王守恩 发表于 2019-1-30 17:52:33

本帖最后由 王守恩 于 2019-1-30 19:38 编辑

chyanog 发表于 2018-6-17 19:07
$$\begin{array}{cc}
\begin{array}{cc}
\cosh \left(\frac{1}{2} n \tanh ^{-1}\left(\frac{\sqrt{7}} ...

       还会有比这更好的通项吗?!!       A005492
15, 52, 151, 372, 799, 1540, 2727, 4516, 7087, 10644, 15415,
21652, 29631, 39652, 52039, 67140, 85327, 106996, 132567, 162484,
197215, 237252, 283111, 335332, 394479, 461140, 535927, 619476, 712447,
815524, 929415, 1054852, 1192591, 1343412, 1508119, 1687540, 1882527, .......
   
\(\D\sum_{k=0}^{\infty}\frac{(k+0)^4}{k!\times e}=15\)
\(\D\sum_{k=0}^{\infty}\frac{(k+1)^4}{k!\times e}=52\)
\(\D\sum_{k=0}^{\infty}\frac{(k+2)^4}{k!\times e}=151\)
\(\D\sum_{k=0}^{\infty}\frac{(k+3)^4}{k!\times e}=372\)
\(\D\sum_{k=0}^{\infty}\frac{(k+4)^4}{k!\times e}=799\)
\(\D\sum_{k=0}^{\infty}\frac{(k+5)^4}{k!\times e}=1540\)
\(\D\sum_{k=0}^{\infty}\frac{(k+6)^4}{k!\times e}=2727\)
\(\D\sum_{k=0}^{\infty}\frac{(k+7)^4}{k!\times e}=4516\)
\(\D\sum_{k=0}^{\infty}\frac{(k+8)^4}{k!\times e}=7087\)

王守恩 发表于 2019-1-31 17:45:58

王守恩 发表于 2019-1-30 17:52
还会有比这更好的通项吗?!!       A005492
15, 52, 151, 372, 799, 1540, 2727, 4516, 708 ...

A052852
1, 4, 21, 136, 1045, 9276, 93289, 1047376, 12975561,
175721140, 2581284541, 40864292184, 693347907421,
12548540320876, 241253367679185, 4909234733857696,
105394372192969489, 2380337795595885156, ....................

\(\D\sum_{k=0}^{0}\frac{0!\times 2!}{\ k!\ (0-k)!\ (k+2)!\ }=1\)
\(\D\sum_{k=0}^{1}\frac{1!\times 3!}{\ k!\ (1-k)!\ (k+2)!\ }=4\)
\(\D\sum_{k=0}^{2}\frac{2!\times 4!}{\ k!\ (2-k)!\ (k+2)!\ }=21\)
\(\D\sum_{k=0}^{3}\frac{3!\times 5!}{\ k!\ (3-k)!\ (k+2)!\ }=136\)
\(\D\sum_{k=0}^{4}\frac{4!\times 6!}{\ k!\ (4-k)!\ (k+2)!\ }=1045\)
\(\D\sum_{k=0}^{5}\frac{5!\times 7!}{\ k!\ (5-k)!\ (k+2)!\ }=9276\)
\(\D\sum_{k=0}^{6}\frac{6!\times 8!}{\ k!\ (6-k)!\ (k+2)!\ }=93289\)
\(\D\sum_{k=0}^{7}\frac{7!\times 9!}{\ k!\ (7-k)!\ (k+2)!\ }=1047376\)


         


   
   
   
   
   

   



王守恩 发表于 2019-1-31 18:00:32

王守恩 发表于 2019-1-31 17:45
A052852
1, 4, 21, 136, 1045, 9276, 93289, 1047376, 12975561,
175721140, 2581284541, 408642921 ...

A028387
1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271,
305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929,
991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805,
1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651, 2755, 2861, ..........

\(\D\sum_{k=0}^{0}\frac{0!\times 3!}{\ k!\ (0-k)!\ (k+3)!\ }=1\)
\(\D\sum_{k=0}^{1}\frac{1!\times 4!}{\ k!\ (1-k)!\ (k+3)!\ }=5\)
\(\D\sum_{k=1}^{2}\frac{2!\times 5!}{\ k!\ (2-k)!\ (k+3)!\ }=11\)
\(\D\sum_{k=2}^{3}\frac{3!\times 6!}{\ k!\ (3-k)!\ (k+3)!\ }=19\)
\(\D\sum_{k=3}^{4}\frac{4!\times 7!}{\ k!\ (4-k)!\ (k+3)!\ }=29\)
\(\D\sum_{k=4}^{5}\frac{5!\times 8!}{\ k!\ (5-k)!\ (k+3)!\ }=41\)
\(\D\sum_{k=5}^{6}\frac{6!\times 9!}{\ k!\ (6-k)!\ (k+3)!\ }=55\)

王守恩 发表于 2019-1-31 18:23:59

王守恩 发表于 2019-1-31 17:45
A052852
1, 4, 21, 136, 1045, 9276, 93289, 1047376, 12975561,
175721140, 2581284541, 408642921 ...

A000290
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256,
289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900,
961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764,
1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, ..........

\(\D\sum_{k=0}^{0}\frac{0!\times 2!}{\ k!\ (0-k)!\ (k+2)!\ }=1\)
\(\D\sum_{k=0}^{1}\frac{1!\times 3!}{\ k!\ (1-k)!\ (k+2)!\ }=4\)
\(\D\sum_{k=1}^{2}\frac{2!\times 4!}{\ k!\ (2-k)!\ (k+2)!\ }=9\)
\(\D\sum_{k=2}^{3}\frac{3!\times 5!}{\ k!\ (3-k)!\ (k+2)!\ }=16\)
\(\D\sum_{k=3}^{4}\frac{4!\times 6!}{\ k!\ (4-k)!\ (k+2)!\ }=25\)
\(\D\sum_{k=4}^{5}\frac{5!\times 7!}{\ k!\ (5-k)!\ (k+2)!\ }=36\)
\(\D\sum_{k=5}^{6}\frac{6!\times 8!}{\ k!\ (6-k)!\ (k+2)!\ }=49\)
\(\D\sum_{k=6}^{7}\frac{7!\times 9!}{\ k!\ (7-k)!\ (k+2)!\ }=64\)


王守恩 发表于 2019-2-1 15:37:05

王守恩 发表于 2019-1-31 18:23
A000290
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256,
289, 324, 361, 40 ...

A001792
1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672,
61440, 131072, 278528, 589824, 1245184, 2621440, 5505024, 11534336,
24117248, 50331648, 104857600, 218103808, 452984832, 939524096, 1946157056,
4026531840, 8321499136, 17179869184, 35433480192, 73014444032, 150323855360,...

\(\D\sum_{k=1}^{2}\frac{2!}{(k-1)!\ (2-k)!\times 4\ }=1\)
\(\D\sum_{k=1}^{3}\frac{3!}{(k-1)!\ (3-k)!\times 4\ }=3\)
\(\D\sum_{k=1}^{4}\frac{4!}{(k-1)!\ (4-k)!\times 4\ }=8\)
\(\D\sum_{k=1}^{5}\frac{5!}{(k-1)!\ (5-k)!\times 4\ }=20\)
\(\D\sum_{k=1}^{6}\frac{6!}{(k-1)!\ (6-k)!\times 4\ }=48\)
\(\D\sum_{k=1}^{7}\frac{7!}{(k-1)!\ (7-k)!\times 4\ }=112\)
\(\D\sum_{k=1}^{8}\frac{8!}{(k-1)!\ (8-k)!\times 4\ }=256\)
\(\D\sum_{k=1}^{9}\frac{9!}{(k-1)!\ (9-k)!\times 4\ }=576\)


王守恩 发表于 2019-2-3 18:17:58

本帖最后由 王守恩 于 2019-2-3 19:07 编辑

王守恩 发表于 2019-2-1 15:37
A001792
1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672,
61440, 131072, 278528 ...

A104144   好心的网友!可以有简单的吗?
1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512,
64960, 129792, 259328, 518145, 1035269, 2068498, 4132920, 8257696, 16499120,
32965728, 65866496, 131603200, 262947072, 525375999, 1049716729, 2097364960,
4190597000, 8372936304, 16729373488, 33425781248, 66785696000, 133439788800,...

\(f(n)=\frac{(n+1)(n-0)!*2^{n-0}}{(n+1)!*0!}-\frac{(n-7)(n-9)!*2^{n-10}}{(n-9)!*1!}+\frac{(n-15)(n-18)!*2^{n-20}}{(n-19)!*2!}-\frac{(n-23)(n-27)!*2^{n-30}}{(n-29)!*3!}+\frac{(n-31)(n-36)!*2^{n-40}}{(n-39)!*4!}-\frac{(n-39)(n-45)!*2^{n-50}}{(n-49)!*5!}+\frac{(n-47)(n-54)!*2^{n-60}}{(n-59)!*6!}-\frac{(n-55)(n-63)!*2^{n-70}}{(n-69)!*7!}+\frac{(n-63)(n-72)!*2^{n-80}}{(n-79)!*8!}-\frac{(n-71)(n-81)!*2^{n-90}}{(n-89)!*9!}+\frac{(n-79)(n-90)!*2^{n-100}}{(n-99)!*10!}-\frac{(n-87)(n-99)!*2^{n-110}}{(n-109)!*11!}\cdots\cdots\)

王守恩 发表于 2019-2-4 11:22:28

王守恩 发表于 2019-2-3 18:17
A104144   好心的网友!可以有简单的吗?
1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 40 ...

      这通项不是挺好吗?         A001296
0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 2431, 3367, 4550,
6020, 7820, 9996, 12597, 15675, 19285, 23485, 28336, 33902, 40250,
47450, 55575, 64701, 74907, 86275, 98890, 112840, 128216, 145112,
163625, 183855, 205905, 229881, 255892, 284050, 314470, ...............

\(\D\sum_{k=1}^{0}\sum_{i=1}^{k}( i*k)=0\)
\(\D\sum_{k=1}^{1}\sum_{i=1}^{k}( i*k)=1\)
\(\D\sum_{k=1}^{2}\sum_{i=1}^{k}( i*k)=7\)
\(\D\sum_{k=1}^{3}\sum_{i=1}^{k}( i*k)=25\)
\(\D\sum_{k=1}^{4}\sum_{i=1}^{k}( i*k)=65\)
\(\D\sum_{k=1}^{5}\sum_{i=1}^{k}( i*k)=140\)
\(\D\sum_{k=1}^{6}\sum_{i=1}^{k}( i*k)=266\)
\(\D\sum_{k=1}^{7}\sum_{i=1}^{k}( i*k)=462\)
\(\D\sum_{k=1}^{8}\sum_{i=1}^{k}( i*k)=750\)
\(\D\sum_{k=1}^{9}\sum_{i=1}^{k}( i*k)=1155\)


王守恩 发表于 2019-2-4 12:40:23

王守恩 发表于 2019-2-4 11:22
这通项不是挺好吗?         A001296
0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 24 ...

A002411
0, 1, 6, 18, 40, 75, 126, 196, 288, 405, 550, 726, 936, 1183,
1470, 1800, 2176, 2601, 3078, 3610, 4200, 4851, 5566, 6348,
7200, 8125, 9126, 10206, 11368, 12615, 13950, 15376, 16896,
18513, 20230, 22050, 23976, 26011, 28158, 30420, 32800, 35301,
37926, 40678, 43560, 46575, 49726, 53016, 56448, 60025, 63750,

\(\D\sum_{k=1}^{0}\sum_{i=1}^{k}( i+k-1)=0\)
\(\D\sum_{k=1}^{1}\sum_{i=1}^{k}( i+k-1)=1\)
\(\D\sum_{k=1}^{2}\sum_{i=1}^{k}( i+k-1)=6\)
\(\D\sum_{k=1}^{3}\sum_{i=1}^{k}( i+k-1)=18\)
\(\D\sum_{k=1}^{4}\sum_{i=1}^{k}( i+k-1)=40\)
\(\D\sum_{k=1}^{5}\sum_{i=1}^{k}( i+k-1)=75\)
\(\D\sum_{k=1}^{6}\sum_{i=1}^{k}( i+k-1)=126\)
\(\D\sum_{k=1}^{7}\sum_{i=1}^{k}( i+k-1)=196\)
\(\D\sum_{k=1}^{8}\sum_{i=1}^{k}( i+k-1)=288\)
\(\D\sum_{k=1}^{9}\sum_{i=1}^{k}( i+k-1)=405\)


王守恩 发表于 2019-2-4 16:21:08

本帖最后由 王守恩 于 2019-2-4 22:41 编辑

王守恩 发表于 2019-2-3 18:17
A104144   好心的网友!可以有简单的吗?
1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 40 ...

             敬祝大家身体健康!新年快乐!

          36楼的通项公式可以化简!!!!!      A104144
1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512,
64960, 129792, 259328, 518145, 1035269, 2068498, 4132920, 8257696, 16499120,
32965728, 65866496, 131603200, 262947072, 525375999, 1049716729, 2097364960,
4190597000, 8372936304, 16729373488, 33425781248, 66785696000, 133439788800,...

\[\D f(n)=\sum_{k=n}^{n}\sum_{m=0}^{k/9}\frac{(k-8m+1)*(k-9m)!*2^{k-10m}\ }{(k-10m+1)!*m!*\cos(m\pi)}\]

王守恩 发表于 2019-2-5 09:51:38

本帖最后由 王守恩 于 2019-2-5 09:53 编辑

王守恩 发表于 2019-2-4 16:21
敬祝大家身体健康!新年快乐!

          36楼的通项公式可以化简!!!!!      A104144
...

A079262
1, 2, 4, 8, 16, 32, 64, 128, 255, 509, 1016, 2028, 4048, 8080,
16128, 32192, 64256, 128257, 256005, 510994, 1019960, 2035872,
4063664, 8111200, 16190208, 32316160, 64504063, 128752121, 256993248,
512966536, 1023897200, 2043730736, 4079350272, 8142510336, 16252704512,

\[\D f(n)=\sum_{k=n}^{n}\sum_{m=0}^{k/8}\frac{(k-7m+1)*(k-8m)!*2^{k-9m}\ }{(k-9m+1)!*m!*\cos(m\pi)}\]
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