数字串的通项公式

王守恩

  有这样一串数：A000045               
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765,
10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269,
2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169,...........

\(a(01)=001=\frac{2}{2^{1}}\)
\(a(02)=002=\frac{2}{2^{2}}+\frac{3}{2^{1}}\)
\(a(03)=003=\frac{2}{2^{3}}+\frac{3}{2^{2}}+\frac{4}{2^{1}}\)
\(a(04)=005=\frac{2}{2^{4}}+\frac{3}{2^{3}}+\frac{4}{2^{2}}+\frac{7}{2^{1}}\)
\(a(05)=008=\frac{2}{2^{5}}+\frac{3}{2^{4}}+\frac{4}{2^{3}}+\frac{7}{2^{2}}+\frac{11}{2^{1}}\)
\(a(06)=013=\frac{2}{2^{6}}+\frac{3}{2^{5}}+\frac{4}{2^{4}}+\frac{7}{2^{3}}+\frac{11}{2^{2}}+\frac{18}{2^{1}}\)
\(a(07)=021=\frac{2}{2^{7}}+\frac{3}{2^{6}}+\frac{4}{2^{5}}+\frac{7}{2^{4}}+\frac{11}{2^{3}}+\frac{18}{2^{2}}+\frac{29}{2^{1}}\)
\(a(08)=034=\frac{2}{2^{8}}+\frac{3}{2^{7}}+\frac{4}{2^{6}}+\frac{7}{2^{5}}+\frac{11}{2^{4}}+\frac{18}{2^{3}}+\frac{29}{2^{2}}+\frac{47}{2^{1}}\)
\(a(09)=055=\frac{2}{2^{9}}+\frac{3}{2^{8}}+\frac{4}{2^{7}}+\frac{7}{2^{6}}+\frac{11}{2^{5}}+\frac{18}{2^{4}}+\frac{29}{2^{3}}+\frac{47}{2^{2}}+\frac{76}{2^{1}}\)
\(a(10)=089=\frac{2}{2^{10}}+\frac{3}{2^{9}}+\frac{4}{2^{8}}+\frac{7}{2^{7}}+\frac{11}{2^{6}}+\frac{18}{2^{5}}+\frac{29}{2^{4}}+\frac{47}{2^{3}}+\frac{76}{2^{2}}+\frac{123}{2^{1}}\)
\(a(11)=144=\frac{2}{2^{11}}+\frac{3}{2^{10}}+\frac{4}{2^{9}}+\frac{7}{2^{8}}+\frac{11}{2^{7}}+\frac{18}{2^{6}}+\frac{29}{2^{5}}+\frac{47}{2^{4}}+\frac{76}{2^{3}}+\frac{123}{2^{2}}+\frac{199}{2^{1}}\)
\(a(12)=233=\frac{2}{2^{12}}+\frac{3}{2^{11}}+\frac{4}{2^{10}}+\frac{7}{2^{9}}+\frac{11}{2^{8}}+\frac{18}{2^{7}}+\frac{29}{2^{6}}+\frac{47}{2^{5}}+\frac{76}{2^{4}}+\frac{123}{2^{3}}+\frac{199}{2^{2}}+\frac{322}{2^{1}}\)
\(a(13)=377=\frac{2}{2^{13}}+\frac{3}{2^{12}}+\frac{4}{2^{11}}+\frac{7}{2^{10}}+\frac{11}{2^{9}}+\frac{18}{2^{8}}+\frac{29}{2^{7}}+\frac{47}{2^{6}}+\frac{76}{2^{5}}+\frac{123}{2^{4}}+\frac{199}{2^{3}}+\frac{322}{2^{2}}+\frac{521}{2^{1}}\)
\(a(14)=610=\frac{2}{2^{14}}+\frac{3}{2^{13}}+\frac{4}{2^{12}}+\frac{7}{2^{11}}+\frac{11}{2^{10}}+\frac{18}{2^{9}}+\frac{29}{2^{8}}+\frac{47}{2^{7}}+\frac{76}{2^{6}}+\frac{123}{2^{5}}+\frac{199}{2^{4}}+\frac{322}{2^{3}}+\frac{521}{2^{1}}+\frac{843}{2^{1}}\)
...................

kastin

王守恩 发表于 2018-12-26 10:10
有这样一串数：A000045               
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584,  ...

数列有递推关系 `a_{n+2}=a_{n+1}+a_n`，特征方程为 `x^2=x+1`，特征根 `x_{1,2}=\frac{1\pm\sqrt{5}}{2}`，通项公式为 `a_n=\frac{5+\sqrt{5}}{10}(\frac{1+\sqrt{5}}{2})^n+\frac{5-\sqrt{5}}{10}(\frac{1-\sqrt{5}}{2})^n`

kastin

设 `b_n` 满足 `b_n=b_{n-1}+b_{n-2}~(n\geqslant 3,~b_0=2,\,b_1=3,\,b_2=4)`，那么算式右边和式可写为 `a_n=\sum_{k=0}^n(\frac 12)^{n-k}b_k`，即 `2^na_n=\sum_{k=0}^n2^kb_k`，而 `b_n` 的特征方程为 `x^2=x+1`，特征根跟上面一样，`b_n` 通项求出后，`a_n` 就不难求了。

王守恩

kastin 发表于 2018-12-27 15:28
设 `b_n` 满足 `b_n=b_{n-1}+b_{n-2}~(n\geqslant 3,~b_0=2,\,b_1=3,\,b_2=4)`，那么算式右边和式可写为 `a ...

谢谢kastin！

\(a_{n}，b_{n}\ \)也可以这样表示
\(a_{1}=1=\sqrt{0.8}\cosh(1\ln(\sqrt{1.25}+0.5))\)
\(a_{2}=1=\sqrt{0.8}\sinh(2\ln(\sqrt{1.25}+0.5))\)
\(a_{3}=2=\sqrt{0.8}\cosh(3\ln(\sqrt{1.25}+0.5))\)
\(a_{4}=3=\sqrt{0.8}\sinh(4\ln(\sqrt{1.25}+0.5))\)
\(a_{5}=5=\sqrt{0.8}\cosh(5\ln(\sqrt{1.25}+0.5))\)
\(a_{6}=8=\sqrt{0.8}\sinh(6\ln(\sqrt{1.25}+0.5))\)
\(a_{7}=13=\sqrt{0.8}\cosh(7\ln(\sqrt{1.25}+0.5))\)
\(a_{8}=21=\sqrt{0.8}\sinh(8\ln(\sqrt{1.25}+0.5))\)
\(a_{9}=34=\sqrt{0.8}\cosh(9\ln(\sqrt{1.25}+0.5))\)
..........
\(b_{2}=3=2\cosh(2\ln(\sqrt{1.25}+0.5))\)
\(b_{3}=4=2\sinh(3\ln(\sqrt{1.25}+0.5))\)
\(b_{4}=7=2\cosh(4\ln(\sqrt{1.25}+0.5))\)
\(b_{5}=11=2\sinh(5\ln(\sqrt{1.25}+0.5))\)
\(b_{6}=18=2\cosh(6\ln(\sqrt{1.25}+0.5))\)
\(b_{7}=29=2\sinh(7\ln(\sqrt{1.25}+0.5))\)
\(b_{8}=47=2\cosh(8\ln(\sqrt{1.25}+0.5))\)
\(b_{9}=76=2\sinh(9\ln(\sqrt{1.25}+0.5))\)
............
看着这些美妙的数字，我不得不想，
我们只是抽出单独的 \(1\) 行，譬如：怎么知道这个和是\(\ 55\)？
\(\frac{2}{2^{9}}+\frac{3}{2^{8}}+\frac{4}{2^{7}}+\frac{7}{2^{6}}+\frac{11}{2^{5}}+\frac{18}{2^{4}}+\frac{29}{2^{3}}+\frac{47}{2^{2}}+\frac{76}{2^{1}}\)

非常感谢kastin！这一年来还真是进步不少！感谢kastin！

   

王守恩

kastin 发表于 2018-12-27 15:28
设 `b_n` 满足 `b_n=b_{n-1}+b_{n-2}~(n\geqslant 3,~b_0=2,\,b_1=3,\,b_2=4)`，那么算式右边和式可写为 `a ...

有这样一串数：A000045               
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765,
10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269,
2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169,..........

谢谢kastin ！我们可以有这样的通项公式。谢谢kastin ！

\(a_{1}=1=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}\)
\(a_{2}=2=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}\)
\(a_{3}=3=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}\)
\(a_{4}=5=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}\)
\(a_{5}=8=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}\)
\(a_{6}=13=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}\)
\(a_{7}=21=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}\)
\(a_{8}=34=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{9}}\)
\(a_{9}=55=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{10}}\)
\(a_{10}=89=\sum_{k=0}^{10}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{11}}\)
\(a_{11}=144=\sum_{k=0}^{11}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{12}}\)
\(a_{12}=233=\sum_{k=0}^{12}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{13}}\)
..............

王守恩

王守恩 发表于 2018-12-22 15:16
好不容易，我在"OEIS"找到下面这串数(A007664)：

1, 3, 5, 9, 13, 17, 25, 33, 41, 49, 65, 81, 97,  ...

进一步明朗 13 楼的 4 柱汉诺塔问题，"OEIS"(A007664)：

1, 3, 5, 9, 13, 17, 25, 33, 41, 49, 65, 81, 97, 113, 129, 161, 193, 225, 257, 289,
321, 385, 449, 513, 577, 641, 705, 769, 897, 1025, 1153, 1281, 1409, 1537, 1665,
1793, 2049, 2305, 2561, 2817, 3073, 3329, 3585, 3841, 4097, 4609, 5121, 5633,.....

通项公式可以是这样：

\(\D a_{n}=n\times 2^{\lbrack \sqrt{2n}-1\rbrack}-\sum_{k=1}^{\lbrack\sqrt{2n}\rbrack}\frac{k!}{(k-2)!\times 2^{3-k}}\)

中括号\(\ \lbrack a\rbrack\ \)是\(\ a\ \)取圆整，即四舍五入。
大家出出主意：这中括号还可以取消吗？

王守恩

kastin 发表于 2018-12-27 15:28
设 `b_n` 满足 `b_n=b_{n-1}+b_{n-2}~(n\geqslant 3,~b_0=2,\,b_1=3,\,b_2=4)`，那么算式右边和式可写为 `a ...

A000110       
1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597,
27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159,
5832742205057, 51724158235372, 474869816156751, 4506715738447323, 44152005855084346,
445958869294805289, 4638590332229999353, 49631246523618756274, 545717047936059989389, .............
\(\D\sum_{n=0}^{\infty}\frac{n^{0}}{n!\times e}=1\)
\(\D\sum_{n=0}^{\infty}\frac{n^{1}}{n!\times e}=1\)
\(\D\sum_{n=0}^{\infty}\frac{n^{2}}{n!\times e}=2\)
\(\D\sum_{n=0}^{\infty}\frac{n^{3}}{n!\times e}=5\)
\(\D\sum_{n=0}^{\infty}\frac{n^{4}}{n!\times e}=15\)
\(\D\sum_{n=0}^{\infty}\frac{n^{5}}{n!\times e}=52\)
\(\D\sum_{n=0}^{\infty}\frac{n^{6}}{n!\times e}=203\)
\(\D\sum_{n=0}^{\infty}\frac{n^{7}}{n!\times e}=877\)
\(\D\sum_{n=0}^{\infty}\frac{n^{8}}{n!\times e}=4140\)

王守恩

王守恩 发表于 2018-7-3 16:29
有这样一串数： 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325,
  ...

A002522
2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226,
257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785,
842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601,
1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501, .................

\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-0)!\times e}=2\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-1)!\times e}=5\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-2)!\times e}=10\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-3)!\times e}=17\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-4)!\times e}=26\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-5)!\times e}=37\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-6)!\times e}=50\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-7)!\times e}=65\)
\(\D\sum_{n=0}^{\infty}\frac{n^2}{(n-8)!\times e}=82\)
         

   
   
   
   
   

   

王守恩

王守恩 发表于 2019-1-30 10:07
A002522
2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226,
257, 290, 325, 362, 40 ...

A002720       
1, 2, 7, 34, 209, 1546, 13327, 130922, 1441729, 17572114, 234662231,
3405357682, 53334454417, 896324308634, 16083557845279, 306827170866106,
6199668952527617, 132240988644215842, 2968971263911288999, 69974827707903049154,
1727194482044146637521, 44552237162692939114282, 1198605668577020653881647, ...............

\(\D\sum_{k=0}^{0}\frac{(0!)^2}{k!\ ((0-k)!)^2}=1\)
\(\D\sum_{k=0}^{1}\frac{(1!)^2}{k!\ ((1-k)!)^2}=2\)
\(\D\sum_{k=0}^{2}\frac{(2!)^2}{k!\ ((2-k)!)^2}=7\)
\(\D\sum_{k=0}^{3}\frac{(3!)^2}{k!\ ((3-k)!)^2}=34\)
\(\D\sum_{k=0}^{4}\frac{(4!)^2}{k!\ ((4-k)!)^2}=209\)
\(\D\sum_{k=0}^{5}\frac{(5!)^2}{k!\ ((5-k)!)^2}=1546\)
\(\D\sum_{k=0}^{6}\frac{(6!)^2}{k!\ ((6-k)!)^2}=13327\)

也可以这样：
\(\D\sum_{k=0}^{0}\frac{(0!)^2}{k!\ k!\ (0-k)!}=1\)
\(\D\sum_{k=0}^{1}\frac{(1!)^2}{k!\ k!\ (1-k)!}=2\)
\(\D\sum_{k=0}^{2}\frac{(2!)^2}{k!\ k!\ (2-k)!}=7\)
\(\D\sum_{k=0}^{3}\frac{(3!)^2}{k!\ k!\ (3-k)!}=34\)
\(\D\sum_{k=0}^{4}\frac{(4!)^2}{k!\ k!\ (4-k)!}=209\)
\(\D\sum_{k=0}^{5}\frac{(5!)^2}{k!\ k!\ (5-k)!}=1546\)
\(\D\sum_{k=0}^{6}\frac{(6!)^2}{k!\ k!\ (6-k)!}=13327\)

王守恩

kastin 发表于 2018-12-27 15:28
设 `b_n` 满足 `b_n=b_{n-1}+b_{n-2}~(n\geqslant 3,~b_0=2,\,b_1=3,\,b_2=4)`，那么算式右边和式可写为 `a ...

A000262
1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, 58941091, 824073141,
12470162233, 202976401213, 3535017524403, 65573803186921, 1290434218669921,
26846616451246353, 588633468315403843, 13564373693588558173, 327697927886085654441,
8281153039765859726341, 218456450997775993367443,6004647590528092507965393,..................

\(\D\sum_{k=0}^{0}\frac{0!\times 1!}{k!\ (0-k)!\ (k+1)!\ }=1\)
\(\D\sum_{k=0}^{1}\frac{1!\times 2!}{k!\ (1-k)!\ (k+1)!\ }=3\)
\(\D\sum_{k=0}^{2}\frac{2!\times 3!}{k!\ (2-k)!\ (k+1)!\ }=13\)
\(\D\sum_{k=0}^{3}\frac{3!\times 4!}{k!\ (3-k)!\ (k+1)!\ }=73\)
\(\D\sum_{k=0}^{4}\frac{4!\times 5!}{k!\ (4-k)!\ (k+1)!\ }=501\)
\(\D\sum_{k=0}^{5}\frac{5!\times 6!}{k!\ (5-k)!\ (k+1)!\ }=4051\)
\(\D\sum_{k=0}^{6}\frac{6!\times 7!}{k!\ (6-k)!\ (k+1)!\ }=37633\)