求数列的通项公式
本帖最后由 yigo 于 2024-7-9 16:09 编辑数列1,2,2,3,3,3,3,4,4,4,4,4,4,4,5...
数n有n*(n-1)/2+1个，
求这个数列的通项公式。
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通项公式为：
\(f(n)=int( \sqrt{6n}+1-\frac{5}{3 \sqrt{6n}})\)
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证明：n按下列方式迭代到1，迭代次数不超过f(n)。
\(n \to n-\frac{f^2(n)-f(n)}{2} \to... \to1\) {1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6}
Table // Flatten 本帖最后由 yigo 于 2024-7-9 15:30 编辑
\(int( \sqrt{6n}+1-\frac{1}{\sqrt{n}})\)
这个不是很精确
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这个验证了2000项没问题。
\(int( \sqrt{6n}+1-\frac{5}{3 \sqrt{6n}})\) 本帖最后由 王守恩 于 2024-7-10 09:24 编辑
先看一串数——OEIS——A360010——Gus Wiseman, Feb 11 2023
1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8,
1, nn=9; First/@Select, 3], GreaterEqual@@#&]
1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8,
2, Table // Flatten
这2个都谈不上是通项公式。楼主可以有一个OEIS——A360010通项公式吗？！哪就厉害了！！！
又: \(\bigg\lfloor\sqrt{\ 6 n\ }+ 1-\frac{ 5}{\ 3\ \sqrt{\ 6 n\ }\ }\bigg\rfloor=\bigg\lceil\sqrt{\ 6 n\ }-\frac{ 5}{\ 3\ \sqrt{\ 6n\ }\ }\bigg\rceil\)
王守恩 发表于 2024-7-10 08:58
先看一串数——OEIS——A360010——Gus Wiseman, Feb 11 2023
1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, ...
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\(f(n)=int( \sqrt{6n}+\frac{1}{3 \sqrt{6n+1}})\)
只验证了300来项，不知道后面还会不会出偏差。 yigo 发表于 2024-7-10 09:19
======================
\(f(n)=int( \sqrt{6n}+\frac{1}{3 \sqrt{6n+1}})\)
只验证了300来项，不知 ...
验证到4500000(a=300)项,没出偏差。
Table - 5/(3 Power)], {a, 8}, {n, (a^3 + 5 a)/6, ((a + 1)^3 + 5*(a + 1))/6 + 1}]
{{1, 2, 2, 3}, {2, 3, 3, 3, 3, 4}, {3, 4, 4, 4, 4, 4, 4, 4, 5}, {4, 5,5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6}, {5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7},{6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8},
{7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9}, {8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10}}
Table
{{{2, 2}, {3, 3, 3, 3}, {4, 4, 4, 4, 4, 4, 4}, {5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5}, {6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6}, {7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7},
{8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8}, {9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9}}}
"解三次方程“思想没悟透，能再来个例子。或给4#——OEIS——A360010——配个通项公式？！谢谢！ 谢谢 yigo ！"解三次方程“，思路很牛很有创造性,试试简单的。
Table - 1/Power], {n, 127}]
{1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9}
a(1)=1,
a(2)=3,
a(3)=7,
a(4)=13,
a(5)=24,
a(6)=40,
a(7)=61,
a(8)=90,
a(9)=127,
得到一串数: 1, 3, 7, 13, 24, 40, 61, 90, 127, 172, 228, 295, 373, 465, 571, 691, 828, 982, 1153, 1344, 1555, 1786, 2040, 2317, 2617, 2943, 3295, 3673, 4080, 4516,...
{1, 3, 7, 13, 24, 40, 61, 90, 127, 172, 228, 295, 373, 465, 571, 691, 828, 982, 1153, 1344, 1555, 1786, 2040, 2317, 2617, 2943, 3295, 3673, 4080, 4516}
2, 4,6, 11, 16,21, 29, 37,45, 56, 67, 78, 92,106,120, 137, 154, 171,191, 211, 231,254,277, 300,326, 352, 378, 407, 436,
2, 2,5,5, 5, 8, 8, 8, 11, 11, 11, 14, 14, 14, 17, 17, 17, 20, 20, 20, 23, 23, 23, 26, 26, 26, 29, 29,
OEIS还没有这串数。 拿你给的那个数列k有k*(k+1)/2项来说吧。
假设f(n)=k，那么从第1项到等于k的最后1项，
一共有\( \displaystyle \sum_{i=1}^k \frac{i(i+1)}{2}=\frac{k(k+1)(k+2)}{6} \)项。
从第1项到等于(k-1)的最后1项，
一共有\( \displaystyle \sum_{i=1}^{k-1} \frac{i(i+1)}{2}=\frac{(k-1)k(k+1)}{6} \)项。
f(n)=k，所以\( \displaystyle \frac{(k-1)k(k+1)}{6}\lt n\leqslant \frac{k(k+1)(k+2)}{6} \)
解这个不等式就可以了，然后忽略一些小量近似，不过这通项公式要证明感觉还是挺麻烦，只是验证了很多项感觉没问题，通项公式也不唯一。 本帖最后由 yigo 于 2024-7-11 15:22 编辑
要证明\(f(n)=\lfloor \sqrt{6n}+\frac{1}{3 \sqrt{6n+1}}\rfloor\)是通项公式，即要证明：
当\(\displaystyle \frac{(k-1)k(k+1)}{6}\lt n\leqslant \frac{k(k+1)(k+2)}{6}\)时，\(f(n)=k\)。
易知f(n)是增函数，所以只要证明：
\(\displaystyle f \left ( \frac{(k-1)k(k+1)}{6}+1\right) =k\)和\(\displaystyle f \left ( \frac{(k(k+1)(k+2)}{6}\right) =k\)，即：
\(\displaystyle \sqrt {(k-1)k(k+1)+6}+\frac{1}{3\sqrt {(k-1)k(k+1)+7}}\geqslant k\)
以及:
\(\displaystyle \sqrt {k(k+1)(k+2)}+\frac{1}{3\sqrt {k(k+1)(k+2)+1}} < k+1\) 谢谢 yigo！接4#。给出——OEIS——A360010——通项公式。请各位查验。 谢谢 yigo！
1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8,
\(a(n)=\bigg\lfloor\sqrt{6n}+\frac{n+1}{(3n+\pi)\sqrt{6n}}\bigg\rfloor\)
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