On sums of binomial coefficients modulo p^2

preprint2010arXivOpen access

Abstract

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum ∑k=0pa−1(hpa−1k)(2kk)/mk\sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}/m^k mod p^2, where h,m are p-adic integers with m\not=0 (mod p). For example, we show that if h\not=0 (mod p) and p^a>3 then sumk=0pa−1(hpa−1k)(2kk)(−h/2)k=(1−2hpa)(1+h((4h−2)p−1/hp−1−1))(modp2), sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}(-h/2)^k =(\frac{1-2h}{p^a})(1+h((4h-2)^{p-1}/h^{p-1}-1)) (mod p^2), where (-) denotes the Jacobi symbol. Here is another remarkable congruence: If p>3 then ∑k=0pa−1(pa−1k)(2kk)(−1)k=3p−1(pa3)(modp2).\sum_{k=0}^{p^a-1}\binom{p^a-1}{k}\binom{2k}{k}(-1)^k =3^{p-1}(\frac{p^a}3) (mod p^2).

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