16 根据$n$%3等于 0,1,2列三个方程然后计算出$a,b,c$的值,$a=1,b=frac{w-1}{3},c=-frac{w+2}{3}$
17 $sum_{0leq k<m}[x+frac{k}{m}]$
$=sum_{j,k}[0leq k<m][1leq j leq x+frac{k}{m}]$
$=sum_{j,k}[0leq k<m][1leq j leq left lceil x ight ceil]-sum_{k}[0leq k <m(left lceil x ight ceil-x)]$
$=mleft lceil x ight ceil-left lceil m(left lceil x ight ceil-x) ight ceil$
$=left lfloor mx ight floor$
18 $S=sum_{0leq j <left lceil nalpha ight ceil}sum_{kgeq n}[jalpha^{-1}leq k <(j+v)alpha^{-1}]$
(1)如果$jleq nalpha -1leq nalpha -v ightarrow (j+v)alpha^{-1}leq n$,那么此时$S=0$
(2)所以现在只需要考虑$j=left lfloor nalpha ight floor$。由于$jalpha^{-1}=frac{left lfloor nalpha ight floor}{alpha}<n$所以$S=sum [nleq k<(left lfloor nalpha ight floor+v)alpha^{-1}]=left lceil (left lfloor nalpha ight floor+v)alpha^{-1} ight ceil-n=left lceil n-Delta +valpha^{-1} ight ceil-nleq left lceil valpha^{-1} ight ceil$,其中$Delta >0$
19 首先若$b$不是整数,那么等式在$x=b$时一定不成立。若$b$为整数,则$log(b)$取整数时必定有$x$为整数。那么根据公式$3.10$,恒成立。
20 $xsum_{k}k[left lceil frac{alpha}{x} ight ceilleq k leq left lfloor frac{eta}{x} ight floor]=frac{x(p+q)(q-p+1)}{2}$
其中$p=left lceil frac{alpha}{x} ight ceil,q=left lfloor frac{eta}{x} ight floor$
21 如果$10^{n}leq 2^M <10^{n+1}$,那么有$n+1$个$m$满足要求。假设$n=4,M=15$,那么满足要求的有$2^{0}=1,2^{4}=16,2^{7}=128,2^{10}=1024,2^{14}=16384$.所以答案为$1+left lfloor Mlog_{10}^{2} ight floor$
22 假设$n=2^{t-1}q$,其中$q$为奇数。那么当$k=t$时,$left lfloor frac{n}{2^{t}}+frac{1}{2} ight floor=frac{q+1}{2},left lfloor frac{n-1}{2^{t}}+frac{1}{2} ight floor=frac{q-1}{2}$.
如果$k eq t$,$left lfloor frac{n}{2^{k}}+frac{1}{2} ight floor=left lfloor frac{n-1}{2^{k}}+frac{1}{2} ight floor$
所以$S_{n}=S_{n-1}+1,T_{n}=T_{n-1}+2^{k}q=T_{n-1}+2n$,所以$S_{n}=n,T_{n}=n(n+1)$
23 假设第$n$个数字是$t$,那么$[1,t-1]$一共有$frac{t(t-1)}{2}$个数字,所以$frac{t(t-1)}{2}<nleq frac{t(t+1)}{2}leftrightarrow t^{2}-t<2nleq t^{2}+t$,进而得到$ t^{2}-t+frac{1}{4}<2n< t^{2}+t+frac{1}{4}leftrightarrow t-frac{1}{2}<sqrt{2n}<t+frac{1}{2}leftrightarrow sqrt{2n}-frac{1}{2}<t<sqrt{2n}+frac{1}{2}Rightarrow t=left lfloor sqrt{2n}+frac{1}{2} ight floor$
24 $N(alpha,n)=left lceil frac{n+1}{alpha} ight ceil-1$
$N(frac{alpha}{alpha + 1},n)=left lceil frac{(n+1)(alpha + 1)}{alpha} ight ceil-1=(n+1)+left lceil frac{n+1}{alpha} ight ceil-1=N(alpha,n)+n+1$
所以数字$m$,其在$Spec(frac{alpha}{alpha +1})$出现的次数比在$Spec(alpha)$出现的次数多1.
25 如果存在$m$满足$K_{m}leq m$,那么当$n=2m+1$时,$K_{n+1}leq n<n+1$
假设这样的$m$存在。那么如果$m$是偶数,同样假设$K_{m/2}leq frac{m}{2}$,否则,需要存在一个数$t=frac{m-1}{2}$,满足$K_{t}leq t$.依次这样下去,可以得到$K_{0}leq 0$
得到矛盾,所以一开始的假设错误,即不存在$m$满足$K_{m}leq m$
所以$K_{n}>n$
26 前半部分很明显成立:$(frac{q}{q-1})^{n}leq D_{n}^{q}$
对于后半部分,由于$(q-1)((frac{q}{q-1})^{n+1}-1)=frac{q^{n+1}}{(q-1)^{n}}-(q-1)<frac{q^{n+1}}{(q-1)^{n}}=q(frac{q}{q-1})^{n}$
所以现在证明$D_{n}^{q}leq (q-1)((frac{q}{q-1})^{n+1}-1)$
当$n=0,1$时成立,假设对于$[0,n-1]$均成立
那么$D_{n}^{q}=left lceil frac{q}{q-1}D_{n-1}^{q} ight ceilleq left lceil frac{q}{q-1}(q-1)((frac{q}{q-1})^{n}-1) ight ceil$
$=left lceil frac{q^{n+1}}{(q-1)^{n}} ight ceil-q<frac{q^{n+1}}{(q-1)^{n}}+1-q$
$=(q-1)((frac{q}{q-1})^{n+1}-1)$
27 首先若第$n$项为偶数,即$D_{n}^{3}=2^{t}q$,$q$为奇数,那么$D_{n+t}^{3}=3^{t}q$为奇数;
若第$n$项为奇数,设为$D_{n}^{3}=2^mq-1$, $q$为奇数。那么$D_{n+1}^{3}=D_{n}^{3}+left lceil frac{D_{n}^{3}}{2} ight ceil=2^{m}q-1+2^{m-1}q=2^{m-1}*3q-1$,所以$D_{n+m}^{3}=3^{m}q-1$为偶数。
28 $a_{n}=m^{2} ightarrow a_{n+2k+1}=(m+k)^{2}+m-k,a_{n+2k+2}=(m+k)^{2}+2m,0leq k leq m$
$ ightarrow a_{n+2m+1}=(2m)^{2}$
29 $s(alpha,n,v)=-s(alpha^{'},left lceil nalpha ight ceil,v^{'})-S+varepsilon +left { (0) or (1) ight }$
$=-s(alpha^{'},left lfloor nalpha ight floor,v^{'})-S+varepsilon +left { (0) or (1) ight }-left { (v^{'}) or(v^{'}-1) ight }$
另外$|-S+varepsilon +left { (0) or (1) ight }-left { (v^{'}) or(v^{'}-1) ight }|leq |-S+varepsilon -v^{'}|+2leq alpha^{-1}+2$
应用公式$a=b+c ightarrow |a|leq |b|+|c|$
可以得到$D(alpha,n)leq D(alpha^{'},left lfloor nalpha ight floor)+alpha^{-1}+2$
交换$s(alpha,n,v),s(alpha^{'},left lfloor nalpha ight floor,v^{'})$的顺序可以得到$D(alpha^{'},left lfloor nalpha ight floor)leq D(alpha,n)+alpha^{-1}+2$
30 可以用数学归纳法证明:$X_{n}=alpha^{2^{n}}+frac{1}{alpha^{2^{n}}}$.而$frac{1}{alpha^{2^{n}}}<1$,同时$X_{n}$是整数,所以$X_{n}=left lceil alpha^{2^{n}} ight ceil$