求数列的通项公式

yigo

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：
\(f(n)=int( \sqrt[3]{6n}+1-\frac{5}{3 \sqrt[3]{6n}})\)

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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)。
\(n \to n-\frac{f^2(n)-f(n)}{2} \to... \to1\)

王守恩

{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}

1. Table[a, {a, 6}, {b, a (a - 1)/2 + 1}] // Flatten

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yigo

\(int( \sqrt[3]{6n}+1-\frac{1}{\sqrt[3]{n}})\)
这个不是很精确

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这个验证了2000项没问题。
\(int( \sqrt[3]{6n}+1-\frac{5}{3 \sqrt[3]{6n}})\)

王守恩

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. 1, nn=9; First/@Select[Tuples[Range[nn], 3], GreaterEqual@@#&]

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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. 2, Table[a, {a, 6}, {b, a(a+1)/2}] // Flatten

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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: \(\bigg\lfloor\sqrt[3]{\ 6 n\ }+ 1-\frac{ 5}{\ 3\ \sqrt[3]{\ 6 n\ }\ }\bigg\rfloor=\bigg\lceil\sqrt[3]{\ 6 n\ }-\frac{ 5}{\ 3\ \sqrt[3]{\ 6n\ }\ }\bigg\rceil\)

yigo

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, ...

======================
\(f(n)=int( \sqrt[3]{6n}+\frac{1}{3 \sqrt[3]{6n+1}})\)
只验证了300来项，不知道后面还会不会出偏差。

王守恩

yigo 发表于 2024-7-10 09:19
======================
\(f(n)=int( \sqrt[3]{6n}+\frac{1}{3 \sqrt[3]{6n+1}})\)
只验证了300来项，不知 ...

验证到4500000(a=300)项,  没出偏差。

1. Table[Ceiling[Power[6 n, (3)^-1] - 5/(3 Power[6 n, (3)^-1])], {a, 8}, {n, (a^3 + 5 a)/6, ((a + 1)^3 + 5*(a + 1))/6 + 1}]

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{{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}}

1. Table[n, {n, a + 1, a + 1}, {a, 8}, {b, n (n - 1)/2 + 1}]

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{{{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！

王守恩

谢谢 yigo ！"解三次方程“，思路很牛很有创造性,  试试简单的。

1. Table[Floor[Power[6 n, (3)^-1] - 1/Power[6 n, (3)^-1]], {n, 127}]

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{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。

yigo

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。

yigo

要证明\(f(n)=\lfloor \sqrt[3]{6n}+\frac{1}{3 \sqrt[3]{6n+1}}\rfloor\)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：

\(\displaystyle \sqrt[3] {(k-1)k(k+1)+6}+\frac{1}{3\sqrt[3] {(k-1)k(k+1)+7}}\geqslant k\)

以及:

\(\displaystyle \sqrt[3] {k(k+1)(k+2)}+\frac{1}{3\sqrt[3] {k(k+1)(k+2)+1}} < k+1\)

王守恩

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,

\(a(n)=\bigg\lfloor\sqrt[3]{6n}+\frac{n+1}{(3n+\pi)\sqrt[3]{6n}}\bigg\rfloor\)