(mathcal{Description})
Link.
积性函数 (f) 满足 (f(p^c)=poplus c~(pinmathbb P,cinmathbb N_+))，求 (sum_{i=1}^n f(i)mod(10^9+7))。
(mathcal{Solution})
首先，考虑 (f) 的素数点值：
[f(p)=egin{cases}
3,&p=2\
p-1,& ext{otherwise}
end{cases}
]
由 (p-1) 联想到 (varphi(p)=p-1)，可惜 (varphi(2)=1)。干脆一点，我们直接强行把 (varphi) 的偶数点值乘上 (3)，令
[g(n)=egin{cases}
varphi(n),&2
otmid n\
3varphi(n),& ext{otherwise}
end{cases}
]
显然它也是积性函数。
接着，求 (g) 的前缀和。其前缀和为 (varphi) 的前缀和加上两倍偶数点的 (varphi) 前缀和。记
[egin{aligned}
S(n)&=sum_{i=1}^nvarphi(2i)\
&=sum_{i=1}^n[2
otmid i]varphi(i)+2sum_{i=1}^n[2mid i]varphi(i)\
&=Sleft(frac{n}{2}
ight)+sum_{i=1}^nvarphi(i)
end{aligned}
]
杜教筛处理 (varphi) 的前缀，(S) 就能在可观（我不会算 qwq）的复杂度内预处理出来，继而也得到了 (g) 的 (mathcal O(sqrt n)) 个前缀和。
此外，我们还需要求 (h(i))，即求 (h(p^c)~(c>1))。考虑 (f(p^c)) 与它的关系：
[f(p^c)=sum_{i=0}^ch(p^i)g(p^{c-i})\
Rightarrow~~~~h(p^c)=f(p^c)-sum_{i=0}^{c-1}h(p^i)g(p^{c-i})
]
顺手把 (mathcal O(sqrt nlnlnsqrt n))（(n) 以内素数的倒数和的规模是 (mathcal O(lnln n))）个 (h(p^c)) 也预处理出来，最后 (mathcal O(sqrt n)) 搜索 Powerful Number 就能求出答案啦！
(mathcal{Code})
/* Clearink */
#include <cmath>
#include <cstdio>
#include <vector>
#include <unordered_map>
#define rep( i, l, r ) for ( int i = l, repEnd##i = r; i <= repEnd##i; ++i )
#define per( i, r, l ) for ( int i = r, repEnd##i = l; i >= repEnd##i; --i )
typedef long long LL;
const int MAXS = 1e7, MAXSN = 1e5, MOD = 1e9 + 7, INV2 = 500000004;
int pn, pr[MAXS + 5], phi[MAXS + 5], phis[MAXS + 5];
bool npr[MAXS + 5];
std::vector<int> gpr[MAXSN + 5];
inline int mul( const long long a, const int b ) { return a * b % MOD; }
inline int sub( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline void subeq( int& a, const int b ) { ( a -= b ) < 0 && ( a += MOD ); }
inline int add( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline void addeq( int& a, const int b ) { ( a += b ) >= MOD && ( a -= MOD ); }
inline void sieve() {
phi[1] = phis[1] = 1;
rep ( i, 2, MAXS ) {
if ( !npr[i] ) phi[pr[++pn] = i] = i - 1;
for ( int j = 1, t; j <= pn && ( t = i * pr[j] ) <= MAXS; ++j ) {
npr[t] = true;
if ( !( i % pr[j] ) ) { phi[t] = phi[i] * pr[j]; break; }
phi[t] = phi[i] * ( pr[j] - 1 );
}
phis[i] = add( phis[i - 1], phi[i] );
}
}
inline int phiSum( const LL n ) {
static std::unordered_map<LL, int> mem;
if ( n <= MAXS ) return phis[n];
if ( mem.count( n ) ) return mem[n];
int ret = mul( mul( n % MOD, ( n + 1 ) % MOD ), INV2 );
for ( LL l = 2, r; l <= n; l = r + 1 ) {
r = n / ( n / l );
subeq( ret, mul( ( r - l + 1 ) % MOD, phiSum( n / l ) ) );
}
return mem[n] = ret;
}
inline int ephiSum( const LL n ) {
if ( !n ) return 0;
return add( ephiSum( n >> 1 ), phiSum( n ) );
}
LL n;
int sn, sum[MAXSN * 2 + 5];
inline void initInvG() {
rep ( i, 1, pn ) {
if ( 1ll * pr[i] * pr[i] > n ) break;
std::vector<int>& curg( gpr[i] );
curg.push_back( 1 ), curg.push_back( 0 );
LL pwr = 1ll * pr[i] * pr[i];
for ( int j = 2; pwr <= n; ++j, pwr *= pr[i] ) {
int g = pr[i] ^ j;
LL pwc = pr[i];
for ( int k = j - 1; ~k; --k, pwc *= pr[i] ) {
subeq( g,
mul( ( pwc / pr[i] * ( pr[i] ^ 1 ) ) % MOD, curg[k] ) );
}
curg.push_back( g );
}
}
}
inline int powerSum( const int pid, LL x, const int g ) {
if ( !g ) return 0;
int ret = 0, p = pr[pid];
if ( pid == 1 || !( x % pr[pid - 1] ) ) {
addeq( ret, mul( g, x > sn ? sum[n / x] : sum[sn + x] ) );
}
if ( ( x *= p ) > n ) return ret;
if ( ( x *= p ) > n ) return ret;
addeq( ret, powerSum( pid + 1, x / ( 1ll * p * p ), g ) );
for ( int i = 2; x <= n; ++i, x *= p ) {
addeq( ret, powerSum( pid + 1, x, mul( g, gpr[pid][i] ) ) );
}
return ret;
}
int main() {
sieve();
scanf( "%lld", &n ), sn = sqrt( 1. * n );
rep ( i, 1, sn ) sum[i] = add( phiSum( i ), mul( 2, ephiSum( i >> 1 ) ) );
rep ( i, 1, sn ) {
sum[i + sn] = add( phiSum( n / i ), mul( 2, ephiSum( n / i >> 1 ) ) );
}
initInvG();
printf( "%d
", powerSum( 1, 1, 1 ) );
return 0;
}