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arXiv 1202.6589math.NTmath.CO
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关于只取素值的函数

On functions taking only prime values

Zhi-Wei Sun

摘要:

对于n= 1,2,3,.将S(n)定义为最小整数m>1,使得对于k=1,.,n是成对不同的;我们证明了S(n)是大于2n-2的最小素数,因此函数S(n)的值集恰好是所有素数的集合.对于每个n= 4,5,.证明了最小素数p> 3 n,其中3| p-1恰好是最小的正整数m,使得18 k(3 k-1)(k=1,.,n)是成对不同的模m。对于d= 4,6,12和n= 3,4,.,我们证明了最小素数p>2n-2,其中p=-1(mod d)是最小整数m,使得那些(2k-1)^d,k=1,.,n是两两不同的模m。我们还提出了几个具有挑战性的命题素数。例如,我们发现了一个令人惊讶的递归素数,即,对于每一个n= 10,11,.第(n+1)个素数p_{n+1}恰好是最小的正整数m,使得2s_k^2(k=1,.,n)是两两不同的模m,其中s_k = sum_{j=1} ^k(-1)^{k-j}p_j.我们还猜想,对于任何正整数m,存在连续的素数p_k,.,p_n(k<n)不超过2 m +2.2*sqrt(m),使得m = p_n- p_{n-1} +.+ (-1)^{n-k}p_k.

英文摘要:

For n=1,2,3,... define S(n) as the smallest integer m>1 such that those 2k(k-1) mod m for k=1,...,n are pairwise distinct; we show that S(n) is the least prime greater than 2n-2 and hence the value set of the function S(n) is exactly the set of all prime numbers. For every n=4,5,... we prove that the least prime p>3n with 3|p-1 is just the least positive integer m such that 18k(3k-1) (k=1,...,n) are pairwise distinct modulo m. For d=4,6,12 and n=3,4,...., we prove that the least prime p>2n-2 with p=-1 (mod d) is the smallest integer m such that those (2k-1)^d for k=1,...,n are pairwise distinct modulo m. We also pose several challenging conjectures on primes. For example, we find a surprising recurrence for primes, namely, for every n=10,11,... the (n+1)-th prime p_{n+1} is just the least positive integer m such that 2s_k^2 (k=1,...,n) are pairwise distinct modulo m where s_k = sum_{j=1}^k(-1)^{k-j}p_j. We also conjecture that for any positive integer m there are consecutive primes p_k,...,p_n (k<n) not exceeding 2m+2.2*sqrt(m) such that m = p_n-p_{n-1}+...+(-1)^{n-k}p_k.

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