Let p 〉 3 be a prime. A p-adic congruence is called a super congruence if it happens to hold modulo some higher power of p. The topic of super congruences is related to many fields including Gauss and Jacobi sums and...Let p 〉 3 be a prime. A p-adic congruence is called a super congruence if it happens to hold modulo some higher power of p. The topic of super congruences is related to many fields including Gauss and Jacobi sums and hypergeometric series. We prove that ∑k=0^p-1（k^2k/2k）≡（-1）^（p-1）/2-p^2Ep-3（modp^3） ∑k=1^（p-1）/2（k^2k）/k≡（-1）^（p＋1）/2 8/3pEp-3（mod p^2）,∑k=0^（p-1）/2（k^2k）^2/16k≡（-1）^（p-1）/2＋p^2Ep-3（mod p^3）,where E0, E1, E2,... are Euler numbers. Our new approach is of combinatorial nature. We also formulate many conjectures concerning super congruences and relate most of them to Euler numbers or Bernoulli numbers. Motivated by our investigation of super congruences, we also raise a conjecture on 7 new series for π2, π-2 and the constant K ：= ∑k=1^∞（k/3）/k^2 （with （-） the Jacobi symbol）, two of which are ∑k=1^∞（10k-3）8k/k2（k^2k）^2（k^3k）=π^2/2and ∑k=1^∞（15k-4）（-27）^k-1/k^3（k^2k）^2（k^3k）=K.展开更多
基金supported by the National Natural Science Foundation of China(GrantNo.10871087)the Overseas Cooperation Fund of China(Grant No.10928101)
文摘Let p 〉 3 be a prime. A p-adic congruence is called a super congruence if it happens to hold modulo some higher power of p. The topic of super congruences is related to many fields including Gauss and Jacobi sums and hypergeometric series. We prove that ∑k=0^p-1（k^2k/2k）≡（-1）^（p-1）/2-p^2Ep-3（modp^3） ∑k=1^（p-1）/2（k^2k）/k≡（-1）^（p＋1）/2 8/3pEp-3（mod p^2）,∑k=0^（p-1）/2（k^2k）^2/16k≡（-1）^（p-1）/2＋p^2Ep-3（mod p^3）,where E0, E1, E2,... are Euler numbers. Our new approach is of combinatorial nature. We also formulate many conjectures concerning super congruences and relate most of them to Euler numbers or Bernoulli numbers. Motivated by our investigation of super congruences, we also raise a conjecture on 7 new series for π2, π-2 and the constant K ：= ∑k=1^∞（k/3）/k^2 （with （-） the Jacobi symbol）, two of which are ∑k=1^∞（10k-3）8k/k2（k^2k）^2（k^3k）=π^2/2and ∑k=1^∞（15k-4）（-27）^k-1/k^3（k^2k）^2（k^3k）=K.
