一个数字问题

N. Y. Wilson
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引用次数: 2

摘要

幂和的一个定理我们将这些结果总结如下。定理。式(2)的解如下:如果p = q, f(x)是ar b i- triv, g(x) = f(x)。如果p ^ q,在p = 2和q = 1时出现唯一的单元解,此时f(x)和g(x)由式(12)定义,其中a是任意实常数,这种情况的非单元解可以用式(13)找到。作为这些结果的一个例子,假设p = 3和q = 4。(14)和(17)我们有13 (n, 4(好几次3 x + 1) J (n = 1, 2, 3,•••)x = x = l l 1 \“\ (4 x 3-6x 2 + 4 x - 1)有无限多的数字财产:如果个位数的一个正整数,M,是6,这是来自它的位置,把剩下的数字的左边,然后一个新的整数,n,将形成,n = 6米。可能的最小M是58位数字(1016949•••677966)。1-4x-x 2 n= 0, x = 0,1,我们得到1,01016949 *••677966,其中周期数(在第一个零后面)是m *
本文章由计算机程序翻译,如有差异,请以英文原文为准。
A Number Problem
A THEOREM ON POWER SUMS 161 We summarize these results in the following. Theorem. The solutions of (2) are as follows. If p = q, f(x) is a r b i-trary and g(x) = f(x). If p ^ q, the only monic solutions occur when p = 2 and q = 1, in which case f(x) and g(x) are defined by (12), where a is an arbitrary real constant Non-monic solutions for that case can be found using (13). As an example of these results suppose that p = 3 and q = 4. By (14) and (17) we have 13 (n , 4 (3x 2-3x + 1) J , (n = 1, 2, 3, • • •) x=l 1 \ x=l ' \ (4X 3-6x 2 + 4x-1) There are infinite many numbers with the property: if units digit of a positive integer, M, is 6 and this is taken from its place and put on the left of the remaining digits of M, then a new integer, N, will be formed, such that N = 6M. The smallest M for which this is possible is a number with 58 digits (1016949 • • • 677966). 1-4x-x 2 n=o with x = 0,1 we have 1,01016949 * • • 677966, where the period number (behind the first zero) is M.*
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