Let p be an odd prime. In this paper, we determine ∑_{k=0}^{p-1}(1/(m^{k}))((2k+d)/k)modp, ∑_{k=0}^{p-1}(k/(m^{k}))((2k+d)/k)modpfor d∈{0,1,...,p} and ∑_{k=0}^{p-1}((B(k,d))/(m^{k}))modp, ∑_{k=0}^{p-1}(k/(m^{k}))B(k,d)modpfor d∈{1,...,p}, where m∈? with p?mΔ. For example, for p≠5 ∑_{k=0}^{p-1}(-1)^{k}((2k+d)/k)≡(-1)^{d+1}F_{d-2(p/5)}(modp),and for p≠3 ∑_{k=0}^{p-1}kB(k,d)≡(-1)^{d}d((?d/2?-p-(-1)^{d})/3)(modp),where F_{n} is the nth Fibonacci number and (-) is the Jacobi symbol.
展开▼
