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} . \end{aligned}
We replace each denominator by the largest one to decrease the sum:
$$\frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} \geq \frac{t^{2}}{m+t} .$$

## MATH223001/MATH223101COURSE NOTES ：

For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}$$
Solving for $a$ and $b$, we get
$$a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}$$
Then
$$L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}$$

# 离散数学与汇编 Discrete Math with Comp MATH223101

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Just as in the treatment of the case $k=200$ above, we subtract the number of primes up to $10^{k-1}$ from the number of primes up to $10^{k}$. By the Prime Number Theorem, this number is about
$$\frac{10^{k}}{k \ln 10}-\frac{10^{k-1}}{(k-1) \ln 10}=\frac{(9 k-10) 10^{k-1}}{k(k-1) \ln 10} .$$
Since
$$\frac{9 k-10}{k-1}=9-\frac{1}{k-1}$$

is very close to 9 if $k$ is large, we get that the number of primes with $k$ digits is approximately
$$\frac{9 \cdot 10^{k-1}}{k \ln 10}$$
Comparing this with the total number of positive integers with $k$ digits, which we know is $10^{k}-10^{k-1}=9 \cdot 10^{k-1}$, we get
$$\frac{9 \cdot 10^{k-1}}{k \ln 10 \cdot 9 \cdot 10^{k-1}}=\frac{1}{(\ln 10) k} \approx \frac{1}{2.3 k} .$$

## MATH223101COURSE NOTES ：

Simplifying, we obtain
$$4 k^{2}-4 n k+n^{2}-n-2<0 .$$
Solving for $k$, we get that the left-hand side is nonpositive between the two roots:
$$\frac{n}{2}-\frac{1}{2} \sqrt{n+2} \leq k \leq \frac{n}{2}+\frac{1}{2} \sqrt{n+2}$$
So the first integer $k$ for which this is nonpositive is
$$k=\left\lceil\frac{n}{2}-\frac{1}{2} \sqrt{n+2}\right]$$