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...( ) ni + ni+1 = ni + k ( n1 + nm n1 ) ∑ m−2 ∏ n1�λ1�···�λm−2�0 i=1 ( λi−1 λi )( ni+1 + ni+2 ni+1 − λi ) , ) , 5 Proof of Theorem 1.4 We need the following theorem of Lucas (see, for example, [6]) and =-=[15]-=- for a recent application. Theorem 5.1 (Lucas’ theorem). Let p be a prime, and let a0, b0, . . .,am, bm ∈ {0, . . .,p− 1}. Then (∑m ) i m∏ ( ) i=0 aip ai ≡ (mod p). ∑m i=0 bipi i=0 Let r + s ≡ 1 (mod ...

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...that any integer n > 1 can be written as k + m with 2k +m prime, where k and m are positive integers. 1. Introduction Let p be a prime and let a be a positive integer. In 2011 the author and W. Zhang =-=[SZ]-=- showed that for each k = pa, pa + 1, . . . , 2pa − 1 the set {(nk) : n = 0, 1, 2, . . .} is dense in the ring of p-adic integers, i.e., it contains a complete system of residues modulo any power of p...