单位根反演

定理

([nmid a] = dfrac1nsum_{k=0}^{n-1}omega_n^{ak})

证明

使用等比数列求和

a ≠ 0 (mod n)

公比不为 1

原式 = (dfrac1n imesdfrac{omega_n^{an}-1}{omega_n^a-1} = dfrac1n imesdfrac{1-1}{omega_n^a-1} = 0)

a = 0 (mod n)

公比为 1

原式 = (dfrac1n imes n imes omega_n^0 = 1)

应用

([aequiv bmod n] = [a-bequiv 0mod n] = [nmid(a-b)] = dfrac1nsum_{k=0}^nomega_n^{(a-b)k})

---

LJJ学二项式定理

[egin{align} &sum_{i=0}^ninom nicdot s^icdot a_{i\% 4}\ &= sum_{i=0}^ninom nicdot s^isum_{j=0}^3a_j[i=jmod4]\ &= sum_{i=0}^ninom nicdot s^isum_{j=0}^3a_jfrac14sum_{k=0}^3omega_4^{k(i-j)}\ &= frac14sum_{j=0}^3a_jsum_{i=0}^ninom nis^isum_{k=0}^3omega_4^{ki}omega_4^{-kj}\ &= frac14sum_{j=0}^3a_jsum_{k=0}^3omega_4^{-kj}sum_{i=0}^ninom nis^iomega_4^{ki}\ &= frac14sum_{j=0}^3a_jsum_{k=0}^3omega_4^{-kj}(1+somega_4^k)^n end{align} ]

提交记录

---

小猪佩奇学数学

(leftlfloordfrac ik ight floor = dfrac {i - imod k}k)

原式等于：

[frac1ksum_{i=0}^ninom ni p^i (i-imod k)\ frac1kleft[sum_{i=0}^ninom ni p^ii-sum_{i=0}^ninom ni p^i(imod k) ight] ]

分开考虑：

[ sum_{i=0}^ninom nip^ii\ npsum_{i=0}^ninom {n-1}{i-1}p^{i-1}\ npsum_{i=0}^{n-1}inom {n-1}ip^i\ np(p+1)^{n-1} ]

[ sum_{i=0}^ninom nip^i(imod k)\ sum_{i=0}^ninom nip^isum_{j=0}^{k-1}j[imod k =j]\ frac1ksum_{i=0}^ninom nip^isum_{j=0}^{k-1}jsum_{t=0}^{k-1}omega_k^{t(i-j)}\ frac1ksum_{j=0}^{k-1}jsum_{t=0}^{k-1}omega_k^{-tj}sum_{i=0}^ninom nip^iomega_k^{ti}\ frac1ksum_{j=0}^{k-1}jsum_{t=0}^{k-1}omega_k^{-tj}(1+pomega_k^t)^n\ frac1ksum_{t=0}^{k-1}(1+pomega_k^t)^nsum_{j=0}^{k-1}j(omega_k^{-t})^j ]

总复杂度是 O(k log n)

附：

[egin{align} S(n,k) &= sum_{i=0}^{n-1}ik^i\ kS(n,k)-S(n,k) &= sum_{i=0}^{n-1}ik^{i+1} - sum_{i=0}^{n-1}ik^i\ &= sum_{i=1}^n(i-1)k^i - sum_{i=1}^{n-1}ik^i\ &= -sum_{i=1}^{n-1}k^i + (n-1)k^n\ &= frac{1-k^n}{k-1} + 1 + (n-1)k^n end{align} ]

于是 (S(n,k) = dfrac{dfrac{1-k^n}{k-1} + 1 + (n-1)k^n}{k-1})

由于 k 总是 n 次单位根， 于是 (S(n,k) = dfrac n{k-1})

若 k = 1， (S(n,1) = sum_{i=0}^{n-1} i = dfrac{n(n-1)}2)

提交记录