Some curious congruences modulo primes

Abstract

Let n be a positive odd integer and let p>n+1 be a prime. We mainly derive the following congruence: $$\sum _{0<i_1<...<i_n<p}(i_1/3)(-1{)}^{i_1}/(i_1...i_n)=0(modp).$$

6 pages

Cited by in corpus (2)

• Congruences involving alternating multiple harmonic sum
• Congruences of multiple sums involving invariant sequences under binomial transform