设p>3为素数.对任何p-adic整数a,我们决定出p−1 X k=0 −k a a−1 k Hk,p−1 X k=0 −k a a−1 k Hk(2),p−1 X k=0 −k a a−1 k H(2)k 2k+1模p 2,其中Hk=P 0<j<k 1/j且Hk(2)=P 0<j>k 1/j2.特别地,我们证明了p−1 X k=0 −k a a−1...设p>3为素数.对任何p-adic整数a,我们决定出p−1 X k=0 −k a a−1 k Hk,p−1 X k=0 −k a a−1 k Hk(2),p−1 X k=0 −k a a−1 k H(2)k 2k+1模p 2,其中Hk=P 0<j<k 1/j且Hk(2)=P 0<j>k 1/j2.特别地,我们证明了p−1 X k=0 −k a a−1 k Hk≡(−1)p 2(Bp−1(a)−Bp−1)(mod p),p−1 X k=0 −k a a−1 k Hk(2)≡−Ep−3(a)(mod p),(2a−1)p−1 X k=0 −k a a−1 k H(2)k 2k+1≡Bp−2(a)(mod p)其中p表示满足a≤r(mod p)的最小非负整数r,Bn(x)与En(x)分别表示次数为n的伯努利多项式与欧拉多项式.展开更多
基金Supported by the National Natural Science Foundation of China(11971222)the initial version was posted to arXiv in 2014 with the ID arXiv:1407.8465.
文摘设p>3为素数.对任何p-adic整数a,我们决定出p−1 X k=0 −k a a−1 k Hk,p−1 X k=0 −k a a−1 k Hk(2),p−1 X k=0 −k a a−1 k H(2)k 2k+1模p 2,其中Hk=P 0<j<k 1/j且Hk(2)=P 0<j>k 1/j2.特别地,我们证明了p−1 X k=0 −k a a−1 k Hk≡(−1)p 2(Bp−1(a)−Bp−1)(mod p),p−1 X k=0 −k a a−1 k Hk(2)≡−Ep−3(a)(mod p),(2a−1)p−1 X k=0 −k a a−1 k H(2)k 2k+1≡Bp−2(a)(mod p)其中p表示满足a≤r(mod p)的最小非负整数r,Bn(x)与En(x)分别表示次数为n的伯努利多项式与欧拉多项式.
