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本帖最后由 王守恩 于 2021-9-17 18:28 编辑
谢谢 OToday! 把奇数(2k+1)规律一并解决了。
\(使得\ n+a\ 能够整除\ 1^{2k+1}+2^{2k+1}+3^{2k+1}+4^{2k+1}+…+n^{2k+1}\ 的最大正整数\ n\ 是什么?\)
\(\displaystyle S(n)=\big(1+\big[\left\{(2n+1)/8\right\}\big]\big)\sum_{j=1}^{\ n-1}\ j^{2k+1}-n\) n=3, 4, 5, ... k=1, 2, 3, 4, ...
\(\big[\ \ \big]\) 表示四舍五入,\(\left\{\ \ \right\}\)表示取小数部分
{15, 32, 95, 444, 875,776,1287,4040,6039, 4344, 6071,16548,22035,14384,18479,46800,58463,...
{63, 272,1295,8844, 24395,29000, 61767,241640, 441639,381864, 630695, 2003988, 3079635,...
{255, 2312, 18695, 193644, 753515, 1200296, 3297447, 16160840, 36160839, 37567584,...
{1023, 20192, 282335, 4470924, 24626315, 52666760, 186884487, 1148609960, 3148609959,...
{4095, 179192, 4373495, 106403244, 831997355, 2393325416, 10983260007, 84728639240,...
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