神仙题。
发现不是 \(2\) 的倍数和不是 \(3\) 的倍数的数之间没有限制，可以分开考虑。
以 \(1\) 为例。
构造一个矩阵，左上角为 \(1\)，每个数右边是祂乘 \(3\),下面是祂乘 \(2\)，原问题为求在这个矩阵里选数，不能选相邻的数的方案数，用状压 \(dp\) 随便做，最后把所有左上角数字不同的矩阵的答案乘起来。要取模。
代码
#include<bits/stdc++.h>
using namespace std;
namespace IO{template<typename T>inline void read(T&x){x=0;char c=getchar();bool f=0;while(!isdigit(c)) c=='-'?f=1:0,c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();f?x=-x:0;}template<typename T>inline void write(T x){if(x==0){putchar('0');return ;}x<0?x=-x,putchar('-'):0;short st[50],top=0;while(x) st[++top]=x%10,x/=10;while(top) putchar(st[top--]+'0');}inline void read(char&c){c=getchar();while(isspace(c)) c=getchar();}inline void write(char c){putchar(c);}inline void read(string&s){s.clear();char c;read(c);while(!isspace(c)&&~c) s+=c,c=getchar();}inline void write(string s){for(int i=0,len=s.size();i<len;i++) putchar(s[i]);}template<typename T>inline void write(T*x){while(*x) putchar(*(x++));}template<typename T,typename...T2> inline void read(T&x,T2&...y){read(x),read(y...);}template<typename T,typename...T2> inline void write(const T x,const T2...y){write(x),putchar(' '),write(y...),sizeof...(y)==1?putchar('\n'):0;}
}using namespace IO;
template<signed mod>struct Modint{signed z;Modint(){z=0;}Modint(signed x){x%=mod;z=x<0?x+mod:x;}Modint(long long x){x%=mod;z=x<0?x+mod:x;}Modint(short x){x%=mod;z=x<0?x+mod:x;}Modint(char x){x%=mod;z=x<0?x+mod:x;}Modint(bool x){x%=mod;z=x<0?x+mod:x;}friend Modint operator+(Modint t,Modint t2){Modint ans;ans.z=(t.z+t2.z)%mod;return ans;}friend Modint operator*(Modint t,Modint t2){Modint ans;ans.z=1ll*t.z*t2.z%mod;return ans;}friend Modint operator-(Modint t,Modint t2){Modint ans;ans.z=(t.z-t2.z)%mod;return ans;}Modint operator<<(const signed t)const{Modint ans;ans.z=(z<<t)%mod;return ans;}Modint operator>>(const signed t)const{Modint ans;ans.z=(z>>t)%mod;return ans;}Modint&operator+=(const Modint t){z=(z+t.z)%mod;return *this;}Modint&operator*=(const Modint t){z=1ll*z*t.z%mod;return *this;}Modint&operator-=(const Modint t){z=(z-t.z)%mod;return *this;}Modint&operator<<=(const signed t){z=(z<<t)%mod;return *this;}Modint&operator>>=(const signed t){z=(z>>t)%mod;return *this;}friend Modint ksm(Modint a,signed b){Modint ans=1;while(b){if(b&1) ans=ans*a;a=a*a,b>>=1;}return ans;}friend void read(Modint&z){signed x=0;char c=getchar();bool f=0;while(!isdigit(c)) c=='-'?f=1:0,c=getchar();while(isdigit(c)) x=(x*10ll+c-'0')%mod,c=getchar();f?x=-x:0;z.z=x;}friend void write(Modint x){x.z<0?x.z+=mod:0;write(x.z);}
};
#define int long long
const int mod=1000000001,maxh=20,maxl=20,maxzt=400000,maxn=100010;
int n,jl=1,lt;
Modint<mod>f[maxh][maxzt],ans=1;
bool flag[maxn];
inline void js(int x){int s=x;f[0][0]=1;Modint<mod> ans=0;lt=0;for(int i=1;;i++){if(s>n) break;int ss=s,l=0;for(;ss<=n;l++) flag[ss]=1,ss*=2;for(int j=0;j<(1<<l);j++){f[i][j]=0;if(j&(j<<1)) continue;for(int k=0;k<(1<<lt);k++){if(j&k) continue;f[i][j]+=f[i-1][k];}}s*=3;if(s>n) for(int j=0;j<1<<l;j++) ans=ans+f[i][j];lt=l;}::ans=::ans*ans;
}
signed main(){read(n);for(int i=1;i<=n;i++){if(flag[i]) continue;js(i);}write(ans);return 0;
}